Prove that, $\int^{\infty}_0\frac{x\sin(rx)}{a^2+x^2}dx=\frac{\pi}{2}e^{-ar}$ [duplicate]

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!

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