Ok, I've got this integral, having slight problems solving it:

 

Int ( [coth(x/2)-1]sin(ax) dx ) from zero to infinity.

 

Now, I've expanded coth(x/2) = (e^(x/2)+e^(-x/2))/(e^(x/2)-e^(-x/2))

 

= (1+e^(-x))/(1-e^(-x))

 

So I get the integral

 

Int( 2e^(-x)/(1-e^(-x))sin(ax) dx).

 

I've thought that this may be siplified using the sum

 

sum_n ( z^n ) = 1/(1-z) when n goes from zero to infinity, thus I get

 

2*int( e^(-x)e^(-nx)sin(ax) dx)

 

Any tips on what to do next? i know the solution of the integral to be

 

pi*coth(pi*a)-1/a, but can't figure out how to get there.

2*int( e^(-x)e^(-nx)sin(ax) dx)

 

Any tips on what to do next?

 

The next logical step would be to combine the two exponentials (e^-(n+1)x) and integrate by parts twice.