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

Any advice and comments will be appreciated

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It might be worth noting that the range for $\sin x$ is $-1\leq \sin x \leq 1$. Therefore $|1−\sin x|$ is always positive and so you can remove the absolute value in the denominator. – E.O. Feb 16 '12 at 10:31
This is the integral on $(0,2\pi)$ of the same function without the absolute values, times $2/(1-\mathrm e^{-2\pi})$. – Did Feb 16 '12 at 11:21
To see what Didier Piau means you might divide the interval into intervals of length $2\pi$ and use periodicity of the denominator together with the fact that $\exp(-x-2n\pi)= \exp(-2n\pi)\exp(-x)$ (giving a geometrical series). – AD. Feb 16 '12 at 12:17
@DidierPiau I see what you mean, but not if it helps.. – AD. Feb 16 '12 at 12:18
Is this homework? For what class? Are you supposed to know complex analysis? – bgins Feb 16 '12 at 14:14

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.

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This is what Mathematica gives for the integral that you got stuck $$-\left(\frac{4}{25}-\frac{28 i}{25}\right) \left(\, _2F_1\left(-\frac{1}{4}+i,1;\frac{1}{4}+i;-i\right)-(1+i) \, _2F_1\left(\frac{3}{4}+i,1;\frac{1}{4}+i;-i\right)\right)+\left(\frac{4}{25}-\fr‌​ac {28 i}{25}\right) e^{-2 \pi } \left(\, _2F_1\left(-\frac{1}{4}+i,1;\frac{1}{4}+i;-i\right)-(1+i) \, _2F_1\left(\frac{3}{4}+i,1;\frac{1}{4}+i;-i\right)\right)+\frac{2^{3/4} e^{-\pi /2} \sqrt{\pi } \Gamma \left(\frac{1}{4}+i\right)}{\Gamma \left(\frac{3}{4}+i\right)}$$ – Norbert Feb 16 '12 at 13:23
I noticed that (1) the Taylor series for $(\frac{1-\cos x}{2})^{-\frac14}/\sqrt x$ converges very nicely and (2) that $x=t^2$ might take the integral slightly closer in your direction. What is $F_1$? – bgins Feb 16 '12 at 13:36
Looks like Barnes integral: – bgins Feb 16 '12 at 14:12
Oh, I forgot to simplify the result $$\frac{1}{425 \Gamma \left(\frac{3}{4}+i\right)}\Biggl(e^{-2 \pi } \left((176+432 i) \left(e^{2 \pi }-1\right) \left(-1+\, _2F_1\left(-\frac{1}{4}+i,1;\frac{1}{4}+i;-i\right)\right) \Gamma \left(\frac{7}{4}+i\right)+425\ 2^{3/4} e^{3 \pi /2} \sqrt{\pi } \Gamma \left(\frac{1}{4}+i\right)\right)\Biggr)$$ – Norbert Feb 16 '12 at 14:22

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