Originally Posted by

**simplependulum** But i find that this method is okay to me .

$\displaystyle \cosh(2z) = \sum_{k\geq 0} \frac{(2z)^{2k}}{(2k)!} $

Sub $\displaystyle z = \sqrt{t} $

$\displaystyle I = \sum_{k\geq 0} \frac{4^k}{(2k)!} \int_0^1 z^{k-1/2}(1-z)^{-1/2}~dz$

$\displaystyle = \sum_{k\geq 0} \frac{4^k}{(2k)!} \frac{ \Gamma(k+1/2) \Gamma(1/2)}{ \Gamma(k+1) } $

$\displaystyle = \sum_{k\geq 0} \frac{4^k}{(2k)!} \frac{ (2k)! \pi }{ 4^k (k!) (k!) } $

$\displaystyle = \pi \sum_{k\geq 0} \frac{1}{(k!)^2} $