Integral of $\int^{\infty}_0 \frac{e^{-x}}{x^s+1}\,dx$ Related information Integral of $\int^{\infty}_0 \frac{x^n}{x^s+1}dx$
This is an integral very similar to the gamma function integral:
$$R(s)=\int^{\infty}_0 (1+x^s)^{-1} e^{-x}\,dx$$
i want to find the function $R$.
I do know some values of $R$:
$$R(0)=1$$
$$R\left(\frac{1}{2}\right)=\frac{-\pi \text{erfi}(1)+\text{Ei}(1)+e \sqrt \pi}{e}$$
$$R(1)= -e\text{Ei}(-1)$$
$$R(2) = \text{Ci}(1)\sin(1)-\text{Si}(1)\cos(1)+\frac{1}{2}\pi\cos(1)$$
Can any of you provide hints or solutions? 
Also, thanks to an answer by Sewer we know that:
$$\lim_{s \to \infty}R(s)=1$$
 A: From  your related question we get 
$$
\begin{split}
R(s) &= \int^{\infty}_0 \frac{\operatorname e^{-x}}{x^s+1}\,\operatorname dx \\
&= \int^{\infty}_0 \sum_{n=0}^{+\infty}(-1)^n\frac{x^n}{n!}\frac{1}{x^s+1}\,\operatorname dx \\
&= \sum_{n=0}^{+\infty}  \frac{(-1)^n}{n!}\int^{\infty}_0\frac{x^n}{x^s+1} \operatorname d x \\
&= \sum_{n=0}^{+\infty} \frac{(-1)^n}{n!}\frac1{n+1}R\left(\frac{s}{n+1};0\right) \\
&= \sum_{n=0}^{+\infty} \frac{(-1)^n}{n!}\frac1{n+1}\frac{n+1}{s}\varGamma\left(\frac{n+1}{s}\right) \varGamma\left(1-\frac{n+1}{s}\right) \\
&=\frac{1}{s}  \sum_{n=0}^{+\infty} \frac{(-1)^n}{n!} \varGamma\left(\frac{n+1}{s}\right) \varGamma\left(1-\frac{n+1}{s}\right)\\
\end{split}
$$
Moreover, if $\frac{n+1}{s} \not \in \mathbb Z$, we can rewrite $R$ as
$$R(s) =\frac{\pi}{s}  \sum_{n=0}^{+\infty} \frac{(-1)^n}{n!} \frac{1}{\sin\left(  \frac{\pi(n+1)}{s}\right)}$$
When we used the property:
$$
\varGamma(1-z) \varGamma(z) = \frac{\pi}{\sin(\pi z)}  \qquad \forall \, z\not\in\mathbb Z
$$
A: Since the Fox-H function may be hard to understand, here is a solution using the Geometric Series. For simplicity, lets do the general integral. Notice the Incomplete Gamma function definition:
$$\int \frac{e^{-x}}{x^s+1}dx= \int \frac{e^{-x}}{1-\left(-{x^s}\right)}dx =\int e^{-x} \sum_{n=0}^\infty \left(-x^s\right)^n dx=\sum_{n=0}^\infty (-1)^n \int x^{ns}e^{-x} dx=C-\sum_{n=0}^\infty (-1)^nΓ(ns+1,x) ,|x^s|<1$$
Even though this may not converge for$ |x^s|\not<1$, this still demonstrates a generalized version of the integral. I thought it might be nice to notice being able to use the linked geometric series. Maybe I will add more possible expansions. Please correct me and give me feedback!
It also may be possible to use the Abel-Plana formula with a complicated real and imaginary part. Note that the sum should have a tricky closed form.
$$\int_0^\infty \frac{e^{-x}}{x^s+1}dx=\sum_{x=0}^\infty \frac{e^{-x}}{x^s+1} -\frac12 \frac{e^{-0}}{0^s+1}+i\int_0^\infty \frac{1}{e^{2\pi x}-1} \frac{e^{-ix}}{(ix)^s+1}-\frac{1}{e^{\pi x}-1}\frac{e^{ix}}{(-ix)^s+1} dx= \sum_{x=0}^\infty \frac{e^{-x}}{x^s+1} -\frac12  +\int_0^\infty \frac{1}{e^{\pi x}-1} \frac{i\cos(x)-\sin(x)}{i^sx^s+1}-\frac{1}{e^{\pi x}-1} \frac{i\cos(x)+\sin(x)}{(-1)^s i^s x^s+1}dx$$
