FOURIER INTEGRAL
Continuum analogue Fourier series. For a function defined on a finite interval real axis, important has a Fourier series representation of it. For the function f(x) . given on the entire axis, a similar role is played by the Fourier expansion of f:
Where 
Expansion (1) can be formally constructed under assumptions that ensure the existence of the written integrals. It is true, for example, for a smooth finite function f(x) . Available great signs, ensuring equality (1) in one sense or another. Substitution of (2) into (1) gives the so-called. integral formula Fourier
the rationale for the cut leads to the mentioned characteristics. It is very useful to represent f(x) by a simple Fourier integral
which is obtained from (3) if we write the outer integral as over the interval (0, N) and change the integrations. IN applied sciences representation (1) is often interpreted as a harmonic expansion: if
then (1) takes the form:
and thus f is represented as a superposition of harmonics, the frequencies of which continuously fill the real semi-axis and the amplitude D and initial phase depend on
In many cases (in particular, for complex-valued functions f), expansion (1) is more convenient to represent in exponential form:
Where
The function is called Fourier transform functions f(in applied sciences WITH(l)
called Provided that f (x) is summable: the function is bounded, uniformly continuous on the axis and at 
The function may turn out to be non-summable and integral (4) non-existent. However, equality (4) allows for a reasonable interpretation if we use the methods of summation of integrals [in this case, we can consider not only pointwise convergence, but also average convergence]. For example, arithmetic averages of truncated f. and.
summable function f(x) converges to f(x) and on average at In the presence of additional restrictions on the function f(x), more specific statements are obtained. For example, if u has limited variation in the vicinity X, That
Applications often use decomposition
true for an absolutely integrable function f(x) that is piecewise smooth in every finite interval, where the integral on the right is understood in the sense of the principal value (6). F. and. is also studied under the assumption of local summability of the function f and under certain requirements that impose restrictions on the behavior of f in Let, for example, then
Where the limit is understood in the sense of convergence in the mean of order 
[however, the limit in (7) also exists in the sense of convergence almost everywhere]. Simple form acquires this result at p = 2 (see. Plancherel's theorem).
Similarly, multiple functions are constructed when we're talking about about the expansion of the function defined in n-dimensional space. The concept of F. and. also applies to generic functions.
Lit.: Titchmarsh E., Introduction to the theory of Fourier integrals, trans. English, M.-L., 1948; Bochner S., Lectures on Fourier integrals, trans. from English, M., 1962; 3igmund A., Trigonometric series, trans. from English, vol. 2, M., 1965.
P. I. Lizorkin.
Mathematical encyclopedia. - M.: Soviet Encyclopedia. I. M. Vinogradov. 1977-1985.
See what "FOURIER INTEGRAL" is in other dictionaries:
Fourier integral, Fourier integral (but: Fourier integral) ... Spelling dictionary-reference book
- (Fourier integral) decomposition of the function f(x), specified on the entire x-axis or on the semi-axes into a superposition of harmonics with frequencies filling the entire semi-axis l of the fl, giving the decomposition of non-periodic. functions on harmonious components, the frequencies of which constitute a continuous set of values ... Big Encyclopedic Polytechnic Dictionary
Problem solving method mathematical physics, based on separation of variables. Proposed for solving problems in the theory of heat conduction by J. Fourier and formulated in full generality by M. V. Ostrogradsky (See Ostrogradsky) in 1828. Solution... ... Great Soviet Encyclopedia
One of the integral transformations, linear operator F, acting in space, the elements of which are functions f(x) of real variables. The minimum domain of definition F is considered to be a set of infinitely differentiable... ... Mathematical Encyclopedia
The Hankel integral, an analogue of the Fourier integral for Bessel functions, having the form Formula (*) can be obtained from the Fourier Bessel series for the interval (0, l) by passing to the limit at G. Hankel (N. Hankel, 1875) established the theorem: if the function f (x)piecewise... ... Mathematical Encyclopedia
The Kurzweil Henstock integral, a generalization of the Riemann integral, allows you to completely solve the problem of reconstructing a differentiable function from its derivative. Neither the Riemann integral (including the improper one) nor the Lebesgue integral give... ... Wikipedia
Fourier integral- - [L.G. Sumenko. English-Russian dictionary on information technology. M.: State Enterprise TsNIIS, 2003.] Topics information Technology in general EN Fourier integral… Technical Translator's Guide
Integral transformation acting in the space of functions of n real variables: For functions Φ L1(Rn) summable over the entire space Rn, the integral (*) correctly determines a certain function F (x) = y(x) Fourier image of function j. Reverse... ... Physical encyclopedia
Books
- Entertaining mathematics. Fourier analysis. Manga, Shibuya Mikio. The girls Rika, Fumika and Erina have organized a rock band and want to perform at the festival, but they can’t find a vocalist. And then there’s the math test, which Fumika has problems with. Clever girl...
Which are already pretty boring. And I feel that the moment has come when it is time to extract new canned goods from the strategic reserves of theory. Is it possible to expand the function into a series in some other way? For example, express a straight line segment in terms of sines and cosines? It seems incredible, but such seemingly distant functions can be
"reunification". In addition to the familiar degrees in theory and practice, there are other approaches to expanding a function into a series.
