TITLE: Prove that $\int_0^{\infty} \frac{\sin(2013 x)}{x(\cos x+\cosh x)}dx=\frac{\pi}{4}$
QUESTION [50 upvotes]: Prove that
$$\int_0^{\infty} \frac{\sin(2013 x)}{x(\cos x+\cosh x)}dx=\frac{\pi}{4}$$

REPLY [9 votes]: $\newcommand{\+}{^{\dagger}}
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$\ds{\int_{0}^{\infty}{\sin\pars{2013 x}\,\dd x
     \over x\bracks{\cos\pars{x} + \cosh\pars{ x}}}={\pi \over 4}:\ {\large ?}}$

Note that
  \begin{align}&\color{#c00000}{\cos\pars{x} + \cosh\pars{x}}
=\cos\pars{x} + \cos\pars{x\ic}
\\[3mm]&=\cos\pars{{x + x\ic \over 2} + {x - x\ic \over 2}}
+\cos\pars{{x + x\ic \over 2} - {x - x\ic \over 2}}
=2\cos\pars{x + x\ic \over 2}\cos\pars{x - x\ic \over 2}
\\[3mm]&=2\cos\pars{\mu x}\cos\pars{\mu^{*}x}
\qquad\mbox{where}\qquad\mu\equiv{1 + \ic \over 2} = \expo{\pi\ic/4}
\end{align}

Then,
\begin{align}&\color{#c00000}{%
\int_{0}^{\infty}{\sin\pars{m x}\,\dd x
\over x\bracks{\cos\pars{x} + \cosh\pars{ x}}}}
={1 \over 4}\int_{-\infty}^{\infty}
{\sin\pars{mx} \over x\cos\pars{\mu x}\cos\pars{\mu^{*}x}}\,\dd x\end{align}
With the identity
$\ds{{\sin\pars{\xi} \over \xi} = \int_{0}^{1}\cos\pars{k\xi}\,\dd k
     =\Re\int_{0}^{1}\expo{\ic\verts{k}\xi}\,\dd k}$ we'll have:
$$
\color{#c00000}{%
\int_{0}^{\infty}{\sin\pars{m x}\,\dd x
\over x\bracks{\cos\pars{x} + \cosh\pars{ x}}}}
={1 \over 4}\,m\,\Re\int_{0}^{1}\color{#00f}{\int_{-\infty}^{\infty}
{\expo{\ic\verts{mk}x}\,\dd x \over \cos\pars{\mu x}\cos\pars{\mu^{*}x}}}\,\dd k
\tag{1}
$$

Zeros of $\ds{\cos\pars{\mu x}}$ are given by
  $\ds{\pars{~\mbox{with}\ n \in {\mathbb Z}~}}$:
  $$
x_{n} = \pars{n + \half}\,{\pi \over \mu}
=\pars{2n + 1}\,\pars{1 - \ic}\,{\pi \over 2}\,,\qquad
\Im\pars{x_{n}} > 0\quad\mbox{if}\quad n=-1,-2,-3,\ldots
$$
  Similarly, zeros of $\ds{\cos\pars{\mu^{*}x}}$ are given by $\ds{\pars{~\mbox{with}\ n \in {\mathbb Z}~}}$:
  $$
x_{n}^{*} = \pars{n + \half}\,{\pi \over \mu^{*}}
=\pars{2n + 1}\,\pars{1 + \ic}\,{\pi \over 2}\,,\qquad
\Im\pars{x_{n}^{*}} > 0\quad\mbox{if}\quad n=0,1,2,\ldots
$$
  $\large\begin{array}{|c|}\hline
\mbox{Note that}\quad \ds{x_{-n - 1} = -x_{n}}\\ \hline\end{array}$.

