2018학년도 2학기
반도체 공학과 공학수학2 (GEDB005) 강의교안 (2018학년도용)
주교재: Erwin Kreyszig, Engineering Mathematics, 10th Edition
부교재: 최신공학수학 I 과 II, 한빛출판사 및 (영문) 다변량 Calculus (이상구, 김응기 et al) (http://www.hanbit.co.kr/EM/sage/)
강의시간: BD1615(화 16:3017:45 / 목 15:0016:15), 반도체관 400102호 담당교수: 김응기 박사
주차 
주교재 
부교재 
2 
11.7: Fourier Integral 11.8: Fourier Cosine and Sine Transforms 
11.4 푸리에 급수의 수렴 11.5 스투름리우빌 정리와 직교함수 
web 
Week 2
11.7. Fourier Integral
Indefinite integral
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
f(x) = x^2 integral(f(x), x) 
1/3*x^3
Definite integral
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
f(x) = sin(x) integral(f(x), x, 0, pi/2) 
1
Fourier series are powerful tools for problems involving functions that are periodic or are of interest on a finite interval only.
Many problems involve functions that are nonperiodic and are of interest on the whole axis.
Definition 
Fourier integrals 
Consider a special function of period and see what happens to its Fourier series if we let . Do the same for an arbitrary function of period . 
Example 1 
Rectangular Wave 
Consider the period rectangular wave of period given by

The left part of Fig. shows this function for as well as the nonperiodic function , which we obtain from if we let ,
.
We now explore what happens to the Fourier coefficients of as increases.
Since is even, for all . For the Euler formulas , Sec. 11.2, give
.
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
var('L, n') assume(n, 'integer') a0 = 1/(2*L)*integral(1, x, 1, 1) print "a0 =", a0 an = 1/L*integral(cos(n*pi*x/L), x, 1, 1) print "an =", an 
a0 = 1/L
an = 2*sin(pi*n/L)/(pi*n)
This sequence of Fourier coefficients is called the amplitude spectrum of because is the maximum amplitude of the wave . Figure shows this spectrum for the periods . We see that for increasing these amplitude because more and more dense on the positive axis, where . Indeed, for we have amplitude per “halfwave” of the function (dashed in the figure). Hence for we have amplitude per halfwave, so that these will eventually be everywhere dense on the positive axis (and will decrease to zero).
From Fourier Series to Fourier Integral
We now consider any periodic function of period that can be represented by a Fourier series
,
.
imply
and find out what happens if we let .
Together with Example the present calculation will suggest that we should expect an integral(series) involving and with no longer restricted to integer multiples of but taking all values. We shall also see what from such an integral might have.
If we insert and from the Euler formulas , Sec. 11.2, and denote the variable of integration , the Fourier series of becomes
.
We now set
The Fourier series in the form
(1)
This representation is valid for any fixed , arbitrarily large, but finite.
We now let and assume that the resulting nonperiodic function
is absolutely integrable on axis; that is, the following(finite) limits exist :
(2) .
Then , and the value of the first term on the right side of approaches zero.
Also and it seems plausible that the infinite series in becomes an integral from to , which represents, namely,
(3) .
If we introduce the notations
(4) ,
we can write this in the form
(5) .
This is a representation of by a Fourier integral.
Theorem 1 
Fourier Integral 
is piecewise continuous in every finite interval. has a righthand derivative and lefthand derivative at every point. If the integral exist, then can be represented by a Fourier integral with and . At a point where is discontinuous the value of the Fourier integral equals the average of the lefthand and righthand limits of at the point. 
Applications of Fourier Integrals
Example 2 
Single Pulse, Sine Integral. Dirichlet’s Discontinuous Factor. Gibbs Phenomenon 
Find the Fourier integral representation of the function .

