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.
We wish to compute $$ \int_{-\infty}^\infty\frac{\mathrm{d}x}{(e^x-x)^2+\pi^2}\tag{1} $$ We will do so by computing the contour integral $$ \int_{\gamma_R}\frac{\mathrm{d}z}{(e^z-z)^2+\pi^2}\tag{2} $$ over a family of contours $\gamma_R$ for suitable values of $R$.
The Singularities
The singularities of the integrand in $(2)$ occur when $(e^z-z)^2+\pi^2=(e^z-z+\pi i)(e^z-z-\pi i)$ vanishes; that is, for $z_k^\pm=x_k^\pm+iy_k^\pm$ where
$$
e^{z_k^\pm}-z_k^\pm\pm\pi i=0\tag{3}
$$
that is
$$
e^{2x_k^\pm}={x_k^\pm}^2+(y_k^\pm\mp\pi)^2\tag{4}
$$
and
$$
y_k^\pm=\mathrm{atan2}(y_k^\pm\mp\pi,x_k^\pm)\tag{5}
$$
In fact, the roots of $(3)$ can be expressed in terms of the multivalued Lambert W function as
$$
z_k^\pm=-W_{-k}(1)\pm\pi i\tag{6}
$$
The negative indices ensure that $z_0^+$ and $z_k^\pm$ for $k>0$ are in the upper half-plane. In Mathematica, these can be computed as -LambertW[-k,1]+Pi I and -LambertW[-k,1]-Pi I.
Note that as $|z_k^\pm|\to\infty$, $(3)$ precludes $x_k^\pm$ from being negative. In fact, as specified in $(6)$, only $x_0^\pm<0$. Equation $(4)$ says that $$ |y_k^\pm\mp\pi|=\sqrt{e^{2x_k^\pm}-{x_k^\pm}^2}\tag{7} $$ As $|z_k^\pm|\to\infty$, $(5)$ and $(7)$ yield $$ |y_k^\pm|\to\frac\pi2\pmod{2\pi}\tag{8} $$
The Contours
We will use the contours, $\gamma_R=\overline{\gamma}_R\cup\overset{\frown}{\gamma}_R$, which circle the upper half plane, $\overline{\gamma}_R$ passing from $-R$ to $R$ on the real axis, then $\overset{\frown}{\gamma}_R$ circling back counter-clockwise along $|z|=R$ in the upper half-plane. To get the desired decay of the integral along $\overset{\frown}{\gamma}_R$, we will also require that $R\equiv\frac{3\pi}{2}\!\!\!\!\pmod{2\pi}$; this is so that $\overset{\frown}{\gamma}_R$ passes well between the singularities.
Let us also define the curves $\rho^\pm$ to be where $|e^z|=|z\mp\pi i|$. Note that all the singularities of the integrand in $(2)$ lie on $\rho^\pm$.
On $\rho^\pm$, we have $R-\pi\le e^x\le R+\pi$, therefore, $$ \log(R-\pi)\le x\le\log(R+\pi)\tag{9} $$ Furthermore, because $R-|y|=\frac{x^2}{R+|y|}<\frac{\log(R+\pi)^2}{R}$, we have $$ R-\frac{\log(R+\pi)^2}{R}\le|y|\le R\tag{10}\\ $$ Therefore, at $\overset{\frown}{\gamma}_R\cap\rho^\pm$ as $R\to\infty$, $$ \text{on }\overset{\frown}{\gamma}_R\text{, }x^2+y^2=R^2\Rightarrow\left|\frac{\mathrm{d}y}{\mathrm{d}x}\right|=\left|\frac xy\right|\sim\frac{\log(R)}{R}\to0 $$ and $$ \text{on }\rho^\pm\text{, }e^{2x}=x^2+(y\mp\pi)^2\Rightarrow\left|\frac{\mathrm{d}y}{\mathrm{d}x}\right|=\left|\frac{e^{2x}-x}{y\mp\pi}\right|\sim R\to\infty $$ Thus, at $\overset{\frown}{\gamma}_R\cap\rho^\pm$ as $R\to\infty$, $\overset{\frown}{\gamma}_R$ becomes horizontal and $\rho^\pm$ becomes vertical. For example, here is the situation when $R=129.5\pi$:
$\hspace{35mm}$
$$ \hspace{-1cm}\small \begin{array}{} z_{64}^+=5.99292081954932666 + 403.67969492855003099 i\\ z_{65}^-=6.00848352082933166 + 403.67988772309602824 i\\ z_{65}^+=6.00848352082933166 + 409.96307303027561472 i\\ z_{66}^-=6.02380774554030566 + 409.96326053797959857 i \end{array} $$ As $R\to\infty$, on $\rho^\pm$ in the upper half-plane, $(9)$ and $(10)$ imply that $\arg(z\mp\pi i)\to\frac\pi2$; thus, as indicated by $(8)$, $e^z$ and $z\mp\pi i$ cancel only when $\mathrm{Im}(z)\approx\frac\pi2\!\!\!\!\pmod{2\pi}$. Likewise, $e^z$ and $z\mp\pi i$ reinforce when $\mathrm{Im}(z)\approx\frac{3\pi}{2}\!\!\!\!\pmod{2\pi}$.
