# Integral/fourier transform

jennyjones
I made a few excercises with fourier series, fourier integrals and fourier transforms.

But i am getting stuck at a few questions,

most of the time a fourier transform needs to be calculated in part a,
and than part b ask to solve an intergal with the help of your aswer by part a.

i made a picture of one of my excersices

I cannot find this sort of problems in my textbook or internet,
does someone have a source?

thanx,

jenny

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## Answers and Replies

MisterX
$\sin(ax)\cos(ax) = \frac{1}{2}\sin(2ax)$

Let $I$ be the integral we want to solve.

$I =\int_{0}^{\infty} \frac{\sin(ax)\cos(ax)}{x}dx = \int_{0}^{\infty} \frac{\sin(2ax)}{2x}dx = \frac{1}{2}\int_{-\infty}^{\infty} \frac{\sin(2ax)}{2x}dx$

$g(x) = \frac{\sin(2ax)}{2x} = f(2x)$

$\tilde{g}(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{\sin(2ax)}{2x}\cos(kx)dx = \frac{1}{2}\tilde{f}(\frac{k}{2})$

Now let $k = 0$ so the cosine becomes 1.

$\tilde{g}(0) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{\sin(2ax)}{2x}dx = \frac{1}{\sqrt{2\pi}}2I = \frac{1}{2}\tilde{f}(0) = \frac{1}{2}\sqrt{\frac{\pi}{2}}$

$\frac{1}{\sqrt{2\pi}}2I = \frac{1}{2}\sqrt{\frac{\pi}{2}}$

$I = \frac{1}{4}\sqrt{\frac{\pi}{2}}\sqrt{2\pi} = \frac{\pi}{4}$

It seems what you did was evaulate the Fourier transform at $\omega = a$, but your description only gives the Fourier transform for $\left|\omega\right| <a$ and $\left|\omega\right| > a$. It happens that at $\omega = a$, the value is 1/2 of what it is when $\left|\omega\right| <a$. That is $\tilde{f}(a) = \frac{1}{2}\tilde{f}(0) = \frac{1}{2}\sqrt{\frac{\pi}{2}}.$ What I wrote above indicates this is correct. This integral can also be performed (tediously) using contour techniques.

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