Question

Represent each of the following functions (a) by a Fourier cosine integral, (b) by a Fourier sine integral. Hint: See the discussion just before theParseval’s theorem.

Short answer

By using the problem 15 and Fourier series cosine transform given function can be proved.

Step-by-step solution

Definition of Fourier series

The for Fourier series formula gives an expansion of a periodic function f (x) in terms of an horizonless sum of sines and cosines. It's used to putrefy any periodic function or periodic signal into the sum of a group of straightforward oscillating functions, videlicet sines and cosines.

Step 2:Given parameters

The given functions are

$\underset{0}{\overset{\infty }{\int }}\frac{1-\cos \alpha }{{\alpha }^2}d\alpha =\frac{\pi }{2}$

There need to prove the given functions using problem 15.

Use problem 15

The problem 15 says that

$\begin{array}{c}f\left(x\right)=\frac{4}{\pi }{\int }_0^{\infty }\frac{1-\cos \left(\alpha x\right)}{{\alpha }^2}\cos \left(\alpha x\right)d\alpha \\ =\left\{\begin{array}{l}\begin{array}{ll}2(a-x)& ,x\in [0,a]\end{array}\\ \begin{array}{ll}2(x+a)& ,x\in [-a,0]\end{array}\end{array}\right.\end{array}$

Set the values of and  used in problem

Set the values that is $a=1$and $x=0$This gives:

$\begin{array}{c}f\left(0\right)=2\\ =\frac{4}{\pi }{\int }_0^{\infty }\frac{1-\cos \alpha }{{\alpha }^2}d\alpha \\ \frac{\pi }{2}={\int }_0^{\infty }\frac{1-\cos \alpha }{{\alpha }^2}d\alpha \end{array}$

Thus, using $x=0$and $a=1$in the problem 15 the given functions are proved.