TITLE: Integral: $\int_0^\infty \tan^{-1}\left(\frac{2ax}{x^2+c^2} \right)\sin(bx) \; dx$
QUESTION [9 upvotes]: 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!

REPLY [4 votes]: $\newcommand{\+}{^{\dagger}}
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$\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}