Adamchik's Integral : Omega Constant

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Problem: Adamchik's Integral representation of the Omega Constant.
$$\int\limits_{-\infty}^{\infty} \dfrac{1}{(e^x - x)^2 + \pi^2}\,dx = \dfrac{1}{1+\Omega}$$

LINKS: Omega Constant(Wiki) related to the Lambet-W-Function.
OEIS and Omega Constant(Wolfram)

Solution links:
(1) Robjohn's detailed solution on M.SE.
(2) WIKI-LINK with a proof (2.5)
(3) AOPS

Solution: Since, $\displaystyle{\frac{1}{{{{\left( {{e^z} - z} \right)}^2} + {\pi ^2}}} = \frac{1}{{\left( {{e^z} - z - i\pi } \right)\left( {{e^z} - z + i\pi } \right)}} = \frac{1}{{2\pi i}}\left( {\frac{1}{{{e^z} - z - i\pi }} - \frac{1}{{{e^z} - z + i\pi }}} \right)}$

We have : $\displaystyle\int\limits_{ - \infty }^{ + \infty } {\frac{1}{{{{\left( {{e^z} - z} \right)}^2} + {\pi ^2}}}dz}$

$\displaystyle = \frac{1}{2\pi i}\left( {\int\limits_{ - \infty }^{ + \infty } {\frac{1}{{{e^z} - z - i\pi }}dz} - \int\limits_{ - \infty }^{ + \infty } {\frac{1}{{{e^z} - z + i\pi }}dz} } \right)$

$\displaystyle = \frac{1}{{2\pi i}}\left( {\int\limits_{ - \infty }^{ + \infty } {\frac{1}{{ - {e^{\left( {z + i\pi } \right)}} - \left( {z + i\pi } \right)}}dz} - \int\limits_{ - \infty }^{ + \infty } {\frac{1}{{ - {e^{\left( {z - i\pi } \right)}} - \left( {z - i\pi } \right)}}dz} } \right)$

$\displaystyle = \frac{1}{{2\pi i}}\left( {\int\limits_{ - \infty + i\pi }^{ + \infty + i\pi } {\frac{1}{{ - {e^w} - w}}dw} - \int\limits_{ - \infty - i\pi }^{ + \infty - i\pi } {\frac{1}{{ - {e^w} - w}}dw} } \right)$

$\displaystyle = \frac{1}{{2\pi i}}\left( {\int\limits_{ - \infty - i\pi }^{ + \infty - i\pi } {\frac{1}{{{e^w} + w}}dw} - \int\limits_{ - \infty + i\pi }^{ + \infty + i\pi } {\frac{1}{{{e^w} + w}}dw} } \right)$

$\displaystyle = \frac{1}{{2\pi i}}\left( {\int\limits_{ - \infty - i\pi }^{ + \infty - i\pi } {\frac{1}{{{e^w} + w}}dw} + \int\limits_{ + \infty + i\pi }^{ - \infty + i\pi } {\frac{1}{{{e^w} + w}}dw} } \right)$.

But, $\displaystyle{\left| {\frac{1}{{{e^w} + w}}} \right| \geqslant \frac{1}{{\left| {{e^w}} \right| + \left| w \right|}}\xrightarrow{{w \to \infty }}0}$
so, $\displaystyle{\int\limits_c {\frac{1}{{{e^w} + w}}dw} = \int\limits_{ - \infty - i\pi }^{ + \infty - i\pi } {\frac{1}{{{e^w} + w}}dw} + \int\limits_{ + \infty + i\pi }^{ - \infty + i\pi } {\frac{1}{{{e^w} + w}}dw} }$

Where, $\displaystyle{c:}$ rectangles with vertices $\displaystyle{A\left( { - M, - \pi } \right),\;B\left( {M, - \pi } \right),\;C\left( {M,\pi } \right)\;\& \;D\left( { - M,\pi } \right)}$ with $\displaystyle{M \to + \infty }$ and a proper direction of rotation.

The function $\displaystyle{f\left( w \right) = \frac{1}{{{e^w} + w}}}$ is meromorphic inside the rectangle $\displaystyle{c}$ (with some poles).

We seek the poles of $\displaystyle{f\left( w \right) = \frac{1}{{{e^w} + w}}}$ inside $\displaystyle{c}$ , therefore the roots of the equation $\displaystyle{{e^w} + w = 0}$ .

Now, $\displaystyle {e^w} + w = 0 \Rightarrow {e^{x + iy}} + x + iy = 0 \Rightarrow {e^x}\cos y + x = 0$ and $\displaystyle {{e^x}\sin y + y = 0}$.

For $\displaystyle{y = 0}$ unique real roots of the equation $\displaystyle{{e^x} + x = 0}$.

For $\displaystyle{y \ne 0}$ we $\displaystyle{{e^x} = - \frac{y}{{\sin y}}}$ which is impossible for $\displaystyle{y \in \left[ { - \pi ,\;\pi } \right]}$ (inside the rectangle $\displaystyle{c}$ ). So if $\displaystyle{\rho :}$ are the roots of the equation $\displaystyle{{e^x} + x = 0}$,

we have, $\displaystyle\int\limits_c {\frac{1}{{{e^w} + w}}dw} = 2\pi i \cdot {\text{Res}}\left( {\frac{1}{{{e^w} + w}},w = \rho } \right)$

$\displaystyle = 2\pi i \cdot \mathop {\lim }\limits_{w \to \rho } \frac{{w - \rho }}{{{e^w} + w}} = \frac{{2\pi i}}{{{e^\rho } + 1}}$,

so, $\displaystyle\int\limits_{ - \infty }^{ + \infty } {\frac{1}{{{{\left( {{e^z} - z} \right)}^2} + {\pi ^2}}}dz}$

$\displaystyle = \frac{1}{{2\pi i}}\left( {\int\limits_{ - \infty - i\pi }^{ + \infty - i\pi } {\frac{1}{{{e^w} + w}}dw} + \int\limits_{ + \infty + i\pi }^{ - \infty + i\pi } {\frac{1}{{{e^w} + w}}dw} } \right)$

$\displaystyle = \frac{1}{{2\pi i}}\int\limits_c {\frac{1}{{{e^w} + w}}dw} = \frac{1}{{{e^\rho } + 1}}$

But, $\displaystyle{e^\rho } + \rho = 0 \Rightarrow {e^\rho } = \left( { - \rho } \right) \Rightarrow \left( { - \rho } \right){e^{\left( { - \rho } \right)}} = 1$, that is $\displaystyle{ - \rho = \Omega }$ where, $\displaystyle{\Omega :}$ the roots of the equation $\displaystyle{x{e^x} = 1}$.

and finally, $\displaystyle\int\limits_{ - \infty }^{ + \infty } {\frac{1}{{{{\left( {{e^z} - z} \right)}^2} + {\pi ^2}}}dz} = \frac{1}{{{e^\rho } + 1}} = \frac{1}{{\left( { - \rho } \right) + 1}} = \frac{1}{{1 + \Omega }}$