i want to prove that:
$\int_0^{\infty}t^{-\frac{3}{2}}(1-e^{-t})dt=2\sqrt{\pi}$
I noticed that the integral can be written in this form:
$\int_0^{\infty}t^{a-1}R(t)dt$ where $R(t)=\frac{1-e^{-t}}{t}$ and $a=\frac{1}{2}$
There is a theorem in my book which says that these integrals can be calculated like this:
$\int_0^{\infty}t^{a-1}R(t)dt=\frac{2\pi}{1-e^{i2\pi a}} \sum_j Res[z_j^{a-1}R(z_j)]$
Where $z_j$ its poles, the problem is that the residue of this function is zero which means that the whole integral calculates to zero, which means means im doing something wrong but i can't figure out what. Also the first integral looks a bit like the Gamma function if it's any help maybe the answer lies there.