TITLE: Need help with $\int_0^\infty\frac{e^{-x}}{\sqrt[3]2+\cos x}dx$
QUESTION [16 upvotes]: Please help me to evaluate this integral:
$$\int_0^\infty\frac{e^{-x}}{\sqrt[3]2+\cos x}dx$$

REPLY [4 votes]: Evaluate the definite integral:
  $$I:=\int_{0}^{\infty}\frac{e^{-x}}{\sqrt[3]{2}+\cos{x}}\mathrm{d}x.$$


First, rewrite the denominator of the integrand in the form,
$$\sqrt[3]{2}+\cos{x}=a(r^2-2r\cos{x}+1).$$
To do this, we must solve the following system of equations in $r$ and $a$ to solve:
$$\begin{cases}
\sqrt[3]{2}=a(r^2+1),\\
1=-2ar.
\end{cases}$$
There are two solutions, $(a_\pm,r_\pm)$, where:
$$\begin{cases}
(a_+,r_+)=(\frac12\left(\sqrt[3]{2}+\sqrt{2^{2/3}-1}\right),-\sqrt[3]{2}+\sqrt{2^{2/3}-1})\approx (1.01317,-0.4935),\\
(a_-,r_-)=(\frac12\left(\sqrt[3]{2}-\sqrt{2^{2/3}-1}\right),-\sqrt[3]{2}-\sqrt{2^{2/3}-1})\approx (0.24675,-2.02634).
\end{cases}$$
It'll be more convenient later on if $|r|<1$, so for that reason we'll use the solution $(a,r)=(a_+,r_+)$ and disregard the other solution. Then, remembering that $\frac{1}{a}=-2r$, we have:
$$\frac{1}{\sqrt[3]{2}+\cos{x}}=\frac{1}{a(r^2-2r\cos{x}+1)}=\frac{-2r}{r^2-2r\cos{x}+1}=\frac{-2r}{1-r^2}\cdot\frac{1-r^2}{r^2-2r\cos{x}+1}.$$
When $r=-\sqrt[3]{2}+\sqrt{2^{2/3}-1}$, a little algebra will show that $\frac{-2r}{1-r^2}=\frac{1}{\sqrt{2^{2/3}-1}}$. Thus, when $r=-\sqrt[3]{2}+\sqrt{2^{2/3}-1}$,
$$\begin{align}
\frac{1}{\sqrt[3]{2}+\cos{x}}&=\frac{1}{\sqrt{2^{2/3}-1}}\cdot\frac{1-r^2}{r^2-2r\cos{x}+1}\\
&=\frac{1}{\sqrt{2^{2/3}-1}}\cdot\frac{1-\left(\sqrt{2^{2/3}-1}-\sqrt[3]{2}\right)^2}{\left(\sqrt{2^{2/3}-1}-\sqrt[3]{2}\right)^2-2\left(\sqrt{2^{2/3}-1}-\sqrt[3]{2}\right)\cos{x}+1}.
\end{align}$$

We shall take for granted that, for $|r|<1$, the following trigonometric series identity holds:
$$1+2\sum_{n=1}^{\infty}r^n\cos{(nx)}=\frac{1-r^2}{r^2-2r\cos{x}+1}.$$
Our work above means that we can take advantage of this series expansion to evaluate the desired definite integral.
$$\begin{align}
I&=\int_{0}^{\infty}\frac{e^{-x}}{\sqrt[3]{2}+\cos{x}}\mathrm{d}x\\
&=\int_{0}^{\infty}\frac{1}{\sqrt[3]{2}+\cos{x}}e^{-x}\,\mathrm{d}x\\
&=\frac{1}{\sqrt{2^{2/3}-1}}\int_{0}^{\infty}\frac{1-r^2}{r^2-2r\cos{x}+1}e^{-x}\,\mathrm{d}x\\
&=\frac{1}{\sqrt{2^{2/3}-1}}\int_{0}^{\infty}\left[1+2\sum_{n=1}^{\infty}r^n\cos{(nx)}\right]e^{-x}\,\mathrm{d}x\\
&=\frac{1}{\sqrt{2^{2/3}-1}}+\frac{2}{\sqrt{2^{2/3}-1}}\int_{0}^{\infty}\left[\sum_{n=1}^{\infty}r^n\cos{(nx)}\right]e^{-x}\,\mathrm{d}x\\
&=\frac{1}{\sqrt{2^{2/3}-1}}+\frac{2}{\sqrt{2^{2/3}-1}}\sum_{n=1}^{\infty}r^n\int_{0}^{\infty}e^{-x}\cos{(nx)}\,\mathrm{d}x\\
&=\frac{1}{\sqrt{2^{2/3}-1}}+\frac{2}{\sqrt{2^{2/3}-1}}\sum_{n=1}^{\infty}\frac{r^n}{n^2+1}\\
&=-\frac{1}{\sqrt{2^{2/3}-1}}+\frac{2}{\sqrt{2^{2/3}-1}}\sum_{n=0}^{\infty}\frac{r^n}{n^2+1}.
\end{align}$$
Hence, if we can find a nice closed form for the series $\sum_{n=1}^{\infty}\frac{r^n}{n^2+1}$, we'll likewise have found a closed form for the definite integral.
Approximately, $\sum_{n=1}^{\infty}\frac{r^n}{n^2+1}\approx -0.207410395311411...$.
WolframAlpha offers a closed form for the series $\sum_{n=0}^{\infty}\frac{r^n}{n^2+1}$ in terms of hypergeometric functions:
$$\sum_{n=0}^{\infty}\frac{r^n}{n^2+1}=\frac12\left(\ _2F_1(i,1;1+i;r)+\ _2F_1(-i,1;1-i;r)\right).$$
This might be the closest one can come to "closed" form.

An equivalent form of the series:
$$\begin{align}
S
&=\sum_{n=1}^{\infty}\frac{r^n}{n^2+1}\\
&=-\sum_{n=1}^{\infty}r^n\sum_{k=1}^{\infty}\frac{(-1)^k}{n^{2k}}\\
&=-\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}(-1)^k\frac{r^n}{n^{2k}}\\
&=-\sum_{k=1}^{\infty}(-1)^k\sum_{n=1}^{\infty}\frac{r^n}{n^{2k}}\\
&=-\sum_{k=1}^{\infty}(-1)^k\operatorname{Li}_{2k}{(r)}\\
&=-\sum_{k=0}^{\infty}(-1)^{k+1}\operatorname{Li}_{2k+2}{(r)}\\
&=\sum_{k=0}^{\infty}(-1)^{k}\operatorname{Li}_{2k+2}{(r)}.
\end{align}$$
Using the integral representation of $\operatorname{Li}_2$, $\operatorname{Li}_{\nu}{(z)}=\frac{z}{\Gamma{(\nu)}}\int_{0}^{\infty}\frac{t^{\nu-1}}{e^t-z}\mathrm{d}t$, we see that,
$$\operatorname{Li}_{2k+2}{(z)}=\frac{z}{(2k+1)!}\int_{0}^{\infty}\frac{t^{2k+1}}{e^t-z}\mathrm{d}t,$$
and using the power series for the sine function, we can formally sum this series for an interesting integral representation of the infinite series:
$$\sum_{k=0}^{\infty}(-1)^{k}\operatorname{Li}_{2k+2}{(r)}=r\int_{0}^{\infty}\frac{\sin{t}}{e^t-r}\mathrm{d}t.$$