FOURIER INTEGRALS Que1. Solve the integral equation

1   , 0    1 f ( x ) cos  x d x   0  1  0, 

Solution: We have given

0

Hence evaluate

sin 2 t 0 t 2 d t

f ( x) cos  x d x  Fc ( )

1   , 0    1 Fc ( )    1  0,

Therefore

(1)

and by inversion formula we have,

2



f ( x) 

0

Fc ( ) cos  x d  

2

1

(1   ) cos  x d   0

1  1 2 sin x sin x ( 1   )   ( 1) d     x 0 0  

2 cosx 2(1  cos x )   x x 0  x2 1

 Now 

0

0

Fc ( )   f ( x ) cos x d x  

2(1  cos x ) cos  x d x  x2

( 2)

From (1) and (2), we have

2



0

Now for

2



1   , 0    1 (1  cos x ) cos x d x   2  1 x  0,

 0

0



 (1  cos x ) (1  cos x )  d x1  dx 2 2 0 x x 2

0

sin 2 x  dx 2 x 2 Dr. Jogendra Kumar

Page 1

FOURIER INTEGRALS Que2. Using Fourier integrals, show that

0

1  sin x, 0  x   sin  sin  x d 2 2  1  x   0,

Solution. Clearly the given problem is Fourier sine integral of some function f (x)

 sin x, 0  x   x   0,

Let f (x)  

Then Fourier sine integral representation of f (x) is given by 

f (x)   B( )sin  x d  0

Where B( ) 

2



0

f (x)sin  x dx 

 2  sin x sin  x dx   0. sin  x dx       0

1  1  2 sin x sin  x dx  [cos (1   ) x  cos (1   ) x]dx  0  0

1  sin(1   ) x sin(1   ) x       1  1   

0

1  2

 (1   )sin(1   ) x  (1   )sin (1   ) x    1 2

1  (1   ) sin x cos  x  cos x sin  x  (1   ) sin x cos  x  cos x sin  x      1 2  1 2sin   2sin      1   2  (1   2 )

0

 f (x) 

1



0

B( )sin  x d  

2



0

0

sin   sin  x d 1  2

Dr. Jogendra Kumar

Page 2

FOURIER INTEGRALS 

0

sin   sin  x    sin x, 0  x   d   f (x)   x  1  2 2 2  0,   sin x , 0 x   2  x   0,

1 for x  1 Que3. Express the function f (x)   as Fourier integral. Hence Evaluate  0 for x  1

0

sin  cos  x

d .

Solution: Clearly the Fourier integral representation of

 1 for x  1 i.e.for  1  x  1 f (x)   is given by  0 for x  1 i.e.for x  1& x  1 

f (x)   [ A( ) cos  x  B( )sin  x] d  where 0

A( ) 

 A( ) 

1





1



f (x) cos  x dx , B( ) 



f (x) cos  x dx 

1



1

1

1





1

1



1



1



1



f (x)sin  x dx

1



1

1

 1

1

1

1.cos  x dx 

1 sin  x



f (x) cos  x dx 

0.cos  x dx 

cos  x dx 

1

1

f (x) cos  x dx 



1

1



1

f (x) cos  x dx

0.cos  x dx

2sin 



Dr. Jogendra Kumar

Page 3

FOURIER INTEGRALS B ( ) 

1

f (x) sin  x dx



1



1

1

1

1



1



1

f (x) sin  x dx  0.sin  x dx 

sin  x dx 

1

1



1

1

1

1.sin  x dx 

1

1 cos  x

f (x) sin  x dx 

1



1

1

1

f (x) sin  x dx

0.sin  x dx

1

0 1

 f (x)   [ A( ) cos  x  B( )sin  x] d  0

  2sin     cos  x  0.sin  x  d  0     2 sin    cos  x d 



0

sin  cos  x

0

d 

 2

   1 for x  1  for x  1 f (x)    2 2  0 for x  1   0 for x  1

Now since f (x) is discontinuous at x  1

 f (x) 

Thus

1 1    f (1  0)  f(1  0)    0   2 2 2  4   2 for x  1   f (x)   for x  1  4  0 for x  1   