Solution
From we obtain
and gives the answer
(6) .
The average of the lefthand and righthand limits of at is equal to .
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
var('w, v') A(w) = 1/pi*integral(cos(w*v), v, 1, 1) print "A(w) =", A(w) B(w) = 1/pi*integral(sin(w*v), v, 1, 1) print "B(w) =", B(w) 
A(w) = 2*sin(w)/(pi*w)
B(w) = 0
From and Theorem we obtain(multiply by )
(7) .
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
var('w') g(x)=integral(cos(w*x)*sin(w)/w, w, 0, +oo) print g(0) print g(1/2) print g(1) print g(2) 
1/2*pi
1/2*pi
1/4*pi
0
We mention that this integral is Dirichlet’s discontinuous factor.
The case is of particular interest. If , then gives
(8*) .
We see that this integral is the limit of the socalled sine integral
(8)
as . The graphs of and of the integrand are shown in Fig. .
In case of a Fourier series the graphs of the partial sums are approximation curves of the curve of the periodic function represented by the series. Similarly, in the case the Fourier integral , approximations are obtained by replacing by numbers .
Hence the integral
(9)
Setting .
, ,
Setting .
, ,
.
approximates the right side in and therefore .
The Figure shows oscillations near the points of discontinuous of .
We might expect that these oscillations disappear approaches infinity.
But this is not true: with increasing , they are shifted closer to the point .
This unexpected behavior, which also occurs in connection with Fourier series, is known as the Gibbs phenomenon.
We can explain it by representing in terms of sine integrals as follows.
Using in App. A3. 1, we have
.
In the first integral on the right we set .
Then , and , corresponds to .
In the last integral we set .
Then , and , corresponds to .
Since , we thus obtain
.
From this and we see that our integral equals
and the oscillations in Fig. 283 result from those in Fig. 282.
The increase of amounts to a transformation of the scale on the axis and cause
the shift of the oscillation toward the point of discontinuity .
Fourier Cosine Integral and Fourier Sine Integral
If is an even function, then in and
(10) .
The Fourier integral then reduces to the Fourier Cosine Integral
(10) ( even).
If is an odd function, then in we have and
(11) .
The Fourier integral then reduces to the Fourier Sine Integral
(11) ( odd).
Example 3 
Laplace integrals 
We shall derive the Fourier cosine and Fourier sine integrals of , where and . (Fig 281) The result will be used to evaluate the socalled Laplace integrals.

Solution
From we have
(12) .
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
var('k, w, v') assume(k > 0) A(w) = 2/pi*integral(exp(k*v)*cos(w*v), v, 0, +oo) print "A(w) =", A(w) 
A(w) = 2*k/((k^2 + w^2)*pi)
By substituting this into the first integral in
we thus obtain the Fourier cosine integral representation
.
From this representation we see that
.
(13)
.
(14) .
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
var('k, w, v') assume(k > 0) B(w) = 2/pi*integral(exp(k*v)*sin(w*v), v, 0, +oo) print "B(w) =", B(w) 
B(w) = 2*w/((k^2 + w^2)*pi)
From we thus obtain the Fourier sine integral representation
.
From this we see that
(15) .
.
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
var('k, w') assume(k > 0) assume(x > 0) integral(w*sin(w*x)/(k^2+w^2), w, 0, +oo) 
1/2*pi*e^(k*x)
The integrals and are called the Laplace integrals.
11.8. Fourier Cosine and Sine Transforms
An integral transformation is a transformation in the form of an integral that produces from given functions new functions depending on a different variable.
The Laplace transform is of this kind and is by far the most important integral transform in engineering.
Definition 
Fourier Integral 
Let us assume that following conditions on 1. and are piecewise continuous in every finite interval. 2. converges i.e. is absolutely integrable in . 
The Fourier integral theorem states that
(*)
is a Fourier integral expansion of .
(**) where
Fourier Cosine Transforms
For an even function , the Fourier cosine integral
(a) , where (b) . ( in (**))
We now set , where suggests “cosine.”
Then from , writing , we have
(1)
This is the Fourier cosine transform of . Similarly, from we have
(2) .
This is the inverse Fourier cosine transform of .
Attention!
In we integrate with respect to and in with respect to . Formula gives from a new function , called the Fourier cosine transform of .
The Formula gives us back from , and we therefore call the inverse Fourier cosine transform of .
The process of obtaining the transform from a given is called the Fourier cosine transform or the Fourier cosine transform method.
Fourier Sine Transforms
For an odd function , the Fourier integral is the Fourier sine integral
(a) , where (b). ( in (**))
, where suggests “sine.”
Hence
Then from , writing , we have
(2) . Fourier sine transform of ,
Similarly, from we have
(2)
inverse Fourier sine transform of .
The process of obtain from is the Fourier sine transform or the Fourier sine transform method.
Remark 

Other notations are , and and for the inverse of and , respectively. 
Example 1 
Fourier Cosine and Sine Transforms 
Find the Fourier cosine and Fourier sine transformation of the function(Fig. 282)

