Integral representation of cosh x On Wolfram math world, there's apparently an integral representation of $\cosh x$ that I'm unfamiliar with. I'm trying to prove it, but I can't figure it out. It goes \begin{equation}\cosh x=\frac{\sqrt{\pi}}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty} \frac{ds}{\sqrt{s}}\,e^{s+\frac{1}{4}\frac{x^{2}}{s}} \gamma >0\end{equation}
The contour is taken along a vertical line with positive real part. I thought at first glance to use the residue theorem but it seems to be of no use here.

Expand the difficult part of the exponential in power series, the integral equals
$$ I = \sqrt\pi \sum_{k\geq0} \frac{(x^2/4)^k}{k!} \frac{1}{2\pi i}\int_{\Re s=\gamma} s^{-k-1/2}e^{s}\,ds. $$
The integral here is the inverse Laplace transform of the function $s^{-k-1/2}$ evaluated at the point $t=1$, given by
$$ \mathcal{L}^{-1}[s^{-k-1/2}](t) = \frac1{2\pi i}\int_{\Re s=\gamma}s^{-k-1/2}e^{st}\,ds. $$
So we can look it up (http://mathworld.wolfram.com/LaplaceTransform.html):
$$ \frac1{2\pi i}\int_{\Re s=\gamma}s^{-k-1/2}e^{s}\,ds = \frac{1}{\Gamma(k+1/2)}, $$
which also satisfies
$$ \frac{\Gamma(1/2)}{\Gamma(k+1/2)} = \frac{1}{\frac12\times\frac32\times\cdots\times(k-\frac12)}, $$
where $\Gamma(1/2)=\sqrt\pi$. Simplifying, we get
$$ \sum_{k\geq0} \frac{(x^2/4)^k}{k!} \frac{\sqrt\pi}{\Gamma(k+1/2)} = \sum_{k\geq0}\frac{x^{2k}}{(2k)!} = \cosh x. $$
