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 .
Check it whether your results are as same as my result
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. .
Thank you very much , halbard
Besides solving contour integral , i have another method to solve it
Then substitute into (1) and
So after equating these two equations , we obtain
Take the real part of the integral ,
Replace by ,