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Suppose that I'm trying to evaluate the following integral:

$$\frac{1}{2\pi i}\ \int_{C} \frac{cos(\pi z)}{z^2-1}dz $$

And further suppose that C is a rectangle going over $ 2+i,2-i,-2+i,and -2-i$. Given that i is a singularity and present ont he boundary, how int he world am I suppose to apply Cauchy's integral formula. I can do this if the singularity is a corner, but what do we do if it's an entire line? I should be able to do this without the residue theorem.

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    $\begingroup$ $i$ is not a singularity, the denominator is $z^2-1 = (z-1)(z+1)$. $\endgroup$ – Daniel Fischer Feb 22 '14 at 16:10
  • $\begingroup$ You can do that. You can also do it in a different way. If you already know the residue theorem, use that. If not, you can for example split the contour by introducing a segment $[-i,i]$ to get $C$ as the "sum" of two closed contours each encircling only one zero of the denominator and apply the integral formula to $\dfrac{\cos \pi z}{z \pm 1}$. $\endgroup$ – Daniel Fischer Feb 22 '14 at 16:25
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    $\begingroup$ @user108149: Effort went into answering your question, and others may benefit from this effort in the future. To remove such a question and the answer would be to have this effort go to waste. If you have a valid reason for deleting this question, please let us know. If you do not, then please do not destroy the question. $\endgroup$ – RghtHndSd Mar 13 '14 at 16:42
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Subdivide your rectangles in the subrectangles

$$C_1 :\;2+i\,,\,i\,,\,-1\,,\,2-i\\ C_2 :\;-2+i\,,\,-2-i\,,\,-i\,,\,i$$

and then apply the CIF:

$$\frac1{2\pi i}\int\limits_C\frac{\cos \pi z}{z^2-1}dz=\frac1{2\pi i}\left[\int\limits_{C_1}\frac{\frac{\cos \pi z}{z+1}}{z-1}dz+\int\limits_{C_2}\frac{\frac{\cos \pi z}{z-1}}{z+1}dz\right]=$$

$$=\left.\frac{\cos \pi z}{z+1}\right|_{z=1}+\left.\frac{\cos \pi z}{z-1}\right|_{z=-1}=\frac{-1}2+\frac{-1}{-2}=0$$

Observe what happens when integrating on the common frontier of both subrectangles...

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  • $\begingroup$ Partial fraction decomposition is for rational functions, @user108149. You have here a transcendental one as cosine is not a polynomial (just try to do it...). And I can't see how could integration be carried on the whole rectangle... $\endgroup$ – DonAntonio Feb 22 '14 at 17:12
  • $\begingroup$ Oh, you meant that...well, it is the very same and still one can't integrate over the whole rectangle...You may prefer to do it that way, but the really important issue here, imo, is that subdivision of the rectangle. $\endgroup$ – DonAntonio Feb 22 '14 at 17:39
  • $\begingroup$ I don't get it, @user108149: didn't you write just two comments above that "...you can't integrate over the whole rectangle"? So what is your method that doesn't require to subdivide the rectangle? Because it certainly isn't the one you call "partial fractions"... $\endgroup$ – DonAntonio Feb 22 '14 at 17:45

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