How to show $ \int_{-\infty}^{\infty} \frac{e^{-(x+1)^2}}{1+e^{-x}}\mathrm{d}x = \frac{\left(2\sqrt[4]{e} -1 \right)\sqrt{\pi}}{2e}$? I was recently looking at this post where the following formula is shown:
$$
\int_{-\infty}^{\infty} \frac{E(x)}{1+\mathcal{E}(x)^{O(x)}}\mathrm{d}x= \int_0^{\infty} E(x) \mathrm{d}x
$$
where $E(x), \mathcal{E}(x)$ are even functions and $O(x)$ is an odd function. One nice application of this formula would be the integral
$$
\int_{-\infty}^{\infty} \frac{e^{-x^2}}{1+e^{-x}}\mathrm{d}x = \frac{\sqrt{\pi}}{2}
$$
where the problem reduces to the evaluation of the Gaussian integral. I then wondered what would happen if I made slight alterations to the above integral, like changing $x^2\to (x+1)^2$. WA evaluates said integral as:

$$
 \int_{-\infty}^{\infty} \frac{e^{-(x+1)^2}}{1+e^{-x}}\mathrm{d}x = \frac{\left(2\sqrt[4]{e} -1 \right)\sqrt{\pi}}{2e}
$$


The even/odd formula can't be applied since the $+1$ makes the function not even anymore. Recalling that $\int^\infty_{-\infty} e^{-(ax^2 + bx+c)}\mathrm{d}x=\sqrt{\frac{\pi}{a}}e^{\frac{b^2}{4a}-c}
$ I attempted to evaluate the integral using geometric series
$$
\int_{-\infty}^{\infty} \frac{e^{-(x+1)^2}}{1+e^{-x}}\mathrm{d}x =  \sum_{n\ge 0}(-1)^n \int_{-\infty}^{\infty}e^{-(x^2+(n+2)x+1)}\, \mathrm{d}x = \frac{\sqrt{\pi}}{e}  \sum_{n\ge 0}(-1)^n e^{\frac{(n+2)^2}{4}}
$$
but the resulting series is divergent, so this method won't work. Does anyone have any ideas on how to evaluate this integral? Thank you!
 A: $$I=\int_{-\infty}^\infty \frac{e^{-(1+x)^2}}{1+e^{-x}}dx=\frac1e\int_{-\infty}^\infty\frac{\color{blue}{e^{-x^2-x}}}{1+e^{x}}dx\overset{x\to -x}=\frac1e\int_{-\infty}^\infty\frac{\color{red}{e^{-x^2+2x}}}{1+e^{x}}dx$$
$$\Rightarrow 2I=\frac1e\int_{-\infty}^\infty \frac{(\color{blue}{1}+\color{red}{e^{3x}})e^{-x^2-x}}{1+e^x}dx$$
$$\frac{1+x^3}{1+x}=1-x+x^2\Rightarrow I=\frac1{2e}\int_{-\infty}^\infty (1-e^x+e^{2x})e^{-x^2-x}dx$$