Then,
\begin{align}&\color{#00f}{\int_{-\infty}^{\infty}
{\expo{\ic\verts{mk}x}\,\dd x \over \cos\pars{\mu x}\cos\pars{\mu^{*}x}}}
\\[3mm]&=2\pi\ic\bracks{%
\sum_{n = -\infty}^{-1}
{\expo{\ic\verts{mk}x_{n}} \over -\mu\sin\pars{\mu x_{n}}\cos\pars{\mu^{*}x_{n}}}
+
\sum_{n = 0}^{\infty}
{\expo{\ic\verts{mk}x_{n}*}\over
\cos\pars{\mu x_{n}^{*}}\pars{-\mu*}\sin\pars{\mu^{*}x_{n}^{*}}}}
\\[3mm]&=2\pi\ic\sum_{n = 0}^{\infty}\bracks{%
{\expo{-\ic\verts{mk}x_{n}} \over \mu\sin\pars{\mu x_{n}}\cos\pars{\mu^{*}x_{n}}}
-
{\expo{\ic\verts{mk}x_{n}*}\over
\mu^{*}\sin\pars{\mu^{*}x_{n}^{*}}\cos\pars{\mu x_{n}^{*}}}}
\\[3mm]&=-4\pi\,\Im
\sum_{n = 0}^{\infty}{\mu^{*} \over \verts{\mu}^{2}}\,
{\expo{-\ic\verts{mk}x_{n}} \over \sin\pars{\mu x_{n}}\cos\pars{\mu^{*}x_{n}}}
\\[3mm]&=-4\pi\,\Im
\sum_{n = 0}^{\infty}{\pars{1 - \ic}/2 \over 1/2}\,
{\expo{-\ic\verts{mk}x_{n}}\over
\sin\pars{\bracks{2n + 1}\pi/2}\cos\pars{\ic\bracks{2n + 1}\pi/2}}
\end{align}

$$\color{#00f}{\int_{-\infty}^{\infty}
{\expo{\ic\verts{mk}x}\,\dd x \over \cos\pars{\mu x}\cos\pars{\mu^{*}x}}}
=-4\pi\,\Im
\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over \cosh\pars{\bracks{2n + 1}\pi/2}}\,
\pars{1 - \ic}\expo{-\ic\verts{mk}x_{n}}
$$

We replace this result in expression $\pars{1}$:
\begin{align}&\color{#c00000}{%
\int_{0}^{\infty}{\sin\pars{m x}\,\dd x
\over x\bracks{\cos\pars{x} + \cosh\pars{ x}}}}
=-\pi\,m\,\Im
\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over \cosh\pars{\bracks{2n + 1}\pi/2}}\,
\pars{1 - \ic}\int_{0}^{1}\expo{-\ic\verts{m}x_{n}k}\,\dd k
\\[3mm]&=-\pi\,m\,\Im
\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over \cosh\pars{\bracks{2n + 1}\pi/2}}\,
\pars{1 - \ic}\,{\expo{-\ic\verts{m}x_{n}} - 1 \over -\ic\verts{m}x_{n}}
\\[3mm]&=-\pi \sgn\pars{m}\Im
\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over \cosh\pars{\bracks{2n + 1}\pi/2}}\,
\,\ic\,{\expo{-\ic\verts{m}\pars{2n + 1}\pi/2}
\expo{-\verts{m}\pars{2n + 1}\pi/2} - 1 \over \pars{2n + 1}\pi/2}
\\[3mm]&=-2\sgn\pars{m}\sum_{n = 0}^{\infty}{\pars{-1}^{n}\over
\pars{2n + 1}\cosh\pars{\bracks{2n + 1}\pi/2}}\times
\\[3mm]&
\phantom{=-2\sgn\pars{m}\sum_{n = 0}^{\infty}}
\bracks{\cos\pars{\verts{m}\pars{2n + 1}\,{\pi \over 2}}
\expo{-\verts{m}\pars{2n + 1}\pi/2} - 1}
\end{align}

Note that
  $$
2\sum_{n = 0}^{\infty}{\pars{-1}^{n}\over
\pars{2n + 1}\cosh\pars{\bracks{2n + 1}\pi/2}} = {\pi \over 4}
$$

\begin{align}&\color{#00f}{\large%
\int_{0}^{\infty}{\sin\pars{m x}\,\dd x
\over x\bracks{\cos\pars{x} + \cosh\pars{ x}}}}=
\\[3mm]&\color{#c00000}{\left\lbrace\begin{array}{lcl}
{\pi \over 4}\,\sgn\pars{m} & \mbox{if} & m\ \mbox{is odd}
\\[3mm]\sgn\pars{m}\braces{%
{\pi \over 4}
-2\pars{-1}^{\verts{m}/2}\sum_{n = 0}^{\infty}\pars{-1}^{n}\,
{\expo{-\verts{m}\pars{2n + 1}\pi/2}\over
\pars{2n + 1}\cosh\pars{\bracks{2n + 1}\pi/2}}}
 & \mbox{if} & m\ \mbox{is even}
\end{array}\right.}
\end{align}