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I have to prove that, $$\int_0^\infty \frac{\sin(\pi\omega)}{\omega}\cdot \sin(\omega x) \,\textrm{d}\omega = \begin{cases}\frac{\pi \sin (\pi x)}{2} & 0 \leq x \leq \pi \\\\ 0 & x \gt \pi \end{cases}$$

What I can see is it is in the form of Fourier sine integral, How should I do it?

I just compared with Fourier Sine integral

$$f(x) = \int_0^\infty B(\omega) \sin(\omega x) \,d\omega$$

where $$B(\omega) = \frac{2}{\pi} \int_0^\pi\frac{\pi}{2} \sin (\pi t) sin(\omega t) \, dt$$

and see If i can get $$B(\omega) = \frac{\sin (\pi \omega)}{\omega}$$

but on integration it yields,

$$B(\omega) = \frac{1 + \cos (\omega\pi)}{2(1 + \omega)}$$

Mark Viola
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Abhilekh Gautam
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  • Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – SoG Aug 06 '22 at 15:36
  • Try to avoid using lots of mathjax in titles because it doesn't display well on the question page. Please also edit your post to add more context, and what you have tried already. – David Sheard Aug 06 '22 at 15:37
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    Did you mean $\sin\pi x$? – FShrike Aug 06 '22 at 15:49

1 Answers1

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The answer you have is not correct. Using the Extended Frullani Integral, we have for $x\ne \pm\pi$

$$\begin{align} I&=\int_0^\infty\frac{\sin(\pi \omega)}{\omega}\sin(\omega x)\,d\omega\\\\ &=\frac12\int_0^\infty\frac{\cos((\pi-x) \omega)-\cos((\pi+x)\omega)}{\omega}\,d\omega\\\\ &=\frac12\int_0^\infty\frac{\cos(|\pi-x| \omega)-\cos(|\pi+x|\omega)}{\omega}\,d\omega\\\\ &=\frac12 \log\left(\left|\frac{\pi+x}{\pi-x}\right|\right) \end{align}$$

Mark Viola
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