On this lesson we'll meet trigonometric series Fourier, we will touch on the issue of its convergence and sum and, of course, we will analyze numerous examples of the expansion of functions in a Fourier series. I sincerely wanted to call the article “Fourier Series for Dummies,” but this would be disingenuous, since solving the problems would require knowledge of other branches of mathematical analysis and some practical experience. Therefore, the preamble will resemble astronaut training =)
Firstly, you should approach the study of page materials in excellent form. Sleepy, rested and sober. Without strong emotions about a broken hamster's paw and obsessive thoughts about the hardships of life aquarium fish. The Fourier series is not difficult to understand, however practical tasks they simply require increased concentration attention - ideally, you should completely detach yourself from external stimuli. The situation is aggravated by the fact that there is no easy way to check the solution and answer. Thus, if your health is below average, then it is better to do something simpler. Is it true.
Secondly, before flying into space you need to study the instrument panel spaceship. Let's start with the values of the functions that should be clicked on the machine:
For any natural value :
1) . Indeed, the sinusoid “stitches” the x-axis through each “pi”:
. In case negative values argument, the result, of course, will be the same: .
2) . But not everyone knew this. The cosine "pi" is the equivalent of a "blinker":
A negative argument does not change the matter: 
.
Perhaps that's enough.
And thirdly, dear cosmonaut corps, you must be able to... integrate.
In particular, confidently subsume the function under the differential sign, integrate piecemeal and be at peace with Newton-Leibniz formula. Let's begin the important pre-flight exercises. I categorically do not recommend skipping it, so as not to squish in weightlessness later:
Example 1
Calculate definite integrals
where takes natural values.
Solution: integration is carried out over the variable “x” and at this stage the discrete variable “en” is considered a constant. In all integrals put the function under the differential sign:
A short version of the solution that would be good to target looks like this:
Let's get used to it:
The four remaining points are on your own. Try to approach the task conscientiously and write the integrals in a short way. Sample solutions at the end of the lesson.
After performing the exercises QUALITY, we put on spacesuits
and getting ready to start!
Expansion of a function into a Fourier series on the interval
Let's consider some function that determined at least for a period of time (and possibly for a longer period). If this function is integrable on the interval, then it can be expanded into trigonometric Fourier series:
, where are the so-called Fourier coefficients.
In this case the number is called period of decomposition, and the number is half-life of decomposition.
It is obvious that in general case The Fourier series consists of sines and cosines: 
Indeed, let’s write it down in detail:
The zero term of the series is usually written in the form .
Fourier coefficients are calculated using the following formulas:
I understand perfectly well that those starting to study the topic are still unclear about the new terms: decomposition period, half-cycle, Fourier coefficients etc. Don’t panic, this is not comparable to the excitement before going out open space. Let’s figure it all out in the following example, before executing which it’s logical to ask some vital questions: practical issues:
What do you need to do in the following tasks?
Expand the function into a Fourier series. Additionally, it is often necessary to depict a graph of a function, a graph of the sum of a series, partial amount and in the case of sophisticated professorial fantasies, do something else.
How to expand a function into a Fourier series?
Essentially, you need to find Fourier coefficients, that is, compose and calculate three definite integral.
Please copy the general form of the Fourier series and the three working formulas into your notebook. I am very glad that some site visitors are realizing their childhood dream of becoming an astronaut right before my eyes =)
Example 2
Expand the function into a Fourier series on the interval. Construct a graph, a graph of the sum of the series and the partial sum.
Solution: The first part of the task is to expand the function into a Fourier series.
The beginning is standard, be sure to write down that:
In this problem, the expansion period is half-period.
Let us expand the function into a Fourier series on the interval: 
Using corresponding formulas, let's find Fourier coefficients. Now we need to compose and calculate three definite integral. For convenience, I will number the points:
1) The first integral is the simplest, however, it also requires eyeballs: 
2) Use the second formula:
This integral is well known and he takes it piece by piece: 
Used when found method of subsuming a function under the differential sign.
In the task under consideration, it is more convenient to immediately use formula for integration by parts in a definite integral 
:
A couple of technical notes. Firstly, after applying the formula the entire expression must be enclosed in large brackets, since there is a constant before the original integral. Let's not lose her! The parentheses can be expanded at any further step; I did this as a last resort. In the first "piece" 
We show extreme care in the substitution; as you can see, the constant is not used, and the limits of integration are substituted into the product. This action highlighted in square brackets. Well, you are familiar with the integral of the second “piece” of the formula from the training task;-)
And most importantly - extreme concentration!
3) We are looking for the third Fourier coefficient:
A relative of the previous integral is obtained, which is also integrates piecemeal:
This instance is a little more complicated, I’ll comment on the further steps step by step: 
(1) The expression is completely enclosed in large brackets. I didn’t want to seem boring, they lose the constant too often.
(2) V in this case I immediately opened those big brackets. Special attention
We devote ourselves to the first “piece”: the constant smokes on the sidelines and does not participate in the substitution of the limits of integration ( and ) into the product . Due to the clutter of the record, it is again advisable to highlight this action with square brackets. With the second "piece" 
everything is simpler: here the fraction appeared after opening the large brackets, and the constant - as a result of integrating the familiar integral;-)
(3) We carry out transformations in square brackets, and in the right integral we substitute the limits of integration.
(4) Take the “flashing light” out of square brackets: , after which we open the inner brackets: .
(5) We cancel 1 and –1 in brackets and make final simplifications.
Finally, all three Fourier coefficients are found: 
Let's substitute them into the formula 
:
At the same time, do not forget to divide in half. At the last step, the constant (“minus two”), which does not depend on “en,” is taken outside the sum.
