Interesting integral related to the Omega Constant/Lambert W Function I ran across an interesting integral and I am wondering if anyone knows where I may find its derivation or proof.  I looked through the site. If it is here and I overlooked it, I am sorry.
$$\displaystyle\frac{1}{\int_{-\infty}^{\infty}\frac{1}{(e^{x}-x)^{2}+{\pi}^{2}}dx}-1=W(1)=\Omega$$
$W(1)=\Omega$ is often referred to as the Omega Constant. Which is the solution to 
$xe^{x}=1$.  Which is $x\approx .567$
Thanks much. 
EDIT:  Sorry, I had the integral written incorrectly. Thanks for the catch.
I had also seen this:
$\displaystyle\int_{-\infty}^{\infty}\frac{dx}{(e^{x}-x)^{2}+{\pi}^{2}}=\frac{1}{1+W(1)}=\frac{1}{1+\Omega}\approx .638$
EDIT:  I do not what is wrong, but I am trying to respond, but can not. All the buttons are unresponsive but this one. I have been trying to leave a greenie and add a comment, but neither will respond. I just wanted you to know this before you thought I was an ingrate.
Thank you. That is an interesting site.

While this is by no means rigorous, but it gives the correct solution.  Any corrections to this are welcome!
Let
$$f(z) := \frac{1}{(e^z-z)^2+\pi^2}$$
Let $C$ be the canonical positively-oriented semicircular contour that traverses the real line from $-R$ to $R$ and all around $Re^{i \theta}$ for $0 \le \theta \le \pi$ (let this semicircular arc be called $C_R$), so 
$$\oint_C f(z)\, dz = \int_{-R}^R f(z)\,dz + \int_{C_R}f(z)\, dz$$
To evaluate the latter integral, we see
$$
\left| \int_{C_R} \frac{1}{(e^z-z)^2+\pi^2}\, dz \right| =
\int_{C_R} \left| \frac{1}{(e^z-z)^2+\pi^2}\right| \, dz \le
\int_{C_R} \frac{1}{(|e^z-z|)^2-\pi^2} \, dz \le
\int_{C_R} \frac{1}{(e^R-R)^2-\pi^2} \, dz
$$
and letting $R \to \infty$, the outer integral disappears.
Looking at the denominator of $f$ for singularities:
$$(e^z-z)^2 + \pi^2 = 0 \implies e^z-z = \pm i \pi \implies z = -W (1)\pm i\pi$$
using this.
We now use the root with the positive $i\pi$ because when the sign is negative, the pole does not fall within the contour because $\Im (-W (1)- i\pi)<0$. 
$$z_0 := -W (1)+i\pi$$
We calculate the beautiful residue at $b_0$ at $z=z_0$:
$$
b_0=
\operatorname*{Res}_{z \to z_0} f(z) = 
\lim_{z\to z_0} \frac{(z-z_0)}{(e^z-z)^2+\pi^2} =
\lim_{z\to z_0} \frac{1}{2(e^z-1)(e^z-z)} =
\frac{1}{2(-W(1) -1)(-W(1)+W(1)-i\pi)} =
\frac{1}{-2\pi i(-W(1) -1)} =
\frac{1}{2\pi i(W(1)+1)} 
$$
using L'Hopital's rule to compute the limit.
And finally, with residue theorem
$$
\oint_C f(z)\, dz = \int_{-\infty}^\infty f(z)\,dz = 2 \pi i b_0 = \frac{2 \pi i}{2\pi i(W(1)+1)} =
\frac{1}{W(1)+1} 
$$

An evaluation of this integral with real methods would also be intriguing.
