${\displaystyle {\int }_0^{\infty }[(\sinh \mathrm{ax})/(\sinh \pi x)]\mathrm{dx}}$

$-\pi <a<\pi$

I am not able to think of a proper contour for this.

love arun

Thanks David!
i got close to the answer but not very close.
i got the answer as 2tan(a/2) but my text says (1/a)*tan(a/2)

I will put my working here.

This is the contour ...

(Sorry for the shabby drawing)
Now since the pole z=0 is on real axis,i had to indent the contour by drawing a semi-circle.

$R(i)=(-i/\pi )\sin (a)$

so the integral along the contour $=2\sin (a)$

Thus the given integral breaks up into six integrals,

${\int }_{-R}^{-r}+{\int }_{-r}^r+{\int }_r^R+{\int }_{\mathrm{AB}}+{\int }_{\mathrm{BC}}+{\int }_{\mathrm{CD}}$

if we let $r\to 0$ and $R\to \infty$, the second, fourth and sixth integrals are zero (if I have not made any mistake).

The first and third then get merged.

Further simplification gives,

$(1+\cos (a)){\int }_{-\infty }^{\infty }[\sinh (\mathrm{ax})/\sinh (\pi x)]\mathrm{dx}+i\sin (a){\int }_{-\infty }^{\infty }[\cosh (\mathrm{ax})/\sinh (\pi x)]\mathrm{dx}$

Then the comparison of real and imaginary parts gives me the answer.

What mistake have I made?

love arun

Demetres

Oops Sorry!!
Thanks Demetres.(Thanks Kerwin and David)

Another Question,
How to do ....

love arun