Need help with $\int_0^\infty\frac{e^{-x}}{\sqrt[3]2+\cos x}dx$

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:

$$\sum_{k=0}^{\infty}(-1)^{k}\operatorname{Li}_{2k+2}{(r)}=r\int_{0}^{\infty}\frac{\sin{t}}{e^t-r}\mathrm{d}t.$$

\begin{align*} \int_{0}^{\infty} \, \frac{e^{-x}}{2^{1/3}+\cos{x}}\, dx &= \int_{0}^{\infty} \, \sum_{n\ge 0} (-1)^n \frac{e^{-x}\, \left(\cos{x}\right)^n}{2^{(n+1)/3}} \, dx \tag 1 \end{align*} Let $\displaystyle I_n=\int_{0}^{\infty} \, e^{-x}\, \left(\cos{x}\right)^n \, dx$, which has a reduction formula:

$\displaystyle I_{n+2}=\frac{\left(n^2+3\, n+2\right)I_n + 1}{n^2+4\, n+5}\; \; , I_0=1, I_1=\frac{1}{2}$

Hence, $(1)$ can be written as the sum: \begin{align*} \int_{0}^{\infty} \, \frac{e^{-x}}{2^{1/3}+\cos{x}}\, dx &= \sum_{n\ge 0} (-1)^{n}\frac{I_n}{2^{(n+1)/3}} \approx 0.763521951811874989 \end{align*} If there exists a closed form for the recurrence, then we can expect a closed form for this integral as well.