How to evaluate $\int \limits_{-\infty}^{\infty}\frac{e^{-|x|}}{|1-\sin x|^{\frac{1}{4}}} \,dx$?

Question

$$\int \limits_{-\infty}^{\infty}\frac{e^{-|x|}}{|1-\sin x|^{\frac{1}{4}}} \,dx$$

Any advice and comments will be appreciated

Answer 1

Here are my thoughts so far -- not an answer, just a possible start.

Observing that $\sin x\le1$, we can easily remove the absolute values: $$ I=\int_{-\infty}^{\infty} \frac{e^{-|x|}}{|1-\sin x|^{\frac{1}{4}}} \,dx =2\int_{0}^{\infty} \frac{e^{-x}}{(1-\sin x)^{\frac{1}{4}}} \,dx. $$ A major challenge of solving this integral with a substitution is that the denominator is periodic, with poles at $x=(2n+\frac12)\pi$. For example, with $z=\tan\frac{x}{2}$, we'd have $dz=\frac12\sec^2\frac{x}{2}dx$ $=$ $\frac12(1+z^2)dx$ or $dx=\frac{2}{1+z^2}dz$ and $x=\frac{2z}{1+z^2}$ which would yield $$ I=2\int_{0}^{?} \frac{e^{-2\tan^{-1}z}}{(1-\sin x)^{\frac{1}{4}}} \,dx, $$ but there would be no way to transform the upper limits integration! However, this problem can be surmounted by integrating separately over each period of $\sin x$ to obtain $$ \eqalign{ I &= 2\sum_{n=0}^{\infty} \int_{2\pi n}^{2\pi(n+1)} \frac{e^{-x}}{(1-\sin x)^{\frac{1}{4}}} \,dx \cr &= 2\sum_{n=0}^{\infty} e^{-2n\pi} \int_{0}^{2\pi} \frac{e^{-x}}{(1-\sin x)^{\frac{1}{4}}} \,dx \cr &= \frac2{1-e^{-2\pi}} \int_{0}^{2\pi} \frac{e^{-x}}{(1-\sin x)^{\frac{1}{4}}} \,dx. } $$ This is an improper integral, but it exists because $1-\sin x$ as $x$ approaches $\frac\pi 2$ behaves the same way that $1-\cos x$ does as $x$ approaches $0$, namely like $x^2$, and $\int_0^1\frac{dx}{x^p}$ exists for $0<p<1.$

At this point we have at least a few options to pursue. We could:

• use a Taylor series expansion for $(1-\sin x)^{-\frac14}=\sum a_nx^n$ and integrate each term $\sum a_nx^ne^{-x}$

• carry out a substitution like the one above, or like $x=t^2$ to get $$ I=\frac2{1-e^{-2\pi}} \int_0^\sqrt{2\pi}\frac{2te^{-t^2}dt}{(1-\sin t^2)^{\frac14}} %\quad\text{or}\quad %I=2\int_0^\infty\frac{2te^{-t^2}dt}{(1-\sin t^2)^{\frac14}} $$

• use $\int e^{-x}(1+\sin x)^{\frac14}(\cosh ix)^{-\frac12}\,dx$ and possibly end up integrating in the complex plane.

There is presumably a way since sage reports $3.2916690469253642$, but I haven't checked how it's getting that.