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- June 19th 2009, 12:21 AMsimplependulumAn interesting integral
solve

- June 20th 2009, 04:48 PMNonCommAlg
- June 20th 2009, 09:54 PMsimplependulum
My answer is in terms of and hyperbolic cosine ( I hope i am correct ) but not in series form .

Actually this theorem inspired me to think of this question . - June 23rd 2009, 03:02 AMsimplependulum
Check it whether your results are as same as my result

- July 3rd 2009, 07:09 AMhalbard
Yes, the substitution transforms the integral to , which can be written as .

We may assume that since cosine is even, and since sine is odd we may replace with as well. Thus , a Fourier transform.

Now consider the integral of the function around the rectangular contour with vertices , , and , where , in the complex plane.

The function is holomorphic inside and on C except at a simple pole , and its residue there is .

By the residue theorem .

Also upon integrating along each side of .

Now , and in these integrals. Thus

, where and similarly for . Letting we have .

Therefore , i.e. . - July 3rd 2009, 10:35 PMsimplependulum
Thank you very much , halbard

Besides solving contour integral , i have another method to solve it

Since (1)

And (2)

Then substitute into (1) and

into (2)

and

But

So after equating these two equations , we obtain

Take the real part of the integral ,

Replace by ,