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Find the Fourier sine integral of the function defined by $$ \,\mathrm{f}\left(x\right) = \begin{cases} \sin\left(x\right) & \text{;}\quad 0 \leq x \leq \pi \\[1mm] 0 & \text{;}\quad \ x\ > \pi \end{cases} $$ and hence evaluate the integral $$ \int_{0}^{\infty}{\sin\left(\pi\omega\right)\sin\left(x\omega\right) \over 1 - \omega^{2}}\,\mathrm{d}\omega\,;\quad x \geq 0 $$

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    $\begingroup$ Done. And you, what have you attempted? $\endgroup$ Commented Jun 3, 2017 at 21:31
  • $\begingroup$ @JackD'Aurizio,I am completely new to fourier.I have tried it on my own,but I failed.If you provide me with an answer,it would be greatful of you. $\endgroup$
    – Asm Arman
    Commented Jun 3, 2017 at 21:39
  • $\begingroup$ @JackD'Aurizio I have tried to solve the first part using sine integral.But failed to solve the integration. $\endgroup$
    – Asm Arman
    Commented Jun 5, 2017 at 16:29

1 Answer 1

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I think we first need $B(\alpha)=\int_{0}^{\infty}f(x)\sin \alpha x \, dx$. Then the Fourier integral (sine) is $\frac{2}{\pi}\int_{0}^{\infty}B(\alpha)\sin \alpha x \, d\alpha$.

Now $B(\alpha)=\int_{0}^{\pi}\sin x \sin \alpha x \, dx$. Note that $2\sin x\sin \alpha x=\cos(x-\alpha x)-\cos(x+\alpha x)$. Hence $B(\alpha)=\frac{1}{2}(\int_{0}^{\pi}\cos(x-\alpha x) \, dx-\int_{0}^{\pi}\cos(x+\alpha x)\, dx)$. So, $2B(\alpha)=\int_{0}^{\pi}\cos(x(1-\alpha))\, dx-\int_{0}^{\pi}\cos(x(1+\alpha))\, dx$. So $2B(\alpha)=\frac{1}{1-\alpha}(\sin(1-\alpha)\pi)-\frac{1}{1+\alpha}(\sin(1+\alpha)\pi)$. So $2B(\alpha)=\frac{1}{1-\alpha}(\sin \pi \alpha)+\frac{1}{1+\alpha}(\sin \pi \alpha)$. So $B(\alpha)=\frac{1}{1-\alpha^2}\sin \pi \alpha$.

Now we need $\frac{2}{\pi}\int_{0}^{\infty}\frac{1}{1-\alpha^2}\cdot \sin \pi\alpha \sin \alpha x \, d\alpha$. Call this integral $I$. I can't find the value of $I$, but probably someone else can delve into this.

Edit: OK, I see the value of $I$ is what is asked to be found in the next part of the question.

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