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The question is: prove that

$$\int^{\infty}_0\frac{x\sin(rx)}{a^2+x^2}dx=\frac{\pi}{2}e^{-ar}$$

This is what I've got so far:

Let $I(r)=\int^{\infty}_0\frac{x\sin(rx)}{a^2+x^2}dx$

$I'(r)=\int^{\infty}_0\frac{x^2\cos(rx)}{a^2+x^2}dx$ = $\int^{\infty}_0\cos(rx)dx-\int^{\infty}_0\frac{a^2\cos(rx)}{a^2+x^2}dx$

$I''(r)=\int^{\infty}_0-x\sin(rx)dx+\int^{\infty}_0\frac{xa^2\sin(rx)}{a^2+x^2}dx$

$I''(r)=\int^{\infty}_0-x\sin(rx)dx+a^2I(r)$

I think I'm supposed to form a differential equation and solve it by letting $I(r)=c_1e^r+c_2e^{-r}$, but the problematic part is this: $\int^{\infty}_0-x\sin(rx)dx$. I don't believe it simplifies to any constant and the differential equation doesn't solve nicely.

Hints/suggestions appreciated!

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  • $\begingroup$ Already on the previous line you have a divergent integral: $\int_0^\infty\cos rx\,dx$. $\endgroup$ – Lord Shark the Unknown Nov 10 '18 at 4:14
  • $\begingroup$ Yes, that's true. Is there another technique to do this? $\endgroup$ – Yip Jung Hon Nov 10 '18 at 4:41
  • $\begingroup$ Contour integration gives it pretty quickly, but I suppose you haven't encountered them? $\endgroup$ – user10354138 Nov 10 '18 at 4:52
  • $\begingroup$ Nope, sadly not, but you can post your answer here, I'm always willing to learn $\endgroup$ – Yip Jung Hon Nov 10 '18 at 4:54
  • $\begingroup$ See the answers to this question: math.stackexchange.com/q/9402/269624 $\endgroup$ – Yuriy S Nov 10 '18 at 8:19

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