Question

In Problems 17to 20,find the Fourier sine transform of the function in the indicated problem, and write f(x)as a Fourier integral [use equation (12.14)]. Verify that the sine integral for f(x)is the same as the exponential integral found previously.

Problem 10.

Short answer

It has been shown that the sine integral for $f\left(x\right)$is the same as the exponential integral found previously. The fourier sine transform of the function in the problem 6 is given below.

$f\left(x\right)=\frac{4}{\pi }\underset{0}{\overset{\infty }{\int }}\frac{a\alpha -\sin \left(a\alpha \right)}{{\alpha }^2}\sin \left(\alpha x\right)d\alpha$

Step-by-step solution

Given Information.

The graph of a given function is shown below.

Definition of fourier transform

The Fourier transform is a mathematical technique for expressing a function as the summation of sines and cosines functions.

Step 3: To find the fourier sine transform of the given function

The fourier sine transform is given below.

$\begin{array}{l}g\left(\alpha \right)=\sqrt{\frac{2}{\pi }}\underset{0}{\overset{\alpha }{\int }}2(\alpha -x)\sin \left(\alpha x\right)dx\\ g\left(\alpha \right)=2\alpha \sqrt{\frac{2}{\pi }}\underset{0}{\overset{\alpha }{\int }}\sin \left(\alpha x\right)dx-2\sqrt{\frac{2}{\pi }}\underset{0}{\overset{\alpha }{\int }}x\sin \left(\alpha x\right)dx\end{array}$

Simplify further

$\begin{array}{l}g\left(\alpha \right)=-\alpha \sqrt{\frac{8}{\pi }}\left[\frac{\cos \left(\alpha x\right)}{\alpha }\right]{|}_0^a-\sqrt{\frac{8}{\pi }}[-\frac{x}{\alpha }\cos \left(\alpha x\right)+\frac{1}{a^2}\sin \left(\alpha x\right)]{|}_0^a\\ g\left(\alpha \right)=-\alpha \sqrt{\frac{8}{\pi }}\left[\frac{\cos \left(\alpha a\right)-1}{\alpha }\right]-\sqrt{\frac{8}{\pi }}[-\frac{a}{\alpha }\cos \left(\alpha a\right)+\frac{1}{a^2}\sin \left(\alpha a\right)]{|}_0^a\\ g\left(\alpha \right)=\sqrt{\frac{8}{\pi }}\left[\frac{\alpha a-\sin \left(\alpha a\right)}{a^2}\right]\end{array}$

Thus the function is$f\left(x\right)=\frac{4}{\pi }\underset{0}{\overset{\infty }{\int }}\frac{\alpha a-\sin \left(\alpha a\right)}{{\alpha }^2}\sin \left(\alpha a\right)da$

The the solution to the problem (12.10) is$f\left(x\right)=\frac{2}{\pi }\underset{-\infty }{\overset{\infty }{\int }}\frac{a\alpha -\sin \left(a\alpha \right)}{{\alpha }^{2}i}{e}^{i\alpha x}d\alpha$

Only consider the $i\sin \left(\alpha x\right)$part of the complex exponential as the function in front of the complex exponential is odd.

$\begin{array}{l}f\left(x\right)=\frac{2}{\pi }\underset{-\infty }{\overset{\infty }{\int }}\frac{a\alpha -\sin \left(\alpha a\right)}{{\alpha }^{2}i}\left(i\left(\sin \left(\alpha x\right)\right)\right)d\alpha \\ f\left(x\right)=\frac{4}{\pi }\underset{0}{\overset{\infty }{\int }}\frac{a\alpha -\sin \left(\alpha a\right)}{{\alpha }^2}\left(\sin \left(\alpha x\right)\right)d\alpha \end{array}$

Therefore, the fourier sine transform of the function in the problem 10 is$f\left(x\right)=\frac{4}{\pi }\underset{0}{\overset{\infty }{\int }}\frac{a\alpha -\sin \left(a\alpha \right)}{{\alpha }^2}\sin \left(\alpha x\right)d\alpha$.