$\int_{-\infty}^{\infty}{e^x+1\over (e^x-x+1)^2+\pi^2}\mathrm dx=\int_{-\infty}^{\infty}{e^x+1\over (e^x+x+1)^2+\pi^2}\mathrm dx=1$

Solution 1:

We have that $\frac{e^x-1}{(e^x-x+1)^2+\pi^2}$ has a simple primitive, given by $\frac{1}{\pi}\arctan\frac{1+e^x-x}{\pi}$. It follows that:

$$ I_1 = \int_{-\infty}^{+\infty}\frac{e^x+1}{(e^x-x+1)^2+\pi^2}\,dx= 2\int_{-\infty}^{+\infty}\frac{dx}{(e^x-x+1)^2+\pi^2}$$ and the residue theorem gives the following

Lemma. If $a>0$ and $b\in\mathbb{R}$, $$ \int_{-\infty}^{+\infty}\frac{a^2\,dx}{(e^x-ax-b)^2+(a\pi)^2}=\frac{1}{1+W\left(\frac{1}{a}e^{-b/a}\right)} $$ where $W$ is Lambert's function.

In our case, by choosing $a=1$ and $b=-1$ we get that $I_1$ depends on $W(e)=1$ and equals $\color{red}{\large 1}$.

$I_2$ is easier to compute: $$ I_2 = \int_{-\infty}^{+\infty}\frac{e^x+1}{(e^x+x+1)^2+\pi^2}\,dx = \frac{1}{\pi}\,\left.\arctan\left(\frac{1+x+e^x}{\pi}\right)\right|_{-\infty}^{+\infty} = \color{red}{1}.$$

Addendum. Due to the identity $$\begin{eqnarray*} I_1 &=& \int_{0}^{+\infty}\frac{u+1}{u}\cdot\frac{du}{(u+1-\log u)^2+\pi^2}\\&=&\frac{1}{\pi}\int_{0}^{+\infty}\int_{0}^{+\infty}(u+1)u^{x-1}e^{-(u+1)x}\sin(\pi x)\,dx\,du\\&=&\frac{2}{\pi}\int_{0}^{+\infty} e^{-x} x^{-x}\Gamma(x)\sin(\pi x)\,dx\\&=&2\int_{0}^{+\infty}\frac{e^{-x} x^{-x}}{\Gamma(1-x)}\,dx\end{eqnarray*}$$ the previous Lemma also proves the highly non-trivial identity $$ \int_{0}^{+\infty}\frac{(ex)^{-x}}{\Gamma(1-x)}\,dx = \frac{1}{2} \tag{HNT}$$ equivalent to the claim $I_1=1$. It would be interesting to find an independent proof of $\text{(HNT)}$, maybe based on Glasser's master theorem, Ramanujan's master theorem or Lagrange inversion. There also is an interesting discrete analogue of $\text{(HNT)}$, $$ \sum_{n\geq 1}\frac{n^n}{n!(4e)^{n/2}}=1$$ that comes from the Lagrange inversion formula.

Solution 2:

Preliminary Notes

This is similar in nature to this answer, where it is shown that $$ \int_{-\infty}^\infty\frac{\mathrm{d}x}{(e^x-x)^2+\pi^2}=\frac1{W_0(1)+1}\tag0 $$ In order to evaluate $$ \int_{-\infty}^\infty\frac{e^x+1}{(e^x+x+1)^2+\pi^2}\,\mathrm{d}x\tag1 $$ we will use $\newcommand{\Res}{\operatorname*{Res}}\newcommand{\W}{\operatorname{W}}\newcommand{\mapsfrom}{\mathrel{\unicode{x21a4}}}$ $$ \Res_{z=\pm\pi i}\left(\frac{e^z+1}{\left(e^z+z+1\right)^2+\pi^2}\right) =\pm\frac1{\pi i}\tag2 $$ and $$ \Res_{\substack{e^z+z+1=\pm i\pi\\z\ne\pm\pi i}}\left(\frac{e^z+1}{\left(e^z+z+1\right)^2+\pi^2}\right) =\pm\frac1{2\pi i}\tag3 $$

Locating the Singularities

In this answer, we are only using the singularities in the upper half plane. Since the integrand is real on the real line, the singularities are reflected in the lower half plane with conjugate residues.

