inverse fourier transform application:

So the question states:
apply the inverse fourier transform to your answer to the previous problem to show that for x>0 and a> 0,
integral (0 to infinity) ((s sin sx) / (a^2+s^2)) ds = (pi/2) e^(-ax).

So the previous problem was:
Find the fourier transform of the function f defined by:
f(x) = -e^(ax) x<0
= e^(-ax) x>0
and my answer for the x>0 part was: 1/(a+ik)

so applying the inverse fourier transform to the f(x) = e^(-ax) bit,
f(x) = 1/2pi . integral (k=-infinity to infinity) f hat (k) e^(ikx) dk

sooooo get:
f(x) = 1/2pi . integral (k=-infinity to infinity) 1/(a+ik) e^(ikx) dk