Integral of $\int^{\infty}_0 \frac{e^{-x}}{x^s+1}\,dx$

Question

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: $$\lim_{s \to \infty}R(s)=1$$

Answer 1

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:

$$ \varGamma(1-z) \varGamma(z) = \frac{\pi}{\sin(\pi z)} \qquad \forall \, z\not\in\mathbb Z $$

Answer 2

I'll just keep here some values of $R$ that my computer generated. If there's a special value you want, tell me and I can try it when I'm free. I'm just using Sympy, the code is simplify(integrate(exp(-t)/t**s+1),(t,0,oo))) (simplify does nothing in everything I tried, but one can hope lol) I know almost nothing about $G$ functions, so these could be saying almost nothing. Regardless:

\begin{align} R(3) &= \displaystyle \frac{\sqrt{3} {G_{1, 4}^{4, 1}\left(\begin{matrix} \frac{2}{3} & \\\frac{2}{3}, 0, \frac{1}{3}, \frac{2}{3} & \end{matrix} \middle| {\frac{1}{27}} \right)}}{6 \pi}, \\ R(4) &= \displaystyle \frac{\sqrt{2} {G_{1, 5}^{5, 1}\left(\begin{matrix} \frac{3}{4} & \\\frac{3}{4}, 0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4} & \end{matrix} \middle| {\frac{1}{256}} \right)}}{8 \pi^{\frac{3}{2}}}, \\ R(5) &= \displaystyle \frac{\sqrt{5} {G_{1, 6}^{6, 1}\left(\begin{matrix} \frac{4}{5} & \\\frac{4}{5}, 0, \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5} & \end{matrix} \middle| {\frac{1}{3125}} \right)}}{20 \pi^{2}},\\ R(6) &= \displaystyle \frac{\sqrt{3} {G_{1, 7}^{7, 1}\left(\begin{matrix} \frac{5}{6} & \\\frac{5}{6}, 0, \frac{1}{6}, \frac{1}{3}, \frac{1}{2}, \frac{2}{3}, \frac{5}{6} & \end{matrix} \middle| {\frac{1}{46656}} \right)}}{24 \pi^{\frac{5}{2}}}, \\ R(7) &= \displaystyle \frac{\sqrt{7} {G_{1, 8}^{8, 1}\left(\begin{matrix} \frac{6}{7} & \\\frac{6}{7}, 0, \frac{1}{7}, \frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7} & \end{matrix} \middle| {\frac{1}{823543}} \right)}}{56 \pi^{3}}, \\ R(8) &= \displaystyle \frac{{G_{1, 9}^{9, 1}\left(\begin{matrix} \frac{7}{8} & \\\frac{7}{8}, 0, \frac{1}{8}, \frac{1}{4}, \frac{3}{8}, \frac{1}{2}, \frac{5}{8}, \frac{3}{4}, \frac{7}{8} & \end{matrix} \middle| {\frac{1}{16777216}} \right)}}{32 \pi^{\frac{7}{2}}} \end{align} This seems to follow the pattern $$ n\in\mathbb Z_{\ge 3} \implies R(n) = \displaystyle \frac{ {G_{1, n+1}^{n+1, 1}\left(\begin{matrix} \frac{n-1}{n} & \\\frac{n-1}{n}, 0, \frac{1}{n}, \frac{2}{n} , \dots , \frac{n-1}n \end{matrix} \middle| {n^{-n}} \right)}}{\sqrt n (2\pi)^{(n-1)/2}}$$ From the comment of metamurphy, $R(s) + R(-s) = 1$ so negative values don't need to be tried, but I'll record here anyway what the computer gives me for a few values:

\begin{align} R(-1) &= \displaystyle e \operatorname{E}_{2}\left(1\right), \\ R(-2) &= \displaystyle \left(- \frac{\pi}{2} + \operatorname{Si}{\left(1 \right)}\right) \cos{\left(1 \right)} - \sin{\left(1 \right)} \operatorname{Ci}{\left(1 \right)} + 1, \\ R(-3) &= \displaystyle \frac{\sqrt{3} {G_{1, 4}^{4, 1}\left(\begin{matrix} - \frac{1}{3} & \\- \frac{1}{3}, 0, \frac{1}{3}, \frac{2}{3} & \end{matrix} \middle| {\frac{1}{27}} \right)}}{6 \pi} \end{align}

Non-integer values: \begin{align}R(1/2) &= \displaystyle \frac{{G_{2, 3}^{3, 2}\left(\begin{matrix} 0, - \frac{1}{2} & \\0, - \frac{1}{2}, 0 & \end{matrix} \middle| {1} \right)}}{\pi}, \\ R(-1/2) &= \displaystyle \frac{{G_{2, 3}^{3, 2}\left(\begin{matrix} - \frac{1}{2}, -1 & \\- \frac{1}{2}, -1, 0 & \end{matrix} \middle| {1} \right)}}{\pi}, \\ R(3/2) &= \displaystyle \frac{\sqrt{3} {G_{2, 5}^{5, 2}\left(\begin{matrix} \frac{2}{3}, \frac{1}{6} & \\\frac{2}{3}, \frac{1}{6}, 0, \frac{1}{3}, \frac{2}{3} & \end{matrix} \middle| {\frac{1}{27}} \right)}}{6 \pi^{2}}\end{align} In particular, it cannot verify your solution for $s=1/2$, but it tells me that your solution is accurate to 124 decimals. Computer doesn't want to give me an answer for $s=1/3,2/3,4/3,e,\pi$.

PS I have sympy installed but you can use it online here https://live.sympy.org

Answer 3

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.

$$\int_0^\infty \frac{e^{-x}}{x^s+1}dx=\sum_{x=0}^\infty \frac{e^{-x}}{x^s+1} -\frac12 \frac{e^{-0}}{0^s+1}+i\int_0^\infty \frac{1}{e^{2\pi x}-1} \frac{e^{-ix}}{(ix)^s+1}-\frac{1}{e^{\pi x}-1}\frac{e^{ix}}{(-ix)^s+1} dx= \sum_{x=0}^\infty \frac{e^{-x}}{x^s+1} -\frac12 +\int_0^\infty \frac{1}{e^{\pi x}-1} \frac{i\cos(x)-\sin(x)}{i^sx^s+1}-\frac{1}{e^{\pi x}-1} \frac{i\cos(x)+\sin(x)}{(-1)^s i^s x^s+1}dx$$