Question

Using problem 17,show that

$\begin{array}{l}\underset{0}{\overset{\infty }{\int }}\frac{1-\cos \pi \alpha }{\alpha }\sin \alpha d\alpha =\frac{\pi }{2}\\ \underset{0}{\overset{\infty }{\int }}\frac{1-\cos \pi \alpha }{\alpha }\sin \alpha d\alpha =\frac{\pi }{4}\end{array}$

Short answer

By using the problem 17 and the Dirichlet theorem the given function can be proved.

Step-by-step solution

Definition of Fourier series

The Dirichlet theorem 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

$\begin{array}{l}\underset{0}{\overset{\infty }{\int }}\frac{1-\cos \pi \alpha }{\alpha }\sin \alpha d\alpha =\frac{\pi }{2}\\ \underset{0}{\overset{\infty }{\int }}\frac{1-\cos \pi \alpha }{\alpha }\sin \alpha d\alpha =\frac{\pi }{4}\end{array}$

There need to prove the given functions using problem 17.

Use problem 17

The problem 17 says that

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

Take different values of used in problem 17

Firstly, For$x=1$:

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

For $x=\pi$Dirichlet theorem says that the series converges to $\frac{1}{2}$, so

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

Thus, using $x=\pi$and $x=1$in the problem 17 the given functions are proved.