Thus, we have obtained the expansion of the function into a Fourier series on the interval: 
Let us study the issue of convergence of the Fourier series. I will explain the theory, in particular Dirichlet's theorem, literally "on the fingers", so if you need strict formulations, please refer to the textbook on mathematical analysis (for example, the 2nd volume of Bohan; or the 3rd volume of Fichtenholtz, but it is more difficult).
The second part of the problem requires drawing a graph, a graph of the sum of a series, and a graph of a partial sum.
The graph of the function is the usual straight line on a plane, which is drawn with a black dotted line: 
Let's figure out the sum of the series. As you know, functional series converge to functions. In our case, the constructed Fourier series 
for any value of "x" will converge to the function, which is shown in red. This function endures ruptures of the 1st kind at points, but is also defined at them (red dots in the drawing)
Thus: 
. It is easy to see that it is noticeably different from the original function, which is why in the entry 
A tilde is used rather than an equals sign.
Let's study an algorithm that is convenient for constructing the sum of a series.
On the central interval, the Fourier series converges to the function itself (the central red segment coincides with the black dotted line of the linear function).
Now let's talk a little about the nature of the trigonometric expansion under consideration. Fourier series 
only periodic functions (constant, sines and cosines) are included, so the sum of the series 
also represents periodic function
.
What does this mean in our specific example? And this means that the sum of the series 
–necessarily periodic and the red segment of the interval must be repeated endlessly on the left and right.
I think the meaning of the phrase “period of decomposition” has now finally become clear. To put it simply, every time the situation repeats itself again and again.
In practice, it is usually sufficient to depict three periods of decomposition, as is done in the drawing. Well, and also “stumps” of neighboring periods - so that it is clear that the graph continues.
Special Interest represent discontinuity points of the 1st kind. At such points, the Fourier series converges to isolated values, which are located exactly in the middle of the “jump” of the discontinuity (red dots in the drawing). How to find out the ordinate of these points? First, let’s find the ordinate of the “top floor”: to do this, we calculate the value of the function at the rightmost point central period decomposition: . To calculate the ordinate of the “lower floor,” the easiest way is to take the leftmost value of the same period: 
. The ordinate of the mean is the average arithmetic sum"top and bottom": . A pleasant fact is that when constructing a drawing, you will immediately see whether the middle is calculated correctly or incorrectly.
Let’s construct a partial sum of the series and at the same time repeat the meaning of the term “convergence.” The motive is also known from the lesson about sum of a number series. Let us describe our wealth in detail:
To compose a partial sum, you need to write zero + two more terms of the series. That is,
The drawing shows the graph of the function green, and, as you can see, it “wraps” the full amount quite tightly. If we consider a partial sum of five terms of the series, then the graph of this function will approximate the red lines even more accurately; if there are one hundred terms, then the “green serpent” will actually completely merge with the red segments, etc. Thus, the Fourier series converges to its sum.
It is interesting to note that any partial amount is continuous function, however, the total sum of the series is still discontinuous.
In practice, it is not so rare to construct a partial sum graph. How to do this? In our case, it is necessary to consider the function on the segment, calculate its values at the ends of the segment and at intermediate points (the more points you consider, the more accurate the graph will be). Then you should mark these points on the drawing and carefully draw a graph on the period, and then “replicate” it into adjacent intervals. How else? After all, approximation is also a periodic function... ...in some ways its graph reminds me of an even heart rhythm on the display of a medical device.
Carrying out the construction, of course, is not very convenient, since you have to be extremely careful, maintaining an accuracy of no less than half a millimeter. However, I will please readers who are not comfortable with drawing - in a “real” problem it is not always necessary to carry out a drawing; in about 50% of cases it is necessary to expand the function into a Fourier series and that’s it.
After completing the drawing, we complete the task:
Answer: 
In many tasks the function suffers rupture of the 1st kind right during the decomposition period:
Example 3
Expand the function given on the interval into a Fourier series. Draw a graph of the function and the total sum of the series.
The proposed function is specified in a piecewise manner (and, note, only on the segment) and endures rupture of the 1st kind at point . Is it possible to calculate Fourier coefficients? No problem. Both the left and right sides of the function are integrable on their intervals, therefore the integrals in each of the three formulas should be represented as the sum of two integrals. Let's see, for example, how this is done for a zero coefficient:
The second integral turned out to be equal to zero, which reduced the work, but this is not always the case.
The other two Fourier coefficients are described similarly.
How to show the sum of a series? On the left interval we draw a straight line segment, and on the interval - a straight line segment (we highlight the section of the axis in bold and bold). That is, on the expansion interval, the sum of the series coincides with the function everywhere except for three “bad” points. At the discontinuity point of the function, the Fourier series will converge to an isolated value, which is located exactly in the middle of the “jump” of the discontinuity. It is not difficult to see it orally: left-sided limit: , right-sided limit: 
and, obviously, the ordinate of the midpoint is 0.5.
Due to the periodicity of the sum, the picture must be “multiplied” into neighboring periods, in particular, the same thing must be depicted on the intervals and . At the same time, at points the Fourier series will converge to the median values.
In fact, there is nothing new here.
Try to cope with this task yourself. Approximate sample final design and drawing at the end of the lesson.
Expansion of a function into a Fourier series over an arbitrary period
For an arbitrary decomposition period, where “el” is any positive number, the formulas of the Fourier series and Fourier coefficients differ in a slightly complicated argument for sine and cosine:
If , then we get the interval formulas with which we started.
The algorithm and principles for solving the problem are completely preserved, but the technical complexity of the calculations increases:
Example 4
Expand the function into a Fourier series and plot the sum. 