Dr. Jogendra Kumar

Page 4

FOURIER INTEGRALS Que4. Using the Fourier integral representation, show that

0

 sin  x  d   e  x (x  0) . 2 1  2

Solution. Clearly the given problem is Fourier sine integral representation of some function f (x) . Let

f (x)  e  x . Therefore the Fourier sine integral representation of 

f (x)   B ( ) sin  x d  ,where B( )  0 We can rewrite the formula as

2

0

f (x) will be

f (x)sin  x dx .

  f (x)   B( )sin  x d  and B( )  2  f (x)sin  x dx 0  0

Now,

B( )  

2

0

f (x)sin  x dx 

0

0

0

e x e sin  x dx    sin  x   cos  x   1 2 0 2

x

2  (1   2 )

 f (x)   B( )sin  x d 



2

2   sin  x d  0 1   2

 sin  x   x d   f (x)  e 1 2 2 2

Note. It means  can be replaced by  in the formula and we will get required solution. Que5. Using the Fourier integral representation, show that

0

cos  x  d   e  x (x  0) . 2 1  2

x

Solution. Let f (x)  e . Therefore the Fourier cosine integral representation of f (x) will be 

f (x)   A( ) cos  x d  , where A( )  0

2



0

f (x) cos  x dx .

Dr. Jogendra Kumar

Page 5

FOURIER INTEGRALS 

2

0

f (x)   A( ) cos  x d and A( ) 

We can rewrite the formula as

0

f (x) cos  x dx

Now,

A( )  

2

0

f (x) cos  x dx 

2

0

e x e cos  x dx    cos  x   sin  x   1 2 0 2

x

2  (1   2 ) 

 f (x)   A( ) cos  x d  0



2



0

cos  x d 1 2

cos  x   d  f (x)  e x 2 1  2 2

0

Que6. Using the Fourier integral representation, show that

0

sin  cos  x  d   , (0  x  1)  2

Solution. Clearly the given problem is Fourier sine integral representation of some function f (x) . Let f (x)  1 for (0  x  1) . Therefore, the Fourier cosine integral representation of f (x) will be given by f (x) 

0

A( ) cos  x d  , where A( ) 

We can rewrite the formula as

2



0

f (x) cos  x dx .

2

0

f (x)   A( ) cos  x d and A( ) 

0

f (x) cos  x dx

Now,

2

2

1

f (x) cos  x dx  0 1 2 1 2 sin  x 2sin    cos  x dx   0    0 

A( ) 

0

f (x) cos  x dx 

f (x) isdefined for 0  x  1only 

Dr. Jogendra Kumar

Page 6

FOURIER INTEGRALS 

2 sin 

0

 f (x)  

sin  x d

0

sin  cos  x

d 

  f (x)  2 2

Que7. Using the Fourier integral representation, show that

0

1  sin  , 0     sin   sin   d  2 2  1    0,

Solution. This is the same question as Que2. In solution of Que2 just replace

 by  and x by  .

Hints: (I) In using Fourier integral and show Type question always suppose f (x) independent from constant and in same conditions which is given for integral as we have taken in above problems. (ii) Observe the term in integral which is defined in term of the variable w.r.t. which function is being integrated and if it is sine term then consider Fourier sine integral and if it is cosine term then consider Fourier cosine integral. (iii) Learn all formulae and definitions.

Dr. Jogendra Kumar

Page 7

## fourier integrals

1 sin(1 )x sin(1 )x. 1 (1 )sin(1 )x (1 )sin(1 )x. 1. 1. 2. 1. (1 ) sin cos cos sin. (1 ) sin cos cos sin. 1. 1. 1 2sin. 2sin. 1. (1. ) x x x x x x x x Ï Ï Ï Î» Î» Î» Î» Î» Î» Ï.

#### Recommend Documents

Integrals-Wikipedia_Pages_on_Integral_Definitions.pdf
Lebesgue. âStieltjes integration 27. Motivic integration 30. Paley. âWiener integral 31. Pfeffer integral 32. Regulated integral 33. Riemann integral 35. Riemann.

Fourier series
Fourier series in an interval of length i2. Even Function. Odd Function. Convergence of Fourier Series: â¢ At a continuous point x = a, Fourier series converges to ...

