Fourier integral problem?

Question

Show that $$ \int_0^{\infty} \frac{\sin \pi \omega \sin x\omega}{1-\omega^2}d\omega= \begin{cases} \frac{\pi}{2}\sin x,&\mbox{ if } 0\leq x\leq\pi\\ \quad\\ 0,&\mbox{ if } x\geq\pi \end{cases} $$

Here is what I did $$ f(x)= \begin{cases} \sin x,&\mbox{ if } 0\leq x\leq\pi\\ \quad\\ 0,&\mbox{ if } x\geq\pi \end{cases} $$ Then I tried to compute the integral

$$ \int_0^{\infty}[A(\omega)\cos(\omega x)+B(\omega)\sin(\omega x)]d\omega $$

But I never reach my target. What is wrong?

Answer 1

I should have defined $f$ as follows $$ f(x)= \begin{cases} \sin x,&\mbox{ if } |x|\leq\pi\\ \quad\\ 0,&\mbox{ if } |x|\geq\pi \end{cases} $$ So, $A(\omega)$ and $B(\omega)$ are going to be evaluated from $-\pi$ to $\pi.$ The rest is straight forward.

Answer 2

Here is an approach.

1)Write the integral as

$$ I = \frac{1}{2}\int_{-\infty}^{\infty} \frac{\sin \pi \omega \sin x\omega}{1-\omega^2}d\omega = \frac{1}{2}\int_{-\infty}^{\infty} \frac{\sin \pi \omega \sin x\omega}{(1-\omega)(1+\omega)}d\omega $$

and then you need to use the Parseval's identity

$$ \int_{-\infty}^{\infty}f(w)g(w)dw = \int_{-\infty}^{\infty}F(t)G(t)dt. $$

See this.