Solution: actually an analogue of Example No. 3 with rupture of the 1st kind at point . In this problem, the expansion period is half-period. The function is defined only on the half-interval, but this does not change the matter - it is important that both pieces of the function are integrable.
Let's expand the function into a Fourier series:
Since the function is discontinuous at the origin, each Fourier coefficient should obviously be written as the sum of two integrals:
1) I will write out the first integral in as much detail as possible:
2) We carefully look at the surface of the Moon:
Second integral take it piece by piece:
What to look for close attention, after we open the continuation of the solution with an asterisk?
Firstly, we do not lose the first integral 
, where we immediately execute subscribing to the differential sign. Secondly, don’t forget the ill-fated constant before the big brackets and don't get confused by the signs when using formula 
. Large brackets are still more convenient to open immediately in the next step.
The rest is a matter of technique; difficulties can only be caused by insufficient experience in solving integrals.
Yes, it’s not for nothing that famous colleagues French mathematician Fourier was indignant - how dare he arrange functions in trigonometric series?! =) By the way, everyone is probably interested in the practical meaning of the task in question. Fourier himself worked on mathematical model thermal conductivity, and subsequently the series named after him began to be used to study many periodic processes, which are visible and invisible in the surrounding world. Now, by the way, I caught myself thinking that it was not by chance that I compared the graph of the second example with the periodic rhythm of the heart. Those interested can familiarize themselves with practical application Fourier transform in third party sources. ...Although it’s better not to - it will be remembered as First Love =)
3) Considering the repeatedly mentioned weak links, let’s look at the third coefficient:
Let's integrate by parts: 
Let us substitute the found Fourier coefficients into the formula 
, not forgetting to divide the zero coefficient in half:
Let's plot the sum of the series. Let us briefly repeat the procedure: we construct a straight line on an interval, and a straight line on an interval. If the “x” value is zero, we put a point in the middle of the “jump” of the gap and “replicate” the graph for adjacent periods: 
At the “junctions” of periods, the sum will also be equal to the midpoints of the “jump” of the gap.
Ready. Let me remind you that the function itself is by condition defined only on a half-interval and, obviously, coincides with the sum of the series on the intervals
Answer:
Sometimes piecewise function happens and is continuous during the period of decomposition. The simplest example: 
. Solution (see Bohan volume 2) the same as in the two previous examples: despite continuity of function at point , each Fourier coefficient is expressed as the sum of two integrals.
On the decomposition interval discontinuity points of the 1st kind and/or there may be more “junction” points of the graph (two, three and generally any final quantity). If a function is integrable on each part, then it is also expandable in a Fourier series. But from practical experience I don’t remember such cruelty. However, there are more difficult tasks than those just considered, and at the end of the article there are links to Fourier series of increased complexity for everyone.
In the meantime, let's relax, lean back in our chairs and contemplate the endless starry expanses:
Example 5
Expand the function into a Fourier series on the interval and plot the sum of the series.
In this problem the function continuous on the expansion half-interval, which simplifies the solution. Everything is very similar to Example No. 2. There's no escape from the spaceship - you'll have to decide =) An approximate design sample at the end of the lesson, a schedule is attached.
Fourier series expansion of even and odd functions
With even and odd functions, the process of solving the problem is noticeably simplified. And here's why. Let's return to the expansion of a function in a Fourier series with a period of “two pi” 
and arbitrary period “two el” 
.
Let's assume that our function is even. The general term of the series, as you can see, contains even cosines and odd sines. And if we are expanding an EVEN function, then why do we need odd sines?! Let's reset the unnecessary coefficient: .
Thus, an even function can be expanded into a Fourier series only in cosines:
Since integrals of even functions along an integration segment that is symmetrical with respect to zero can be doubled, then the remaining Fourier coefficients are simplified.
For the gap: 
For an arbitrary interval: 
Textbook examples that can be found in almost any textbook on mathematical analysis include expansions of even functions 
. In addition, they have been encountered several times in my personal practice:
Example 6
The function is given. Required:
1) expand the function into a Fourier series with period , where is an arbitrary positive number;
2) write down the expansion on the interval, construct a function and graph the total sum of the series.
Solution: in the first paragraph it is proposed to solve the problem in general view, and it's very convenient! If the need arises, just substitute your value.
1) In this problem, the expansion period is half-period. During further actions, in particular during integration, "el" is considered a constant
The function is even, which means it can be expanded into a Fourier series only in cosines: 
.
We look for Fourier coefficients using the formulas 
. Pay attention to their unconditional advantages. Firstly, the integration is carried out over the positive segment of the expansion, which means we safely get rid of the module 
, considering only “X” of the two pieces. And, secondly, integration is noticeably simplified.
Two: 
Let's integrate by parts:
Thus:
, while the constant , which does not depend on “en”, is taken outside the sum.
