The integral form of the $n$-th modified Bessel function of the first kind is $$ I_n(z)=\frac{1}{\pi}\int_0^{\pi}e^{z\cos\theta}\cos(n\theta)\;d\theta. $$ However, I found an integral $$ L(a)=\frac{2}{\pi}\int_0^{\pi/2}\cos^2k~\textrm{sech}^2(a\cos k)~dk,\;\;a>0 $$ I don't know whether the integral $L(a)$ relates to the Bessel function $I_n(z)$ or not. I just want to know the closed-form of the integral $L(a)$. So, if there exist a closed-form or not for $L(a)$?

In Mathematica, I found that $L(1)=0.263272$, $L(2)=0.0768269$.


There is no closed form of this integral that can be readily determined. It can be determined that $\textrm{sech}^{2}(x)$ has coefficients related to Bernoulli numbers and are similar in form to those of $\tanh(x)$ and has the form $$\textrm{sech}^{2}(x) = \sum_{n=0}^{\infty} \frac{(-1)^{n} \, \theta_{n}}{n!} \, x^{2n}.$$
Now the integral in question is $$ L(a)=\frac{2}{\pi}\int_0^{\pi/2}\cos^2k~\textrm{sech}^2(a\cos k)~dk,\;\;a>0 $$ for which the following is determined. \begin{align} L(a) &= \frac{2}{\pi} \, \int_{0}^{\pi/2} \cos^{2}t \, \textrm{sech}^2(a\cos t) \, dt \\ &= \frac{2}{\pi} \, \sum_{n=0}^{\infty} \frac{(-1)^{n} \, \theta_{n} \, a^{2n}}{n!} \, \int_{0}^{\pi/2} \cos^{2n+2}t \, dt \end{align} Since $$\int_{0}^{\pi/2} \cos^{2n+2}t \, dt = \frac{\pi \, (2n+3) \, \left(\frac{1}{2}\right)_{n}}{4 \, (n+1)!}$$ then $$L(a) = \frac{1}{2} \, \sum_{n=0}^{\infty} \frac{(-1)^{n} \, (2n+3) \, \left(\frac{1}{2}\right)_{n} \, \theta_{n}}{n! \, (n+1)! } \, a^{2n},$$ where $(a)_{n}$ is the Pochhammer symbol and $\theta_{n}$ is related to $$\theta_{n} \sim \frac{4^{n} \, (4^{n} -1)}{(2n)!} \, B_{2n}$$ where $B_{2n}$ are the even Bernoulli numbers.

  • $\begingroup$ Thank you for your answer. According to your analysis, shall we get the asymptotic behavior of $L(a)$ as $a\rightarrow\infty$? $\endgroup$ – Roger209 Aug 24 '15 at 2:55
  • $\begingroup$ That is an interesting proposal for an expansion. It may be possible, but I am not sure currently. $\endgroup$ – Leucippus Aug 24 '15 at 23:56

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