What if one of the singularity gives infinity to the residue. Consider this contour;

$$X=\int_{\gamma} e^{i(\frac{z^{2}+1}{2z})}\frac{{(z^{2}-1)}^4}{2z^2(z-i)^{3}(z+i)^{3}}dz$$

I have singularities at z=0,i,-i. But when z=0, exponential function goes to infinity. How can this kind of countour be evaluated?

Note: Initial integration is

$$\int_{0}^{\pi} e^{i\cos(\theta)}\frac{\sin^{4}(\theta)}{\cos^{3}(\theta)}d\theta.$$

My attempt to solve this by substituting $cos(\theta)=z$ yields; $$\int_{-1}^{1} e^{iz}\frac{(1-z^2)^{3/2}}{z^3}dz.$$

  • $\begingroup$ This means at z=0 function is not analytic. In that case is there any method to solve the integration? (may be using a branch cut) $\endgroup$ – MaxQuantum Feb 21 '15 at 1:54
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    $\begingroup$ How did you get something so complicated? Why not try $z=\cos(\theta)$? Also, for what it's worth $e^{1/z}$ has an essential singularity at $z=0$, which is ok, it still has a power series expansion around $z=0$. There's no branch cut involved. $\endgroup$ – Alex R. Feb 21 '15 at 3:40
  • $\begingroup$ @AlexR. This thing comes with quantum field theory. I first tried z=cos(\theta) but it contains square root of function (1-z^2) and could not get an answer for that. Then I tried using residue theorem. What do you mean by " Also, for what it's worth e^1/z has an essential singularity at z=0"? I did not get that exactly.Any suggestion on this? $\endgroup$ – MaxQuantum Feb 21 '15 at 4:13
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    $\begingroup$ en.wikipedia.org/wiki/Essential_singularity Also, given that there's a singularity at $\pi/2$ in the original integral, are you interpreting it as a principal value? $\endgroup$ – Alex R. Feb 21 '15 at 4:18
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    $\begingroup$ @MaxQuantum : You are right. I retracted it. Sorry for the inconvenience. $\endgroup$ – JJacquelin Feb 21 '15 at 6:38

With change of variable $\theta=t+\frac{\pi}{2}$ $$I=\int_{0}^{\pi} e^{i\cos(\theta)}\frac{\sin^{4}(\theta)}{\cos^{3}(\theta)}d\theta=-\int_{-\pi/2}^{\pi/2} e^{-i\sin(t)}\frac{\cos^{4}(t)}{\sin^{3}(t)}dt$$ $x=\sin(t)$ $$I=-\int_{-1}^{1} e^{-ix}\frac{(1-x^2)^{3/2}}{x^3}dx$$ $$e^{-ix}\frac{(1-x^2)^{3/2}}{x^3}=\frac{1}{x^3}+\frac{i}{x^2}-\frac{2}{x}-\frac{5i}{3}+O(x)$$ The function to be integrated around $x=0$ includes an odd part $\frac{1}{x^3}-\frac{2}{x}$ which integral is finite in the sens of Cauchy principal value.

The function also includes an even part $\frac{i}{x^2}$ which integral is not convergent in terms of Cauchy principal value. So, the whole integral is not convergent, even in the sens of Cauchy principal value. This concerns the imaginary part of the integral, since the non-convergent term $\frac{i}{x^2}$ is imaginary. The real part of the integral is convergent and has a finite Cauchy principal value.

Now, if we consider only the real part, the function to be integrated is odd. So, the result is (in sens of Cauchy principal value) : $$\Re e(I)=-\int_{-1}^{1} \cos(x)\frac{(1-x^2)^{3/2}}{x^3}dx=0$$ The imaginary part of the integral is infinite : $$\Im m(I)=\infty$$

  • $\begingroup$ Fantastic.Thank you soo much. I was about to conclude this thinking the value is non convergent.But did not get that the real part is convergent. I took the series expansion inside the residue which is not convergent. And after your answer I noticed only the imaginary part doesnot converge. Thanks again $\endgroup$ – MaxQuantum Feb 21 '15 at 7:46
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    $\begingroup$ $O(x)$ is the Landau'symbol for the part of the series expension which tends to $0$ when $x$ tends to $0$ : mathworld.wolfram.com/LandauSymbols.html $\endgroup$ – JJacquelin Feb 21 '15 at 8:01
  • $\begingroup$ I found that (specially in theoretical physics) there are many methods to avoid singularities and obtain an answer. Do you have any idea on getting a value to the imaginary part here? or can we conclue that the answer to the imaginary part here is infinite since it is an even function integration? $\endgroup$ – MaxQuantum Feb 23 '15 at 10:52
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    $\begingroup$ I can answer only on the mathematical side of the problem, not in the physical side and not on the modelisation linking both. On the physical viewpoint, all depends how the modelization is made and what is the signifiance of the complex number in the model. Sometimes, the complex numbers act only as a kind of trick to simplify the mathematical solving. Sometimes, the imaginary part disappears because conjugates complex number are involves. In such cases, it doesn't matter if there is no convergence for the imaginary part. $\endgroup$ – JJacquelin Feb 23 '15 at 11:33
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    $\begingroup$ In the present case, the imaginary part of the integral is infinite. There is no miraculous method which could render it non-infinite. That's the mathematical result and a mathematical result do not depends on the methods of solving, except if there is a mistake in them. But changing something in the physical model might avoid the problem. $\endgroup$ – JJacquelin Feb 23 '15 at 11:48

It may be a good start with the initial integral is to go this way

$$ I = \int_{0}^{\pi} e^{i\cos(\theta)}\frac{\sin^{4}(\theta)}{\cos^{3}(\theta)}d\theta. = \frac{1}{2}\int_{-\pi}^{\pi}e^{i\cos(\theta)}\frac{\sin^{4}(\theta)}{\cos^{3}(\theta)}d\theta. $$

Using the substitution $z=e^{i\theta}$ transforms the integral to

$$ I = \int_{|z|=1}\frac{(z^2-1)^4e^{\frac{iz}{2}}e^{\frac{i}{2z}}}{z(z^2+1)^3}dz. $$

You can see that $z=0$ is an essential singularity and $z=i,-i$ are singularities which lie on the unit circle.

  • $\begingroup$ Thanks a lot.Do you have any idea how to deal with essential singularities. Im investigating on that now $\endgroup$ – MaxQuantum Feb 21 '15 at 6:06
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    $\begingroup$ @MaxQuantum: You are very welcome! There are many problems on this website that dealt with integrands have essential singularities. Try to make some search. It is really getting late now. By the way, where did this problem come from? $\endgroup$ – science Feb 21 '15 at 6:11
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    $\begingroup$ I am trying my best on this :). This comes from photon behavior in confined geomatries,related to quantum field theory $\endgroup$ – MaxQuantum Feb 21 '15 at 6:14
  • $\begingroup$ @MaxQuantum: Thank you. $\endgroup$ – science Feb 21 '15 at 6:15

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