Answer: 
2) Let us write the expansion on the interval, for this purpose in general formula substitute desired value half cycle:
One of the powerful tools for studying problems in mathematical physics is the method of integral transformations. Let the function f(x) be given on an interval (a, 6), finite or infinite. An integral transformation of a function f(x) is a function where K(x, w) is a function fixed for a given transformation, called the kernel of the transformation (it is assumed that the integral (*) exists in its proper or improper sense). §1. Fourier integral Any function f(x), which on the interval [-f, I] satisfies the conditions of expansion into a Fourier series, can be represented on this interval by a trigonometric series. Coefficients a*, and 6„ of series (1) are determined by the Euler-Fourier formulas : FOURIER TRANSFORM Fourier Integral Complex form integral Fourier transform Cosine and sine transform Amplitude and phase spectra Properties Applications The series on the right side of equality (1) can be written in a different form. For this purpose, we will enter into it from formulas (2) the values of the coefficients a" and op, and put under the signs of the integrals cos ^ x and sin x (which is possible, since integration variable is m) O) and use the formula for the cosine of the difference. We will have If the function /(g) was initially defined on the interval number axis, larger than the segment [-1,1] (for example, on the entire axis), then expansion (3) will reproduce the values of this function only on the segment [-1,1] and continue throughout the entire numerical axis as a periodic function with a period of 21 (Fig. 1). Therefore, if the function f(x) (generally speaking, non-periodic) is defined on the entire number line, in formula (3) one can try to go to the limit at I +oo. In this case, it is natural to require that the following conditions be met: 1. f(x) satisfies the conditions of decomposability in a Fourier series at any final segment axis Ox\ 2. The function f(x) is absolutely integrable on the entire number axis. When condition 2 is satisfied, the first term on the right side of equality (3) tends to zero as I -* +oo. In fact, Let us try to establish what the sum on the right side of (3) turns into in the limit at I +oo. Let us assume that Then the sum on the right side of (3) takes the form Due to absolute convergence integral, this sum for large I differs little from the expression which resembles the integral sum for a function of variable £ composed for the interval (0, +oo) of change. Therefore, it is natural to expect that for sum (5) will go into the integral. On the other hand, for fixed) from the formula ( 3) it follows that we also obtain the equality. The sufficient condition for the validity of formula (7) is expressed by the following theorem. Theorem 1. If the function f(x) is absolutely integrable on the entire real line and has, together with its derivative, final number points of discontinuity of the first kind on any interval [a, 6], then the equality is true. Moreover, at any point xq, which is a discontinuity point of the first kind of function /(x), the value of the integral on the right side of (7) is equal to Formula (7) is called by the Fourier integral formula, and the integral on its right side is the Fourier integral. If we use the formula for the cosine of the difference, then formula (7) can be written in the form The functions a(ξ), b(ζ) are analogues of the corresponding Fourier coefficients an and bn of a 2m-periodic function, but the latter are defined for discrete values n, while a(0> BUT are defined for continuous values£ G (-oo, +oo). Complex form of the Fourier integral Assuming /(x) is absolutely integrable on the entire Ox axis, consider the integral This integral converges uniformly for, since and therefore represents a continuous and, obviously, odd function of But then On the other hand, the integral is even function variable so that Therefore, the Fourier integral formula can be written as follows: Multiply the equality by imaginary unit i and add to equality (10). We get from where, by virtue of Euler's formula, we will have This is the complex form of the Fourier integral. Here, outer integration over £ is understood in the sense of the Cauchy principal value: §2. Fourier transform. Cosine and sine Fourier transforms Let the function f(x) be piecewise smooth on any finite segment of the Ox axis and absolutely integrable on the entire axis. Definition. The function from which, by virtue of Euler's formula, we will have is called the Fourier transform of the function /(r) (spectral function). This is the integral transformation of the function /(r) on the interval (-oo,+oo) with the kernel. Using the Fourier integral formula we obtain This is the so-called inverse conversion Fourier, which gives the transition from F(ξ) to f(x). Sometimes direct conversion The Fourier transform is defined as follows: Then the inverse Fourier transform is determined by the formula The Fourier transform of the function /(x) is also defined as follows: FOURIER TRANSFORM Fourier integral Complex form of the integral Fourier transform Cosine and sine transforms Amplitude and phase spectra Properties Applications Then, in turn, At this position The factor ^ is quite arbitrary: it can be included either in formula (1") or in formula (2"). Example 1. Find the Fourier transform of the function -4 We have This equality allows differentiation with respect to £ under the integral sign (the integral obtained after differentiation converges uniformly when ( belongs to any finite segment): Integrating by parts, we will have The out-of-integral term vanishes, and we we obtain from where (C is the constant of integration). Setting £ = 0 in (4), we find C = F(0). By virtue of (3) we have It is known that In particular, for) we obtain that Example 2 (discharge of the codemsetor through the copropylene ). Let us consider the function 4 For the spectra of the function F(ξ), we obtain Hence (Fig. 2). The condition for the absolute integrability of the function f(x) on the entire number line is very strict. It excludes, for example, such elementary functions, as) = cos x, f(x) = e1, for which the Fourier transform (in the classical form considered here) does not exist. Only those functions that quickly tend to zero as |x| have a Fourier transform. -+ +oo (as in examples 1 and 2). 