usind Residue theorem prove that :

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- Aug 14th 2009, 11:30 AMflower3Residue theorem
usind Residue theorem prove that :

- Aug 14th 2009, 11:32 AMflower3
i using but the answer not the same

- Aug 14th 2009, 06:05 PMputnam120
First off, I think that you mean .

I would suggest try using . I haven't tried it out yet but it usually helps. After this you should have a normal rational function which is usually easier to integrate. - Aug 15th 2009, 12:39 AMMauritzvdworm
The redisue theorem will only work when you are integrating over some closed positively oriented contour where the function you want to integrate is analytic inside and on the contour except at say points

we can then write the answer in the following way

Now since the function you want to integrate is analytic everywhere the answer to your integral should be zero but there might be trick here, try decomposing the function in the following way

- Aug 15th 2009, 01:28 AMOpalg
- Aug 15th 2009, 01:31 PMflower3
- Aug 15th 2009, 02:29 PMOpalg
First, by a bit of elementary algebra, simplify the integral to . Next, identify the poles inside the unit circle by factorising the denominator. It is a quadratic in , with roots . As luck would have it, , so the denominator has four roots .

Now for a quotient of the form , the residue at a simple root of is given by . In this case and . So the residue at is . The poles inside the unit circle are given by . For both of these, and so the residue is .

Finally, the integral is equal to times the sum of the two residues, namely .