What is the Fourier transform of \frac{\sin(ax)\cos(ax)}{x}?

[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

 

[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.