How to evaluate $\int \limits_{-\infty}^{\infty}\frac{e^{-|x|}}{|1-\sin x|^{\frac{1}{4}}} \,dx$? $$\int \limits_{-\infty}^{\infty}\frac{e^{-|x|}}{|1-\sin x|^{\frac{1}{4}}} \,dx$$
Any advice and comments will be appreciated
 A: 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.
