Guest sAke Posted February 9, 2005 Posted February 9, 2005 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.
Tom Mattson Posted February 9, 2005 Posted February 9, 2005 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.
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