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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

cand
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mike
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  • Did you mean the Riemann $\zeta$ function? – robjohn May 30 '14 at 23:05
  • Changing the variable of integration from $u$ to $t$ seems like a meaningless edit. – robjohn May 30 '14 at 23:07
  • @robjohn I mean Riemann $\xi(s)$ function, which is related to the Riemann $\zeta(s)$ function. The reason I changed $u$ to $t$ is because I tried to unify the notation of this question with that of another question of mine "a question about theorem 2 of de Bruijn's 1950 paper...". Thanks – mike May 30 '14 at 23:18

1 Answers1

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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.

Start wearing purple
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  • I think that $\cosh^{m} t$ in the text should be $\cosh^{-m} t$ – mike Aug 22 '14 at 07:39
  • @mike Indeed, thanks! I will correct in a moment. – Start wearing purple Aug 22 '14 at 07:45
  • I may also need another integral where $(\cosh t)^{-m}$ is replaced by $(2+\cosh t)^{-m}$. This would add a factor $\exp(-2u)$ in $J_n$ of (2). Can it still be expressed in terms of hypergeometric functions? Thanks- – mike Aug 22 '14 at 07:55
  • It looks like not that easy. Mathematica 7.0 can work out the integration of $J_n$ as shown in (2). When I added an extra factor $\exp(-2u)$, it gave up. – mike Aug 22 '14 at 08:10
  • @mike I've checked Prudnikov-Brychkov-Marychev, unfortunately they have no formula that could help in the modified case. – Start wearing purple Aug 22 '14 at 09:30
  • Thanks a lot for the effort. Best regards! – mike Aug 22 '14 at 09:32