2.1. Cosine and sine Fourier transforms Using the cosine and difference formula, we rewrite the Fourier integral formula in the following form: Let f(x) be an even function. Then we have equality (5). In the case of odd f(x), we similarly obtain If f(x) is given only on (0, -foo), then formula (6) extends f(x) to the entire Ox axis in an even manner, and formula (7) - odd. (7) Definition. The function is called the Fourier cosine transform of f(x). From (6) it follows that for an even function f(x) This means that f(x), in turn, is a cosine transform for Fc(£). In other words, the functions / and Fc are mutual cosine transformations. Definition. The function is called the Fourier sine transform of f(x). From (7) we obtain that for odd function f(x) i.e. f and Fs are mutual sine transformations. Example 3 (rectangular pulse). Let f(t) be an even function defined as follows: (Fig. 3). Let us use the obtained result to calculate the integral. By virtue of formula (9), we have Fig. 3 0 0 At the point t = 0, the function f(t) is continuous and equal to unity. Therefore, from (12") we obtain 2.2. Amplitude and phase spectra of the Fourier integral Let the periodic function /(x) with a period of 2m be expanded into a Fourier series. This equality can be written in the form where is the amplitude of the oscillation with frequency n, is the phase. On this path we We come to the concepts of amplitude and phase spectra of a periodic function. For a non-periodic function f(x), given on (-oo, +oo), under certain conditions it turns out to be possible to represent it by a Fourier integral that expands this function over all frequencies (expansion over a continuous frequency spectrum). ). Definition. Spectral function, or spectral density of the Fourier integral, is called the expression (the direct Fourier transform of the function f is called the amplitude spectrum, and the function Ф«) = -аggSfc) is called the phase spectrum of the function f(«). The amplitude spectrum A(ξ) serves as a measure of the contribution of frequency ζ to the function f(x). Example 4. Find the amplitude and phase spectra of function 4 Find the spectral function From here The graphs of these functions are shown in Fig. 4. §3. Properties of the Fourier transform 1. Linearity. If and G(0 are the Fourier transforms of the functions f(x) and d(x), respectively, then for any constant a and p the Fourier transform of the function a f(x) + p d(x) will be the function a Using the property of linearity of the integral, we have Thus, the Fourier transform is a linear operator. Denoting it by we will write. If F(ξ) is the Fourier transform of a function f(x) that is absolutely integrable on the entire numerical axis, then F(()) is bounded for all. Let the function f(x) be absolutely is integrable on the entire axis - the Fourier transform of the function f(x). f(x). Using the definition of the Fourier transform, show that Problem. Let the function f(z) have the Fourier transform F(0> h - real number. Show that 3. Fourier transform and differentiation operations. Let an absolutely integrable function f(x) have a derivative f"(x), which is also absolutely integrable on the entire Ox axis, so that f(x) tends to zero as |x| -» +oo. Considering f"(x) smooth function, we write Integrating by parts, we will have the out-integral term vanishes (since, and we get Thus, the differentiation of the function /(x) corresponds to the multiplication of its Fourier image ^Π/] by the factor If the function f(x) has a smooth absolutely intefiable derivatives up to order m inclusive and all of them, like the function f(x) itself, tend to zero; then, integrating by parts the required number of times, we obtain The Fourier transform is very useful precisely because it replaces the operation of differentiation with the operation of multiplication by a value and thereby simplifies the problem of integrating certain types of differential equations, since the Fourier transform of an absolutely integrable function f^k\x) is. limited function from (property 2), then from relation (2) we obtain the following estimate for: FOURIER TRANSFORM Fourier integral Complex form of the integral Fourier transform Cosine and sine transformations Amplitude and phase spectra Properties Applications From this estimate it follows: than more function f(x) has absolutely integrable derivatives, the faster its Fourier transform tends to zero at. Comment. The condition is quite natural, since the usual theory of Fourier integrals deals with processes that in one sense or another have a beginning and ends, but do not continue indefinitely with approximately the same intensity. 4. Relationship between the rate of decrease of the function f(x) as |z| -» -f oo and the smoothness of its Fourm transformation. Let us assume that not only f(x), but also its product xf(x) is an absolutely integrable function on the entire Ox axis. Then the Fourier transform) will be a differentiable function. Indeed, formal differentiation with respect to the parameter £ of the integrand leads to an integral that is absolutely and uniformly convergent with respect to the parameter. Therefore, differentiation is possible, and Thus, i.e., the operation of multiplying f(x) by the argument x goes after the Fourier transform into the operation t . If, together with the function f(x), the functions are absolutely integrable on the entire Ox axis, then the differentiation process can be continued. We obtain that the function has derivatives up to order m inclusive, and Thus, the faster the function f(x) decreases, the smoother the function becomes. Theorem 2 (about the drill). Let be the Fourier transforms of the functions f,(x) and f2(x), respectively. Then where does double integral converges absolutely on the right side. Let's put - x. Then we will have or, changing the order of integration, The function is called the convolution of functions and is denoted by the symbol Formula (1) can now be written as follows: This shows that the Fourier transform of the convolution of the functions f\(x) and f2(x) is equal to y/2x multiplied by product of Fourier transforms of convolvable functions. Remark. Easy to install following properties convolution: 1) linearity: 2) commutativity: §4. Applications of the Fourier transform 1. Let P(^) be a linear differential operator order m s constant coefficients, Using the formula for the Fourier transform of derivatives of the function y(x), we find "Consider the differential equation where P is the differential operator introduced above. Let us assume that the desired solution y(x) has the Fourier transform y (O. and the function f(x) has the transformation /(£) Applying the Fourier transform to equation (1), we obtain instead of the differential algebraic equation on the axis relative to where so formally where the symbol denotes the inverse Fourier transform. The main limitation of the applicability of this method is due to the following fact. Ordinary solution differential equation with constant coefficients contains functions of the form eL*, eaz cos fix, eax sin px. They are not absolutely integrable on the -oo axis< х < 4-оо, и преобразование Фурье для них не определено, так что, строго говоря, применятьданный метод нельзя. Это ограничение можно обойти, если ввести в рассмотрение так называемые обобщенные функции. Однако в ряде случаев преобразование Фурье все же применимо в своей классической форме. Пример. Найти решение а = а(х, t) уравнения (а = const), при начальных условиях Это - задача о free vibrations of an infinite homogeneous string when the initial deviation is given<р(х) точек сгруны, а начальные скорости отсутствуют. 4 Поскольку пространственная переменная х изменяется в пределах от -оо до +оо, подвергнем уравнение и начальные условия преобразованию Фурье по переменной х. Будем предполагать, что 1) функции и(х, t) и
I. Fourier transforms.
Definition 1. Function
Called Fourier transform functions
The integral here is understood in the sense of the principal value
and it is believed that it exists.