Solution
From the definitions and ,
we have the following Fourier cosine transform
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
var('k, w, a') assume(k > 0) assume(a > 0) sqrt(2/pi)*integral(k*cos(w*x), x, 0, a) 
sqrt(2)*k*sin(a*w)/(sqrt(pi)*w)
We also will have the following Fourier sine transform
.
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
var('k, w, a') assume(k > 0) assume(a > 0) sqrt(2/pi)*integral(k*sin(w*x), x, 0, a) 
(cos(a*w)/w  1/w)*sqrt(2)*k/sqrt(pi)
This agrees with formulas in the first two tables in Sec. 11.10 (where ).
Note that , these transform do not exist.(Why?)
Example 2 
Fourier Cosine Transform of the Exponential Function 
Find . 
Solution
By integration by parts and recursion,
.
This agrees with formula in Table I, Sec. 11. 10, with .
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
var('w') sqrt(2/pi)*integral(exp(x)*cos(w*x), x, 0, +oo) 
sqrt(2)/((w^2 + 1)*sqrt(pi))
Linearity, Transforms of Derivatives
If is absolutely integrable on the positive axis and piecewise continuous on every finite interval, then the Fourier cosine and sine transforms of exists.
Theorem 
Fourier Cosine Transforms 
If and have Fourier cosine transforms, then for any constants and has Fourier cosine transforms. 
Proof
Theorem 
Fourier Sine Transforms 
If and have Fourier sine transforms, then for any constants and has Fourier sine transforms. 
Proof
Shows that the Fourier cosine and sine transforms are linear operation,
(3) .
Theorem 1 
Cosine and Sine Transforms of Derivatives 
Let be continuous and absolutely integrable on the axis, let be piecewise continuous on every finite interval, and let as . Then (4) 
Proof
,
.
The Formula , say , with instead of
(when , satisfy the respective assumptions for , in Theorem)
instead of
(5a) .
Similarly
(5b)
Example 3 
An Application of the Operational Formula (5) 
Find the Fourier cosine transform of , where . 
Solution
, thus
.
From this, , and the linearity ,
.
Hence
.
Answer is (see. Table I, Sec.11.10)
.
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
var('w, a') assume(a > 0) f(x) = exp(a*x) sqrt(2/pi)*integral(f(x)*cos(w*x), x, 0, +oo) 
sqrt(2)*a/((a^2 + w^2)*sqrt(pi))
http://matrix.skku.ac.kr/sglee/
[한빛 아카데미] Engineering Mathematics with Sage:
[저자] 이상구, 김영록, 박준현, 김응기, 이재화
Contents
A. 공학수학 1 – 선형대수, 상미분방정식+ Lab
Chapter 01 벡터와 선형대수 http://matrix.skku.ac.kr/EMSage/EMathChapter1.html
Chapter 02 미분방정식의 이해 http://matrix.skku.ac.kr/EMSage/EMathChapter2.html
Chapter 03 1계 상미분방정식 http://matrix.skku.ac.kr/EMSage/EMathChapter3.html
Chapter 04 2계 상미분방정식 http://matrix.skku.ac.kr/EMSage/EMathChapter4.html
Chapter 05 고계 상미분방정식 http://matrix.skku.ac.kr/EMSage/EMathChapter5.html
Chapter 06 연립미분방정식, 비선형미분방정식 http://matrix.skku.ac.kr/EMSage/EMathChapter6.html
Chapter 07 상미분방정식의 급수해법, 특수함수 http://matrix.skku.ac.kr/EMSage/EMathChapter7.html
Chapter 08 라플라스 변환 http://matrix.skku.ac.kr/EMSage/EMathChapter8.html
B. 공학수학 2  벡터미적분, 복소해석 + Lab
Chapter 09 벡터미분, 기울기, 발산, 회전 http://matrix.skku.ac.kr/EMSage/EMathChapter9.html
Chapter 10 벡터적분, 적분정리 http://matrix.skku.ac.kr/EMSage/EMathChapter10.html
Chapter 11 푸리에 급수, 적분 및 변환 http://matrix.skku.ac.kr/EMSage/EMathChapter11.html
Chapter 12 편미분방정식 http://matrix.skku.ac.kr/EMSage/EMathChapter12.html
Chapter 13 복소수와 복소함수, 복소미분 http://matrix.skku.ac.kr/EMSage/EMathChapter13.html
Chapter 14 복소적분 http://matrix.skku.ac.kr/EMSage/EMathChapter14.html
Chapter 15 급수, 유수 http://matrix.skku.ac.kr/EMSage/EMathChapter15.html
Chapter 16 등각사상 http://matrix.skku.ac.kr/EMSage/EMathChapter16.html
Made by Prof. SangGu LEE sglee at skku.edu
http://matrix.skku.ac.kr/sglee/ with Dr. Jae Hwa LEE