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!
 A: 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.
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\int_{0}^{\infty}\arctan\pars{2ax \over x^{2} + c^{2}}\sin\pars{bx}\,\dd x
     ={\pi \over \verts{b}}
     \expo{-\verts{b}\root{\vphantom{\Large A}a^{2} + c^{2}}}\sinh\pars{ab}:
     \ {\large ?}}$

\begin{align}&\color{#c00000}{\int_{0}^{\infty}\arctan\pars{2ax \over x^{2} + c^{2}}\sin\pars{bx}\,\dd x}
=\half\,\sgn\pars{ab}\Im\int_{-\infty}^{\infty}
\arctan\pars{2\verts{a}x \over x^{2} + c^{2}}\expo{\ic\verts{b}x}\,\dd x
\\[3mm]&=\half\,\sgn\pars{ab}\,\Im\int_{-\infty}^{\infty}{\ic \over 2}\,\ln\pars{%
1 - 2\verts{a}x\ic/\bracks{x^{2} + c^{2}}\over
1 + 2\verts{a}x\ic/\bracks{x^{2} + c^{2}}}\expo{\ic\verts{b}x}\,\dd x
\\[3mm]&={1 \over 4}\,\sgn\pars{ab}\,\Re\int_{-\infty}^{\infty}\ln\pars{%
x^{2}- 2\verts{a}\ic x + c^{2} \over x^{2} + 2\verts{a}\ic x + c^{2}}
\ \underbrace{\expo{\ic\verts{b}x}\,\dd x}
_{\ds{\dd\pars{\expo{\ic\verts{b}x} \over \ic\verts{b}}}}
\end{align}
  $$\begin{array}{|c|}\hline
\\
\quad\mbox{Here, we used the identity}\quad
\arctan\pars{x} = {\ic \over 2}\,\ln\pars{1 - x\ic \over 1 + x\ic}\quad
\\
\\ \hline
\end{array}
$$

Integrating by parts:
\begin{align}&\color{#c00000}{%
\int_{0}^{\infty}\arctan\pars{2ax \over x^{2} + c^{2}}\sin\pars{bx}\,\dd x}
\\[3mm]&=-\,{1 \over 4}\,\sgn\pars{ab}\,\Re\int_{-\infty}^{\infty}
\pars{{2x - 2\verts{a}\ic \over x^{2} - 2\verts{a}\ic x + c^{2}}-
      {2x + 2\verts{a}\ic \over x^{2} + 2\verts{a}\ic x + c^{2}}}
\expo{\ic\verts{b}x}\,{\dd x \over \ic\verts{b}}
\\[3mm]&=-\,{\sgn\pars{a} \over 2b}\,\Im\int_{-\infty}^{\infty}
\pars{{x - \verts{a}\ic \over x^{2} - 2\verts{a}\ic x + c^{2}}-
      {x + \verts{a}\ic\over x^{2} + 2\verts{a}\ic x + c^{2}}}
\expo{\ic\verts{b}x}\,\dd x
\end{align}

\begin{align}
&\mbox{Zeros of}\quad x^{2} - 2\verts{a}\ic x + c^{2} =0
\quad\mbox{are given by}\quad
\left\lbrace\begin{array}{rcl}
\phantom{-}x_{1} & = &\pars{\verts{a} + \root{a^{2} + c^{2}}}\ic
\\[2mm]
\phantom{-}x_{2} & = & \pars{\verts{a} - \root{a^{2} + c^{2}}}\ic
\end{array}\right. 
\\[3mm]&\mbox{Zeros of}\quad x^{2} + 2\verts{a}\ic x + c^{2} =0
\quad\mbox{are given by}\quad
\left\lbrace\begin{array}{rcl}
-x_{1} & = &\pars{-\verts{a} - \root{a^{2} + c^{2}}}\ic
\\[2mm]
-x_{2} & = & \pars{-\verts{a} + \root{a^{2} + c^{2}}}\ic
\end{array}\right. 
\end{align}
  Note that $\ds{\Im\pars{x_{1}} > 0}$ and $\ds{\Im\pars{x_{2}} < 0}$.

Therefore,
\begin{align}
&\color{#c00000}{%
\int_{0}^{\infty}\arctan\pars{2ax \over x^{2} + c^{2}}\sin\pars{bx}\,\dd x}
\\[3mm]&=-\,{\sgn\pars{a} \over 2b}\,\Im\pars{%
2\pi\ic\,{\pars{x_{1} - \verts{a}\ic}\expo{\ic\verts{b}x_{1}}\over
2x_{1} - 2\verts{a}\ic}
-
2\pi\ic\,{\pars{-x_{2} + \verts{a}\ic}\expo{-\ic\verts{b}x_{2}}\over
-2x_{2} + 2\verts{a}\ic}}
\\[3mm]&=-\,{\pi\sgn\pars{a} \over 2b}\braces{%
\exp\pars{-\verts{b}\bracks{\verts{a} + \root{a^{2} + c^{2}}}}
-
\exp\pars{\verts{b}\bracks{\verts{a} - \root{a^{2} + c^{2}}}}}
\\[3mm]&=-\,{\pi\sgn\pars{a} \over 2b}\bracks{%
\expo{-\verts{b}\root{\vphantom{\Large A}a^{2} + c^{2}}}
\pars{\expo{-\verts{ab}} - \expo{\verts{ab}}}}
\\[3mm]&={\pi\sgn\pars{a} \over b}
\expo{-\verts{b}\root{\vphantom{\Large A}a^{2} + c^{2}}}\sinh\pars{\verts{ab}}
={\pi \over \verts{b}}
\expo{-\verts{b}\root{\vphantom{\Large A}a^{2} + c^{2}}}\sinh\pars{ab}
\end{align}

\begin{align}
&\color{#66f}{\large%
\int_{0}^{\infty}\arctan\pars{2ax \over x^{2} + c^{2}}\sin\pars{bx}\,\dd x
={\pi \over \verts{b}}
\expo{-\verts{b}\root{\vphantom{\Large A}a^{2} + c^{2}}}\sinh\pars{ab}}
\end{align}

