It is known that the modified Bessel Function $K_z(a)$ ($a>0$)can be expressed as a Fourier transform $$K_z(a)=\frac{1}{2}\int_{-\infty}^{\infty}\exp(-a\cosh t)\cosh(zt){\rm d}t=K_{-z}(a)$$
Can the following integral ($m=1,2,3...$) be carried out in terms of functions like $K_z(a)$?
$$\frac{1}{2}\int_{-\infty}^{\infty}\frac{\exp(-a\cosh t)}{(\cosh t)^m}\cosh(zt){\rm d}t=?$$
We encountered this problem when we follow Polya's lead and try to find an approximation to the Riemann $\xi$ function.
Thanks- mike
Let us denote $$I_m=\frac{1}{2}\int_{-\infty}^{\infty}\frac{\exp(-a\cosh t)}{\cosh^m t}\cosh zt\,{d}t.$$ After replacing $\displaystyle\cosh^{-m} t=\frac{1}{(m-1)!}\int_0^{\infty}s^{m-1}e^{-s\cosh t}ds$ and exchanging the orders of integration we find that $$I_m=\frac{1}{(m-1)!}\int_0^{\infty}s^{m-1}K_z(a+s)\,ds.$$ Clearly, (after the change of variables $s\to s-a$) the last integral can be represented as a linear combination of $m$ integrals: \begin{align} &I_m=\sum_{n=1}^m\frac{(-a)^{m-n}J_n}{(n-1)!(m-n)!},\tag{1}\\ &J_n=\int_a^{\infty}u^{n-1}K_z(u)\,du,\qquad n=1,\ldots,m.\tag{2}\end{align}
Finally, according to Prudnikov-Brychkov-Marychev (vol. 2), the integrals (2) can all be expressed in terms of the hypergeometric function $_1F_2$: \begin{align}J_n=2^{n-2}\Gamma\left(\frac{n+z}{2}\right)\Gamma\left(\frac{n-z}{2}\right)-\frac{2^{z-1}\Gamma(z)}{n-z}a^{n-z}{}_1F_2\left(\frac{n-z}{2};1-z,\frac{n-z}{2}+1;\frac{a^2}{4}\right)\;\\ -\frac{2^{-z-1}\Gamma(-z)}{n+z}a^{n+z}{}_1F_2\left(\frac{n+z}{2};1+z,\frac{n+z}{2}+1;\frac{a^2}{4}\right). \end{align} This answer does not seem to be expressible in terms of simpler special functions.