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I am trying to solve this integration $$\int_{0}^{\pi} e^{cos(\theta)} \tan^{3}(\theta)d\theta$$

putting $$z=e^{i\theta}$$ $$\int_{\gamma} e^{(\frac{z^{2}+1}{2z})}\frac{{(z^{2}-1)}^3}{z(z-i)^{3}(z+i)^{3}}d\theta$$

I tried to solve the expression like this $$=\pi i[\frac{1}{2!}\lim_{z \to i} \frac{d^2}{dz^2}(z-i)^3f_{2}(z)+\frac{1}{2!}\lim_{z \to -i} \frac{d^2}{dz^2}(z+i)^3f_{2}(z)]$$ where $$f_{2}(z)=e^{(\frac{z^{2}+1}{2z})}\frac{{(z^{2}-1)}^3}{z(z-i)^{3}(z+i)^{3}}$$ If the integrand is symmetric around $\pi$, we change limits from $0$ to $2\pi$. (Answer to my previous question on this Residue theorem:When a singularity on the circle (not inside the circle)). However in this particular scenario integrand is not symmetric about $\pi$. Therefore I have a doubt whether I can derive the third equation as written in the form above. Any help on this is highly appreciated. Thanks

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MaxQuantum
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    You don't need a branch cut at zero. Zero is a pole. – dustin Feb 20 '15 at 06:23
  • @dustin Okay thanks. But can I even formulate that contour as a full circle, since this integrand is not an even function? Bacically can we change the limit to 2 pi? – MaxQuantum Feb 20 '15 at 06:29
  • You can because $\int_0^{\pi}f(\theta)d\theta = \int_0^{2\pi}f(\theta)d\theta$ where $f(\theta) = \tan^3 (\theta)$. – dustin Feb 20 '15 at 06:30
  • but say for example f(theta) is not equals to f(-theta) from 0 to pi, so is your above equation correct? (f(theta) should be symmetric about pi to get it as I understood,am I right?) – MaxQuantum Feb 20 '15 at 06:43
  • in 6th page of http://www.phys.ufl.edu/~pjh/teaching/phz3113/notes/week13.pdf has a similar example and they say function should be even to double the limit from 0 to 2 pi. @dustin Thanks for the help on this – MaxQuantum Feb 20 '15 at 06:49
  • @dustin I made a mistake in writing the equation. I have an extra exponental term. So this gives me the trouble on finding the contour and also taking pole at z=0 (which gives me exp^{infinity} at z=0) – MaxQuantum Feb 20 '15 at 08:14
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    I think that any method to solve this integral will lead to the same conclusion : the integral is not convergent. Even if we consider the integral in the sens of Cauchy principal value, it is not convergent. – JJacquelin Feb 20 '15 at 10:07
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    Yes. We need a convergent integral on the domain. – dustin Feb 20 '15 at 21:30

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