Integral: $\int_0^\infty \tan^{-1}\left(\frac{2ax}{x^2+c^2} \right)\sin(bx) \; dx$ Please help me in proving the following result:
$$\displaystyle \int_0^\infty \tan^{-1}\left(\frac{2ax}{x^2+c^2} \right)\sin(bx) \; dx=\frac{\pi}{b}e^{-b\sqrt{a^2+c^2}}\sinh (ab)$$

I found this integral from here: http://integralsandseries.prophpbb.com/post2652.html?sid=d6641d4d4a3726f1b27bbb4b98ca840a and the solution uses contour integration. I am wondering if there is a way to solve it without using contour integration. I tried differentiating wrt $a$ and $c$ but in both cases, the resulting expression was dirty which made me reluctant to proceed further. I am out of ideas for this one.
Any help is appreciated. Thanks!

In order to prove the final result I will need to state a lemma that will be used later.

Lemma$\require{autoload-all}$ $1$:
$$\int_0^\infty \! \frac{\cos(bx)}{x^2+\alpha} \mathrm{d}x = \frac{\pi e^{-b\sqrt{\alpha}}}{2b\sqrt{\alpha}}\tag{1}$$
Proof here.

Consider
$$I = \int_0^\infty\!\! \tan^{-1}\left(\frac{2ax}{x^2+c^2} \right)\sin(bx) \; \mathrm{d}x$$
Integrate by parts
$$I = \int_0^\infty \!\!\frac{2 a \left(c^2-x^2\right) \cos (b x)}{x^4 +(4 a^2+2 c^2) x^2+c^4} \; \mathrm{d}x$$
Decompose this function by partial fractions $$\frac{2 a \left(c^2-x^2\right) }{x^4 +(4 a^2+2 c^2) x^2+c^4} = \frac{a_-}{x^2+x_0} + \frac{a_+}{x^2+x_1}$$
It so happens that
$$x_0 = 2 a^2+2 a\sqrt{a^2+c^2}+c^2,\quad x_1 = 2 a^2-2a \sqrt{a^2+ c^2}+c^2$$
$$a_- =\frac{2a(c^2+x_0)}{x_1-x_0}, \quad a_+ = \frac{2a(c^2+x_1)}{x_0-x_1}$$
Note that both $x_0$ and $x_1$ are greater than $0$.

Re-write the integral
$$\begin{align} I &= a_-\int_0^{\infty} \!\! \frac{\cos(bx)}{x^2+x_0} \, \mathrm{d}x + a_+\int_0^{\infty} \!\! \frac{\cos(bx)}{x^2+x_1} \, \mathrm{d}x\\[.3cm] &= \frac{2a(c^2+x_0)}{x_1-x_0}\int_0^{\infty} \!\! \frac{\cos(bx)}{x^2+x_0} \, \mathrm{d}x +\frac{2a(c^2+x_1)}{x_0-x_1}\int_0^{\infty} \!\! \frac{\cos(bx)}{x^2+x_1} \, \mathrm{d}x\end{align}$$
Using $(1)$:
$$\begin{align} I &= \frac{2a(c^2+x_0)}{x_1-x_0}\cdot\frac{\pi e^{-b\sqrt{x_0}}}{2b\sqrt{x_0}} - \frac{2a(c^2+x_1)}{x_1-x_0}\cdot\frac{\pi e^{-b\sqrt{x_1}}}{2b\sqrt{x_1}}\\[.3cm] &= \left(\frac{a\pi}{b(x_1-x_0)}\right)\left(\frac{(c^2+x_0)e^{-\sqrt{x_0}}}{\sqrt{x_0}} - \frac{(c^2+x_1)e^{-\sqrt{x_1}}}{\sqrt{x_1}}\right)\end{align}$$

I will digress here to state (without proof but easily verified) that $$\frac{c^2+x_0}{\sqrt{x_0}}= \frac{c^2+x_1}{\sqrt{x_1}} = \frac{x_1-x_0}{2a}.$$ This allows us a tremendous simplification so that we can write
$$\begin{align} I = \left(\frac{\pi}{2b}\right)\left(e^{-b\sqrt{x_1}}-e^{-b\sqrt{x_0}} \right). \end{align}$$
It can also be shown that 
$$\sqrt{x_1} = -a+\sqrt{a^2+c^2}$$ 
$$\sqrt{x_0} = a+\sqrt{a^2+c^2}$$ 
Simply square each side to find the desired equality.
We can now complete the proof:
$$\begin{align} I &= \frac{\pi}{2b}\left(e^{-b\sqrt{x_1}}-e^{-b\sqrt{x_0}} \right) \\[.2cm]
&=  \frac{\pi}{2b}\left(\exp\left(ab-b\sqrt{a^2+c^2}\right)-\exp\left(-ab-b\sqrt{a^2+c^2}\right) \right) \\[.2cm]
&=  \frac{\pi}{b}\exp\left(-b\sqrt{a^2+c^2}\right)\frac12(\exp(ab)-\exp(ab)) \\[.2cm]
&= \dfrac{\pi}{b}\exp\left(-b\sqrt{a^2+c^2}\right)\sinh{ab}\end{align}$$
If you are interested in working through the simplifications that I did not prove, I recommend that you begin by squaring each side after verifying that each side shares the same sign.
