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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 helpful?

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up vote 1 down vote accepted

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, 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.

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Thank you very much! But if you can think the procedure that can without need to change the integrand to the algebraic form first this should be better. – Harry Peter Nov 9 '13 at 4:06
Can the similar approach applicable to the integral like – Harry Peter Nov 9 '13 at 4:09

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