$$I=\frac{1}{2e}\int_{-\infty}^\infty \left(e^{-x^2-x}+\color{blue}{e^{-x^2+x}}\right)dx-\frac{1}{2e}\int_{-\infty}^\infty e^{-x^2}dx$$
$$\overset{\color{blue}{x\to -x}}=\frac{1}{e}\int_{-\infty}^\infty e^{-x^2-x}dx-\frac{\sqrt \pi}{2e}=\boxed{\frac{\sqrt[4]e\sqrt \pi}{e}-\frac{\sqrt\pi}{2e}}$$
A: Combining J.G.'s and Zacky's answers, I've found a generalization for a family of special cases which includes the integral in question. Let $I = \int_{-\infty}^{\infty}  \frac{e^{-(ax+b)^2}}{{1+e^{-x}}}\, \mathrm{d}x$ then
\begin{align}
2I& = \int_{-\infty}^{\infty} \frac{e^{-(ax+b)^2}}{{1+e^{-x}}}\, \mathrm{d}x + \int_{-\infty}^{\infty} \frac{e^{-(ax+b)^2}}{{1+e^{-x}}}\, \mathrm{d}x \\
& \overset{\color{darkblue}{x \to -x}}{=}\int_{-\infty}^{\infty} \frac{e^{-(ax+b)^2}}{{1+e^{-x}}}\, \mathrm{d}x + \int_{-\infty}^{\infty} \frac{\color{darkblue}{e^{-x}}e^{-(\color{darkblue}{-}ax+b)^2}}{{1+e^{-x}}}\, \mathrm{d}x\\
& = e^{-b^2} \int_{-\infty}^{\infty} e^{-a^2x^2-(2ab-1)x} \frac{1+e^{(4ab-1)x} }{1+e^{x}}\, \mathrm{d}x
\end{align}
So if $2ab \in \mathbb{N} \setminus \{0\}$, recalling that
$$
\int_{-\infty}^{\infty} e^{-\alpha x^2 -\beta x}\mathrm{d}x \overset{\color{darkblue}{x\to  \frac{x}{\sqrt{\alpha}} -\frac{\beta}{2\alpha}}}{=}\frac{e^{\frac{\beta^2}{4\alpha}}}{\sqrt{\alpha}}\int_{-\infty}^{\infty} e^{-x^2}\mathrm{d}x =\sqrt{\frac{\pi}{\alpha}}e^{\frac{\beta^2}{4\alpha}}
$$
we get
\begin{align}
2I& = e^{-b^2} \int_{-\infty}^{\infty} e^{-a^2x^2-(2ab-1)x}\sum_{k=0}^{4ab-2}(-1)^k e^{kx}\, \mathrm{d}x\\
& = e^{-b^2} \sum_{k=0}^{4ab-2}(-1)^k\int_{-\infty}^{\infty} e^{-a^2x^2-(2ab-k-1)x}\, \mathrm{d}x\\
 & = \frac{\sqrt{\pi}e^{-b^2}}{|a|} \sum_{k=0}^{4ab-2}(-1)^ke^{\frac{(2ab-k-1)^2}{4a^2}}
\end{align}
Thus

$$
\int_{-\infty}^{\infty}  \frac{e^{-(ax+b)^2}}{{1+e^{-x}}}\, \mathrm{d}x =\frac{\sqrt{\pi}}{2|a|e^{b^2}} \sum_{k=0}^{4ab-2}(-1)^ke^{\frac{(2ab-1-k)^2}{4a^2}}, \quad \ 2ab \in \mathbb{N} \setminus \{0\}
$$

And after substitution $x\to -x$, this gives the complementary formula

$$
\int_{-\infty}^{\infty}  \frac{e^{-(ax-b)^2}}{{1+e^{-x}}}\, \mathrm{d}x =\frac{\sqrt{\pi}}{2|a|e^{b^2}} \sum_{k=0}^{4ab}(-1)^ke^{\frac{(2ab-k)^2}{4a^2}}, \quad \ 2ab \in \mathbb{N} \cup \{0\}
$$


Some other interesting integrals that can be evaluated using the previous formula are

*

*$$\int_{-\infty}^{\infty}  \frac{e^{-(\frac{x}{a}+a)^2}}{{1+e^{-x}}}\, \mathrm{d}x = \frac{\sqrt{\pi} |a|  }{2 e^{a^2}} \left(2 e^{\frac{a^2}{4}} - 1\right), \qquad a \neq 0$$

*$$\int_{-\infty}^{\infty}  \frac{e^{-(x-1)^2}}{{1+e^{-x}}}\, \mathrm{d}x =  \frac{\sqrt{\pi}}{2e}\left(1-2\sqrt[4]{e} +2e \right)
$$

*$$\int_{-\infty}^{\infty}  \frac{e^{-(\frac{x}{2}+3)^2}}{{1+e^{-x}}}\, \mathrm{d}x = \sqrt{\pi}\left(e^{-9} -2 e^{-8} +2 e^{-5} \right)$$