Integrals-Wikipedia_Pages_on_Integral_Definitions.pdf
Page 2 of 54. Contents. Articles. Bochner integral. 1. Daniell integral. 3. Darboux integral. 5. Henstock–Kurzweil integral. 8. Homological integration 11. Itō calculus 12. Lebesgue integration 19. Lebesgue. –Stieltjes integration 27. Motivic integra

Integrals-CÃ lcul arees.pdf
... de calcular aquestes Ã rees. Ãrea de figures planes. â¢ Ãrea limitada per la grÃ fica d'una funciÃ³ contÃ­nua, l'eix d'abscisses i les rectes x = a i x = b. Page 1 of 5 ...

THE FOURIER-STIELTJES AND FOURIER ALGEBRAS ...
while Cc(X) is the space of functions in C(X) with compact support. The space of complex, bounded, regular Borel measures on X is denoted by. M(X).

fourier transformation
1. (x). (s). 2. 1. 2sin sin. (x). (x). 2. Now putting x 0 both sides, we get sin. (0). [ f(0) 1by definition of f(x)] sin sin sin. 2. 2 isx isx isx f. F. e d s s s f e ds e ds f s s s. d s.

lecture 15: fourier methods - GitHub
LECTURE 15: FOURIER METHODS. â¢ We discussed different bases for regression in lecture. 13: polynomial, rational, spline/gaussianâ¦ â¢ One of the most important basis expansions is ... dome partial differential equations. (PDE) into ordinary diffe

Computing multiple integrals involving matrix ...
Aug 3, 2006 - Key words: multiple integrals, matrix exponential. ... the context of filtering theory, as the system noise covariance matrix of the extended ...

Fascinating Fourier Series
Nov 30, 2007 - by M.R. Spiegel. Using these formulae, any periodic function can be expressed in terms of its Fourier series expansion. We use these definitions to deduce some interesting mathematical series in the following sections. Using the Fourie

A Study on Double Integrals
This paper uses the mathematical software Maple for the auxiliary tool to study two types of ... The computer algebra system (CAS) has been widely employed in ...

Complex Integrals (final).pdf
HMK We say that the arc L is rectifiable if the least upper bounds (or supremum) of the. sums. |GK â GN. | + |GO â GK. | + â¯ + |GL â GLJK|.....(i) taken over all ...

Choquet Integrals for Symmetric Ca
published online January 17, 2002 ... (i) the decision maker respects (Ak), (ii) f is a polynomial of degree k, (iii) the weight of all coalitions with ..... We then get: Tk.

Ellis, Rosen, Laplace's Method for Gaussian Integrals with an ...
Ellis, Rosen, Laplace's Method for Gaussian Integrals with an Application to Statistical Mechanics.pdf. Ellis, Rosen, Laplace's Method for Gaussian Integrals with ...

Convergence of Wigner integrals to the tetilla law
disjoint subsets of [m], called blocks, such that their union is equal to [m]. A partition Ï of [m] is said to be non-crossing if ..... are all equal to zero is a direct consequence of the symmetry of the density h, see (2.8). Recall next that Îº2k+

Chen's theory of iterated integrals and the algebraic ...
Chen gives meaning to. A(PM) through the notion of a differentiable space. The n-iterated integral map / : A(M)ân â A(PM) is constructed as an integral along ...

Fourier Transformation for Pedestrians
casting, reproduction on microfilm or in any other way, and storage in data banks. Duplication of ... This English edition is based on the third, enlarged edition in German. [4]. ...... smaller and smaller, provided only we make N big enough.

Fourier Transformation for Pedestrians
Springer is a part of Springer Science+Business Media. ... From the host of available material we'll ..... serve as a small introduction to dealing with complex numbers. ..... As frequency Ï = 0 â a frequency as good as any other frequency Ï =0â

FOURIER-MUKAI TRANSFORMATION ON ...
Also, since Âµ and p1 : X Ã ËX â X are transverse morphisms, we have p12â(Âµ Ã id ËX)âP = Âµâp1âP. But, in K0(X Ã ËX), p1âP = âi(â1)i[Rip1âP]. From [13, Â§ 13] we have that Rip1âP = 0 for i = g and Rgp1âP = k(0) where k(0)