If is an absolutely integrable function on ℝ, then, since 
at , for any such function the Fourier transform (1) makes sense, and the integral (1) converges absolutely and uniformly along the entire straight line ℝ.
Definition 2. If 
– Fourier transform of the function
, then the comparable integral
Understood in the sense of the main meaning, it is called Fourier integral of the function .
Example 1. Find the Fourier transform of a function
The given function is absolutely integrable on , indeed,
Definition 3. Integrals understood in the sense of principal value
Named accordingly cosine- And sine-Fourier transforms of the function .
Believing , 
, 
, we obtain the relation that is partly already familiar to us from the Fourier series
As can be seen from relations (3), (4),
Formulas (5), (6) show that Fourier transforms are completely defined on the entire line if they are known only for non-negative values of the argument.
Example 2. Find cosine - and sine - Fourier transforms of functions
As shown in Example 1, the given function is absolutely integrable on .
Let's find its cosine - the Fourier transform using formula (3):
Similarly, it is not difficult to find the sine - Fourier transform of the function f(x) according to formula (4):
Using examples 1 and 2, it is easy to verify by direct substitution that for f(x) relation (5) is satisfied.
If the function is real-valued, then from formulas (5), (6) in this case it follows
Since in this case and are real functions on R, as can be seen from their definitions (3), (4). However, equality (7) provided 
is also obtained directly from definition (1) of the Fourier transform, taking into account that the conjugation sign can be inserted under the integral sign. The latest observation allows us to conclude that for any function the equality
It is also useful to note that if is a real and even function, i.e. , That
if is a real and odd function, i.e. , That
And if it is a purely imaginary function, i.e. . 
, That
Note that if is a real-valued function, then the Fourier integral can also be written in the form
Where 
Example 3. 
(counting 
)
since we know the value of the Dirichlet integral
The function considered in the example is not absolutely integrable on and its Fourier transform has discontinuities. The following shows that the Fourier transform of absolutely integrable functions has no discontinuities:
Lemma 1. If the function locally integrable and absolutely integrable on , That
a) its Fourier transform defined for any value
b) 
Let us recall that if– a real or complex-valued function defined on an open set, then the function called locally integrable on, if any dot has a neighborhood in which the function is integrable. In particular, if , the condition for local integrability of the function is obviously equivalent to the fact that 
for any segment.
Example 4. Let's find the Fourier transform of the function 
:
Differentiating the last integral with respect to the parameter and then integrating by parts, we find that
or
Means, 
, where is a constant, which, using the Euler-Poisson integral, we find from the relation
So, we found that , and at the same time showed that , and 
.
Definition 4. They say that the function 
, defined in a punctured neighborhood of the point , satisfies the Dini conditions at the point if
a) both one-sided limits exist at a point
b) both integrals
They agree absolutely.
Absolute convergence of the integral 
means the absolute convergence of the integral at least for some value.
Sufficient conditions for a function to be representable by a Fourier integral.
Theorem 1.If absolutely integrable on and locally piecewise continuous function satisfies at the point Dini conditions, then its Fourier integral converges at this point, and to the value
equal to half the sum of the left and right limits of the function values at this point.
Corollary 1.If the function continuous, has at every point finite one-sided derivatives and absolutely integrable on , then she appears on by its Fourier integral
Where 
Fourier transform of a function .
The representation of a function by the Fourier integral can be rewritten as:
Comment. The conditions on the function formulated in Theorem 1 and Corollary 1 are sufficient, but are not necessary for the possibility of such a representation.
Example 5. Represent the function as a Fourier integral if
This function is odd and continuous on ℝ, except for the points , , .
Due to the fact that the function is odd and real, we have:
and from equalities (5) and (10) it follows that
At the points of continuity of the function we have:
But the function is odd, so
since the integral is calculated in the sense of the principal value.
The function is even, so
If , . When the equality must be satisfied
Assuming, from here we find
If we put in the last expression for , then
Assuming here, we will find
If a real-valued function is piecewise continuous on any segment of the real line, is absolutely integrable on and has finite one-sided derivatives at each point, then at points of continuity the function is represented as a Fourier integral
and at the points of discontinuity of the function, the left side of equality (1) should be replaced by
If a continuous, absolutely integrable function has finite one-sided derivatives at each point, then in the case when this function is even, the equality is true
and in the case when is an odd function, the equality
Example 5'. Represent the function as a Fourier integral if:
Since is a continuous even function, then, using formulas (13.2), (13.2’), we have
Let us denote by the symbol the integral understood in the sense of the principal value
Corollary 2.For any function , satisfying the conditions of Corollary 1, there exist all transformations , , , and the equalities take place
With these relations in mind, transformation (14) is often called inverse Fourier transform and instead they write , and the equalities themselves (15) are called formula for inverting the Fourier transform.
Example 6. Let it be
Note that if 
, then for any function
Let's now take the function . Then
If we take a function that is an odd continuation of the function 
, on the entire numerical axis, then
Using Theorem 1, we obtain that
All integrals here are understood in the sense of the principal value,
Separating the real and imaginary parts in the last two integrals, we find the Laplace integrals
Definition . Function
we will call it the normalized Fourier transform.
