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.
 A: 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
$$ \sum_{k\geq0} \frac{(x^2/4)^k}{k!} \frac{\sqrt\pi}{\Gamma(k+1/2)} = \sum_{k\geq0}\frac{x^{2k}}{(2k)!} = \cosh x. $$
A: It will almost certainly help to know that 
$$
\mathcal L^{-1}\{F(s)\}= \frac1{2\pi i} 
\int_{\gamma - i\infty}^{\gamma + i\infty} e^{st}F(s)\,ds \quad \gamma>0
$$
Gives you the Inverse Laplace transform of $F(s)$ (where $\gamma$ is taken to be in the region of convergence of $F(s)$).  Your formula looks very much like some permutation of the above.
