Krdytkyu

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.

$$displaystylefrac{1}{int_{-infty}^{infty}frac{1}{(e^{x}-x)^{2}+{pi}^{2}}dx}-1=W(1)=Omega$$

$W(1)=Omega$ is often referred to as the Omega Constant. Which is the solution to

$xe^{x}=1$. Which is $xapprox .567$

Thanks much.

EDIT: Sorry, I had the integral written incorrectly. Thanks for the catch.

I had also seen this:

$displaystyleint_{-infty}^{infty}frac{dx}{(e^{x}-x)^{2}+{pi}^{2}}=frac{1}{1+W(1)}=frac{1}{1+Omega}approx .638$

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
$$
int_{-infty}^inftyfrac{mathrm{d}x}{(e^x-x)^2+pi^2}tag{1}
$$
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
$$
e^{z_k^pm}-z_k^pmpmpi i=0tag{3}
$$
that is
$$
e^{2x_k^pm}={x_k^pm}^2+(y_k^pmmppi)^2tag{4}
$$
and
$$
y_k^pm=mathrm{atan2}(y_k^pmmppi,x_k^pm)tag{5}
$$
In fact, the roots of $(3)$ can be expressed in terms of the multivalued Lambert W function as
$$
z_k^pm=-W_{-k}(1)pmpi itag{6}
$$
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 $|z_k^pm|toinfty$, $(3)$ precludes $x_k^pm$ from being negative. In fact, as specified in $(6)$, only $x_0^pm<0$. Equation $(4)$ says that
$$
|y_k^pmmppi|=sqrt{e^{2x_k^pm}-{x_k^pm}^2}tag{7}
$$
As $|z_k^pm|toinfty$, $(5)$ and $(7)$ yield
$$
|y_k^pm|tofracpi2pmod{2pi}tag{8}
$$

The Contours

We will use the contours, $gamma_R=overline{gamma}_Rcupoverset{frown}{gamma}_R$, 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 $Requivfrac{3pi}{2}!!!!pmod{2pi}$; this is so that $overset{frown}{gamma}_R$ passes well between the singularities.

Let us also define the curves $rho^pm$ to be where $|e^z|=|zmppi i|$. Note that all the singularities of the integrand in $(2)$ lie on $rho^pm$.

On $rho^pm$, we have $R-pile e^xle R+pi$, therefore,
$$
log(R-pi)le xlelog(R+pi)tag{9}
$$
Furthermore, because $R-|y|=frac{x^2}{R+|y|}<frac{log(R+pi)^2}{R}$, we have
$$
R-frac{log(R+pi)^2}{R}le|y|le Rtag{10}\
$$
Therefore, at $overset{frown}{gamma}_Rcaprho^pm$ as $Rtoinfty$,
$$
text{on }overset{frown}{gamma}_Rtext{, }x^2+y^2=R^2Rightarrowleft|frac{mathrm{d}y}{mathrm{d}x}right|=left|frac xyright|simfrac{log(R)}{R}to0
$$
and
$$
text{on }rho^pmtext{, }e^{2x}=x^2+(ymppi)^2Rightarrowleft|frac{mathrm{d}y}{mathrm{d}x}right|=left|frac{e^{2x}-x}{ymppi}right|sim Rtoinfty
$$
Thus, at $overset{frown}{gamma}_Rcaprho^pm$ as $Rtoinfty$, $overset{frown}{gamma}_R$ becomes horizontal and $rho^pm$ becomes vertical. For example, here is the situation when $R=129.5pi$:

$hspace{35mm}$

$$
hspace{-1cm}small
begin{array}{}
z_{64}^+=5.99292081954932666 + 403.67969492855003099 i\
z_{65}^-=6.00848352082933166 + 403.67988772309602824 i\
z_{65}^+=6.00848352082933166 + 409.96307303027561472 i\
z_{66}^-=6.02380774554030566 + 409.96326053797959857 i
end{array}
$$
As $Rtoinfty$, on $rho^pm$ in the upper half-plane, $(9)$ and $(10)$ imply that $arg(zmppi i)tofracpi2$; thus, as indicated by $(8)$, $e^z$ and $zmppi i$ cancel only when $mathrm{Im}(z)approxfracpi2!!!!pmod{2pi}$. Likewise, $e^z$ and $zmppi i$ reinforce when $mathrm{Im}(z)approxfrac{3pi}{2}!!!!pmod{2pi}$.

Therefore, when $Requivfrac{3pi}{2}!!!!pmod{2pi}$ and $zinoverset{frown}{gamma}_Rcaprho^pm$, we have that $|e^z-zpmpi i|sim2R$. As $zinoverset{frown}{gamma}_R$ moves to the right, by even just $1$, $e^z$ more than doubles, and dominates $zmppi i$; thus, $|e^z-zpmpi i|ge2R-R$. As $zinoverset{frown}{gamma}_R$ moves to the left, by even just $1$, $e^z$ decreases by more than half, and $zmppi i$ dominates; thus, $|e^z-zpmpi i|ge R-R/2$. Therefore, when $Requivfrac{3pi}{2}!!!!pmod{2pi}$, $|e^z-zpmpi i|ge R/2$, and $|(e^z-z)^2+pi^2|ge R^2/4$. This guarantees that, as $Rtoinfty$, the integral over $overset{frown}{gamma}_R$ vanishes.

Thus, the integral along the real axis is $2pi i$ times the sum of the residues in the upper half-plane.

The Residues

Let $z_k^pm=-W_{-k}(1)pmpi i$. We will use the fact that $e^{z_k^pm}-z_k^pm=mppi i$.

The residue of $displaystylefrac1{(e^z-z)^2+pi^2}$ at $z_k^pm=-W_{-k}(1)pmpi i$ is
$$
begin{align}
lim_{zto z_k^pm}frac{z-z_k^pm}{(e^z-z)^2+pi^2}
&=frac1{2(mppi i)(e^{z_k^pm}-1)}\
&=frac1{2pi i}frac1{mp(z_k^pm-1mppi i)}\
&=frac1{2pi i}frac1{pm(W_{-k}(1)+1)}\
end{align}
$$
Thus, the residues at the pairs of singularities cancel each other, leaving us with the residue at $z_0^+$ which is $dfrac1{2pi i}dfrac1{W_0(1)+1}$. Thus, the integral is
$$
int_{-infty}^inftyfrac{mathrm{d}x}{(e^x-x)^2+pi^2}=frac1{W_0(1)+1}
$$

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:

$$(e^z-z)^2 + pi^2 = 0 implies e^z-z = pm i pi implies z = -W (1)pm ipi$$

using this.

We now use the root with the positive $ipi$ because when the sign is negative, the pole does not fall within the contour because $Im (-W (1)- ipi)<0$.

$$z_0 := -W (1)+ipi$$

We calculate the beautiful residue at $b_0$ at $z=z_0$:

$$
b_0=
operatorname*{Res}_{z to z_0} f(z) =
lim_{zto z_0} frac{(z-z_0)}{(e^z-z)^2+pi^2} =
lim_{zto z_0} frac{1}{2(e^z-1)(e^z-z)} =
frac{1}{2(-W(1) -1)(-W(1)+W(1)-ipi)} =
frac{1}{-2pi i(-W(1) -1)} =
frac{1}{2pi i(W(1)+1)}
$$

using L'Hopital's rule to compute the limit.

And finally, with residue theorem

$$
oint_C f(z), dz = int_{-infty}^infty f(z),dz = 2 pi i b_0 = frac{2 pi i}{2pi i(W(1)+1)} =
frac{1}{W(1)+1}
$$

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.