To solve $$ e^z+z+1=\pm i\pi\tag4 $$ we will write $(4)$ as $$ e^x\cos(y)+ie^x\sin(y)+x+iy+1=\pm i\pi\tag5 $$ The real part of $(5)$ can be solved as $$ x=\log\left(\W\left(\frac{\cos(y)}e\right)\sec(y)\right)\tag6 $$ and the imaginary part of $(5)$ can be solved as $$ x=\log\left(\frac{\pm\pi-y}{\sin(y)}\right)\tag7 $$ Combining $(6)$ and $(7)$, we get $$ \W_k\left(\frac{\cos(y)}e\right)\tan(y)+y=\pm\pi\tag8 $$ Since $y$ in $(8)$ is real, we will only be interested in $k\in\{-1,0\}$.

For $k=0$, the only solutions to $(8)$ are $y=\pm\pi$. Since these are solutions for any $k$, we need only consider $k=-1$.

Approximating the Singularities

Using $y=\frac{4k+3}2\pi-\delta$, and $\sin(y)\approx-1$, $\cos(y)\approx-\delta$, and $\tan(\delta)\approx\frac1\delta$, we can approximate solutions to $$ \W\left(\frac{\cos(y)}e\right)\tan(y)+y=\pm\pi\tag9 $$ by $$ \delta=\frac{1+\log\left(\frac{4k+3\mp2}2\pi\right)}{\frac{4k+3\mp2}2\pi}\tag{10} $$ Thus, we have $$ y_k^{\pm}\approx\frac{4k+3}2\pi-\frac{1+\log\left(\frac{4k+3\mp2}2\pi\right)}{\frac{4k+3\mp2}2\pi}\tag{11} $$ and we can use $(7)$ to get $$ x_k^{\pm}\approx\log\left(\frac{4k+3\mp2}2\pi\right)\tag{12} $$ Using $(11)$ and $(12)$ to approximate $z_k^{\pm}=x_k^{\pm}+iy_k^{\pm}$, we can refine the approximation with the Newton's method recursion $$ z\mapsfrom z-\frac{e^z+z+1\mp\pi i}{e^z+1}\tag{13} $$

Furthermore, for $k\ge1$, $(11)$ and $(12)$ lead to $$ \left|\,z_k^+-z_k^-\,\right|\le\frac1k\tag{14} $$

Contour Integration

For $k\ge1$, $z_k^{\pm}$ occur in pairs whose residues cancel and whose imaginary parts are approximately $\frac32\pi$ mod $2\pi$. Therefore, $$ \int_{-\infty}^\infty\frac{e^x+1}{\left(e^x+x+2\pi i+1\right)^2+\pi^2}\,\mathrm{d}x=\int_{-\infty}^\infty\frac{e^x+1}{\left(e^x+x+2k\pi i+1\right)^2+\pi^2}\,\mathrm{d}x\tag{14} $$ The right side tends to $0$ as $k\to\infty$, therefore, the integral on the left is $0$.

The only two residues that do not cancel are those for $$ z_0^+=\pi i\quad\text{with residue}\quad\frac1{\pi i}\tag{15a} $$ and $$ z_0^-=2.0888430156+4.3198966321i\quad\text{with residue}\quad-\frac1{2\pi i}\tag{15b} $$ Applying $(14)$ and $(15)$ and using the contour (as $R\to\infty$) $$ \overbrace{[-R,R]}^{\to(1)}\cup\overbrace{[R,R+2\pi i]}^{\to0}\cup\overbrace{[R+2\pi i,-R+2\pi i]}^{\to0\text{ by }(14)}\cup\overbrace{[-R+2\pi i,-R]}^{\to0} $$ we get $$ \begin{align} \int_{-\infty}^\infty\frac{e^x+1}{(e^x+x+1)^2+\pi^2}\,\mathrm{d}x &=\int_\gamma\frac{e^z+1}{(e^z+z+1)^2+\pi^2}\,\mathrm{d}z\\ &=2\pi i\left(\frac1{\pi i}-\frac1{2\pi i}\right)\\[6pt] &=1 \end{align} $$