*$$\int_{-\infty}^{\infty}  \frac{e^{-2(x+1)^2}}{{1+e^{-x}}}\, \mathrm{d}x =\sqrt{\frac{\pi}{8}} \left(2e^{-\frac{7}{8}} - 2e^{-\frac{3}{2}} + 2 e^{-\frac{15}{8}} -e^{-2}\right)$$
A: Note that
$$\frac{e^{-(x+1)^2}}{1+e^{-x}}=e^{-x^2-1}\left(e^{-x}-\frac1{1+e^x}\right)
$$
Then, with $\int_{-\infty}^{\infty} \frac{e^{-x^2}}{1+e^{x}}dx=\frac{\sqrt{\pi}}2$
\begin{align}
\int_{-\infty}^{\infty} \frac{e^{-(x+1)^2}}{1+e^{-x}}{d}x 
&= \frac1e\int_{-\infty}^{\infty} e^{-x^2-x}dx
-\frac1e \int_{-\infty}^{\infty} \frac{e^{-x^2}}{1+e^{x}}dx 
= \frac{\sqrt{\pi}}e{e^{\frac14}}- \frac{\sqrt{\pi}}{2e}
\end{align}
A: $\begin{align}
\int_{-\infty}^{\infty}\frac{e^{-(x+1)^2}}{1+e^{-x}}dx
&=\int_{0}^{\infty}\frac{e^{-(x+1)^2}+e^{-x}e^{-(-x+1)^2}}{1+e^{-x}}dx\\
&=\int_{0}^{\infty}\frac{e^{-x^2-1}(e^{-2x}+e^{x})}{1+e^{-x}}dx\\
&=\int_{0}^{\infty}\frac{e^{-x^2-1}(e^{\frac{3}{2}x}+e^{-\frac{3}{2}x})}{e^{\frac{1}{2}x}+e^{-\frac{1}{2}x}}dx\\
&=\int_{0}^{\infty}e^{-x^2-1}(e^x-1+e^{-x})dx\\
&=\int_{0}^{\infty}e^{-x^2-x-1}dx+\int_{0}^{\infty}e^{-x^2+x-1}dx-\int_{0}^{\infty}e^{-x^2-1}dx\\
&=\int_{0}^{\infty}e^{-x^2-x-1}dx+\int_{-\infty}^{0}e^{-x^2-x-1}dx-\frac{\sqrt{\pi}}{2e}\\
&=\int_{-\infty}^{\infty}e^{-x^2-x-1}dx-\frac{\sqrt{\pi}}{2e}\\
&=\sqrt{\pi}e^{\frac{1}{4}-1}-\frac{\sqrt{\pi}}{2e}\\
&=\frac{\sqrt{\pi}}{2e}(2\sqrt[4]{e}-1)\\
\end{align}$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Hereafter, $\ds{\on{H}}$ is the $\ds{Heaviside\ Step\ Function}$.
\begin{align}
& \bbox[5px,#ffd]{\int_{-\infty}^{\infty}{\expo{-\pars{x + 1}^{\,\,\,2}} \over 1 + \expo{-x}}\,\dd x} =
\int_{-\infty}^{\infty}\expo{-\pars{x + 1}^{\,\,\,2}}
\bracks{\on{H}\pars{x} - {\on{sgn}\pars{x} \over
\expo{\verts{x}} + 1}}\dd x
\\[5mm] = & \
\int_{0}^{\infty}\bracks{\expo{-\pars{x + 1}^{\,\,\,2}} -
{\expo{-\pars{x + 1}^{\,\,\,2}} -
\expo{-\pars{-x + 1}^{\,\,\,2}}\over \expo{x} + 1}}\dd x
\\[5mm] = & \
\int_{0}^{\infty}\expo{-x^{2}\ -\ 1}
\pars{-1 + \expo{x} + \expo{-x}}\dd x =
{1 \over 2\expo{}}\int_{-\infty}^{\infty}\expo{-x^{2}}
\pars{-1 + \expo{x} + \expo{-x}}\dd x
\\[5mm] = & \
{1 \over 2\expo{}}\bracks{-\int_{-\infty}^{\infty}\expo{-x^{2}}
\dd x + \int_{-\infty}^{\infty}\expo{-\pars{x - 1/2}^{\,\,\,2}\,\, + 1/4}
\,\,\dd x \int_{-\infty}^{\infty}\expo{-\pars{x + 1/2}^{\,\,\,2}\,\, + 1/4}\,\,
\dd x}
\\[5mm] = & \
{1 \over 2\expo{}}\pars{-1 + \expo{1/4} + \expo{1/4}}
\int_{-\infty}^{\infty}\expo{-x^{2}}\,\,\dd x =
\bbx{{2\expo{1/4} - 1 \over 2\expo{}}\root{\pi}} \approx 0.5112
\end{align}
