Prove $\frac{\sqrt\pi z^v}{2^v~\Gamma\left(v+\frac{1}{2}\right)}\int_0^\infty e^{-z\cosh t}\sinh^{2v}t~dt=\int_0^\infty e^{-z\cosh t}\cosh vt~dt$? How to prove $\dfrac{\sqrt\pi z^v}{2^v~\Gamma\left(v+\dfrac{1}{2}\right)}\int_0^\infty e^{-z\cosh t}\sinh^{2v}t~dt=\int_0^\infty e^{-z\cosh t}\cosh vt~dt$ ?
Does some formulae in http://dlmf.nist.gov/5.12 helpful?
 A: Let $K_{\nu}(z)$ the modified Bessel function of the second kind (or Macdonald function). The function $K_{\nu}(z)$ admits the Sommerfeld integral representation
$$
K_{\nu}(z)=\int_0^{\infty}\operatorname{e}^{-z\cosh t}\cosh(\nu t)\operatorname{d}t
$$
for $|\arg z|<\frac{\pi}{2}$ and $\nu\in\Bbb C$.
If we take $\operatorname{e}^{-u}=t$, then 
$$
K_{\nu}(z)=\int_0^{\infty}\operatorname{e}^{-z\cosh t}\cosh(\nu t)\operatorname{d}t=\frac{1}{2}\int_0^{\infty}\operatorname{e}^{-\frac{z}{2}\left(u+\frac{1}{u}\right)}u^{-\nu-1}\operatorname{d}u
$$
that can also be written as
$$
K_{\nu}(2\sqrt{z\lambda})=\frac{1}{2}\left(\frac{\lambda}{z}\right)^{\nu/2}\int_0^{\infty}\operatorname{e}^{-zt-\frac{\lambda}{t}}t^{-\nu-1}\operatorname{d}t
$$
Using the simple result
$$
u^{-\nu-\frac{1}{2}}=\frac{1}{\Gamma\left(\nu+\frac{1}{2}\right)}\int_0^{\infty}\operatorname{e}^{-ux}x^{\nu-\frac{1}{2}}\operatorname{d}x \qquad \text{for}\;\frak{Re}(\nu)>-\frac{1}{2}
$$
the integral becomes
$$
K_{\nu}(z)=\frac{1}{2\Gamma\left(\nu+\frac{1}{2}\right)}\int_0^{\infty}x^{\nu-\frac{1}{2}}\left[ \int_0^{\infty}\operatorname{e}^{-u\left(x+\frac{z}{2}\right)-\frac{z/2}{u}}u^{-\frac{1}{2}}\operatorname{d}u \right]\operatorname{d}x
$$
where the order of integration may be interchanged.
The inner integral is a special case of $K_{\nu}$ for $\nu=-1/2$ and observing that $K_{1/2}(x)=K_{-1/2}(x)=\sqrt{\frac{\pi}{2x}}\operatorname{e}^{-x}$ (see http://dlmf.nist.gov/10.39#2), the integral is
$$
K_{1/2}(2\sqrt{y\lambda})=\sqrt{\frac{\pi/2}{2\sqrt{y\lambda}}}\operatorname{e}^{-2\sqrt{y\lambda}}
$$
with $y=\frac{z}{2}$ and $\lambda=x+\frac{z}{2}$, that is
$$
K_{1/2}\left(2\sqrt{\frac{z}{2}\left(x+\frac{z}{2}\right)}\right)=\frac{1}{2}
\left(\frac{z/2}{x+z/2}\right)^{-1/4} \sqrt{\frac{\pi}{x+z/2}}\operatorname{exp}\left(-2\sqrt{\frac{z}{2}\left(x+\frac{z}{2}\right)}\right)
$$
With the substitution $\theta=\sqrt{\frac{x+z/2}{z/2}}$ in the remaining integral in $x$, one has
$$
K_{\nu}(z)=\frac{\sqrt\pi (z/2)^{\nu}}{\Gamma\left(\nu+\frac{1}{2}\right)}\int_1^{\infty} 
\operatorname{e}^{-z\theta}(\theta^2-1)^{\nu-\frac{1}{2}}\operatorname{d}\theta
$$
The change of variables $\theta=\cosh t$ yields
$$
K_{\nu}(z)=\frac{\sqrt\pi (z/2)^{\nu}}{\Gamma\left(\nu+\frac{1}{2}\right)}\int_0^{\infty} 
\operatorname{e}^{-z\cosh t}\sinh^{2\nu}t\operatorname{d}t.
$$
This ends the proof.