Therefore, when $R\equiv\frac{3\pi}{2}\!\!\!\!\pmod{2\pi}$ and $z\in\overset{\frown}{\gamma}_R\cap\rho^\pm$, we have that $|e^z-z\pm\pi i|\sim2R$. As $z\in\overset{\frown}{\gamma}_R$ moves to the right, by even just $1$, $e^z$ more than doubles, and dominates $z\mp\pi i$; thus, $|e^z-z\pm\pi i|\ge2R-R$. As $z\in\overset{\frown}{\gamma}_R$ moves to the left, by even just $1$, $e^z$ decreases by more than half, and $z\mp\pi i$ dominates; thus, $|e^z-z\pm\pi i|\ge R-R/2$. Therefore, when $R\equiv\frac{3\pi}{2}\!\!\!\!\pmod{2\pi}$, $|e^z-z\pm\pi i|\ge R/2$, and $|(e^z-z)^2+\pi^2|\ge R^2/4$. This guarantees that, as $R\to\infty$, the integral over $\overset{\frown}{\gamma}_R$ vanishes.
Thus, the integral along the real axis is $2\pi i$ times the sum of the residues in the upper half-plane.
The Residues
Let $z_k^\pm=-W_{-k}(1)\pm\pi i$. We will use the fact that $e^{z_k^\pm}-z_k^\pm=\mp\pi i$.
The residue of $\displaystyle\frac1{(e^z-z)^2+\pi^2}$ at $z_k^\pm=-W_{-k}(1)\pm\pi i$ is $$ \begin{align} \lim_{z\to z_k^\pm}\frac{z-z_k^\pm}{(e^z-z)^2+\pi^2} &=\frac1{2(\mp\pi i)(e^{z_k^\pm}-1)}\\ &=\frac1{2\pi i}\frac1{\mp(z_k^\pm-1\mp\pi i)}\\ &=\frac1{2\pi i}\frac1{\pm(W_{-k}(1)+1)}\\ \end{align} $$ Thus, the residues at the pairs of singularities cancel each other, leaving us with the residue at $z_0^+$ which is $\dfrac1{2\pi i}\dfrac1{W_0(1)+1}$. Thus, the integral is $$ \int_{-\infty}^\infty\frac{\mathrm{d}x}{(e^x-x)^2+\pi^2}=\frac1{W_0(1)+1} $$
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.
I also considered this integral in another site, but it is only imperfect and non-rigorous one.
It seems that the Formelsammlung Mathematik is rendering a complete solution. It is written in German, but your bona fide translater Google may read this for you.
The identity is due to Victor Adamchik, see
http://mathworld.wolfram.com/OmegaConstant.html
You may want to contact Dr Adamchik himself via the e-mail at
http://www.cs.cmu.edu/~adamchik/research.html
because this particular paper doesn't seem to be in the list, as far as I can see.