Definition . If is the normalized Fourier transform of the function, then the comparable integral
We will call the function the normalized Fourier integral.
We will consider the normalized Fourier transform (16).
For convenience, we introduce the following notation:
(those. 
).
In comparison with the previous notation, this is just a renormalization: This means, in particular, relations (15) allow us to conclude that
or, in shorter notation,
Definition 5. We will call the operator the normalized Fourier transform, and the operator will be called the inverse normalized Fourier transform.
In Lemma 1 it was noted that the Fourier transform of any absolutely integrable function tends to zero at infinity. The next two statements state that, like Fourier coefficients, the Fourier transform tends to zero faster the smoother the function from which it is taken (in the first statement); A related fact will be that the faster the function from which the Fourier transform is taken tends to zero, the smoother its Fourier transform (second statement).
Statement 1(on the connection between the smoothness of a function and the rate of decrease of its Fourier transform). If and all functions 
absolutely integrable on , That:
A) at any 
b) 
Statement 2(about the connection between the rate of decrease of a function and the smoothness of its Fourier transform). If a locally integrable function : is such that the function absolutely integrable A , That:
A) Fourier transform of a function belongs to class 
b) there is inequality
Let us present the main hardware properties of the Fourier transform.
Lemma 2. Let there be a Fourier transform for functions (respectively, an inverse Fourier transform), then, whatever the numbers and , there is a Fourier transform (respectively, an inverse Fourier transform) for the function 
, and
(respectively ).
This property is called the linearity of the Fourier transform (respectively, the inverse Fourier transform).
Consequence. 
.
Lemma 3. The Fourier transform, like the inverse transform, is a one-to-one transformation on a set of continuous functions that are absolutely integrable on the entire axis and have one-sided derivatives at each point.
This means that if and are two functions of the specified type and if 
(respectively, if 
), then on the entire axis.
From the statement of Lemma 1 we can obtain the following lemma.
Lemma 4. If the sequence of absolutely integrable functions 
and absolutely integrable function are such that
then the sequence converges uniformly on the entire axis to the function .
Let us now study the Fourier transform of convolutions of two functions. For convenience, let's modify the definition of convolution by adding an additional factor
Theorem 2. Let the functions be bounded, continuous and absolutely integrable on the real axis, then
those. the Fourier transform of the convolution of two functions is equal to the product of the Fourier transforms of these functions.
Let's compile a summary table No. 1 of the properties of the normalized Fourier transform, useful in solving the problems given below.
Table No. 1
| Function | Normalized Fourier transform |
 |
|
 | |
 |
|
 |
|
 |  |
 |  |
 |  |
 |  |
 | |
 | |
 |
Using properties 1-4 and 6, we get
Example 7. Find the normalized Fourier transform of a function
In example 4 it was shown that
because if
Therefore, by property 3 we have:
Similarly, you can create table No. 2 for the normalized inverse Fourier transform:
Table No. 2
| Function | Normalized inverse Fourier transform |
 |
|
| | |
 |
|
 |
|
| |  |
| |  |
| |  |
| | |
| | |
| | |
 |
Just as before, using properties 1-4 and 6 we get that
Example 8. Find the normalized inverse Fourier transform of a function
As follows from example 6
When we have:
Representing the function as
use property 6 when
Options for tasks for calculation and graphic work
1. Find the sine – Fourier transform of the function
2. Find the sine – Fourier transform of the function
3. Find the cosine – Fourier transform of the function
4. Find the cosine – Fourier transform of the function
5. Find the sine – Fourier transform of the function
6. Find the cosine - Fourier transform of the function
7. Find the sine - Fourier transform of the function
8. Find cosine – Fourier transform of a function
9. Find cosine – Fourier transform of a function
10. Find the sine – Fourier transform of the function
11. Find the sine – Fourier transform of the function
12. Find sine - transformation of a function
13. Find sine - transformation of a function
14. Find cosine - transforming a function
15. Find cosine - transformation of a function
16. Find the Fourier transform of the function if:
17. Find the Fourier transform of the function if:
18. Find the Fourier transform of the function if:
19. Find the Fourier transform of the function if:
20. Find the Fourier transform of the function if:
21. Find the Fourier transform of the function if:
22. Find the normalized inverse Fourier transform of the function
using formula
24. Find the normalized inverse Fourier transform of the function
using formula
26. Find the normalized inverse Fourier transform of the function
using formula
28. Find the normalized inverse Fourier transform of the function
using formula
30. Find the normalized inverse Fourier transform of the function
using formula
23. Find the normalized inverse Fourier transform of the function
using formula
25. Find the normalized inverse Fourier transform of the function
using formula
27. Find the normalized inverse Fourier transform of the function
using formula
29. Find the normalized inverse Fourier transform of the function
using formula
31. Find the normalized inverse Fourier transform of the function
using formula
32. Represent a function by a Fourier integral
33. Represent a function by a Fourier integral
34. Represent a function by a Fourier integral
35. Represent a function by a Fourier integral
36. Represent a function by a Fourier integral
37. Represent a function by a Fourier integral
38. Represent a function by a Fourier integral
39. Represent a function by a Fourier integral
40. Represent a function by a Fourier integral
41. Represent a function by a Fourier integral
42. Represent a function by a Fourier integral
43. Represent the function as a Fourier integral, extending it in an odd way to the interval if:
44. Represent the function as a Fourier integral, extending it in an odd way to the interval if:
