Expected value of the sech of a normal random variable

Question

I am curious if there is any closed form for the following integral.

$$ H(k) = \frac{k}{\sqrt{2\pi}}\int_\mathbb{R} sech(t) e^{-\frac{k^2 t^2}{2}}dt $$

So far, what I've got is the following result: If we define $F(k):= H(\sqrt{\frac{2}{\pi}}k)$, then

$$ F(k) = k F(\frac{1}{k}) $$

but I don't think it is that useful. Also, I have tried the following idea: $H(k)$ can be written in terms of the pdf (at 0) of the sum of two independent variables, one is a normal with mean 0 and the other one follows a sech distribution.

Answer 1

As @Bob Dobbs commented, using $$\text{sech}(t)=\sum_{n=0}^\infty \frac{E_{2 n}}{(2 n)!} t^{2n}$$ using the exponential integral function $$\int t^{2n} e^{-\frac{k^2 t^2}{2}}\,dt=-\frac{1}{2} t^{2 n+1} E_{\frac{1}{2}-n}\left(\frac{k^2 t^2}{2}\right)$$ $$\int_0^\infty t^{2n} e^{-\frac{k^2 t^2}{2}}\,dt=2^{n-\frac{1}{2}} k^{-(2 n+1)} \Gamma \left(n+\frac{1}{2}\right)$$

$$H(k)=\frac{1}{2 \sqrt{\pi }}\sum_{n=0}^\infty \frac{2^n\, E_{2 n} \, \Gamma \left(n+\frac{1}{2}\right)}{(2 n)!}\,k^{-2 n}$$ which does not seem to have a closed form even inn terms of special function.

Converting the series expansion into a simple Padé approximant, using $t=\frac 1 {4k ^2}$, a rather good approximation seems to be $$H(k)\sim\frac 12 -\frac{t \left(4014355439 t^2+768501000 t+26097105\right)}{12729407095 t^3+7448005794 t^2+898986525 t+26097105}$$ which is $O\left(\frac{1}{k^{14}}\right)$

which seems to be decent. $$\left( \begin{array}{ccc} k & \text{approximation} &\text{solution} \\ 0.5 & 0.2721144728 & 0.2534350510 \\ 1.0 & 0.3718590167 & 0.3706321367 \\ 1.5 & 0.4239836019 & 0.4238919207 \\ 2.0 & 0.4509710566 & 0.4509710480 \\ 2.5 & 0.4661656602 & 0.4661646055 \\ 3.0 & 0.4754077398 & 0.4754075871 \\ 3.5 & 0.4813910557 & 0.4813910350 \\ 4.0 & 0.4854631489 & 0.4854634468 \\ 4.5 & 0.4883490124 & 0.4883490641 \\ 5.0 & 0.4904633764 & 0.4904633785 \end{array} \right)$$

For sure, we could do much better using Padé approximants of higher order.

Edit

Consider the coefficient $$a_n=\frac{2^n\, E_{2 n} \, \Gamma \left(n+\frac{1}{2}\right)}{(2 n)!}$$ which grow extremely fast $(a_{20}\sim 1.03\times 10^{16})$

$$\left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{2} \left|\frac{E_{2 (n+1)}}{(n+1) E_{2 n}}\right|$$ Using the good approximation (see here) $$E_{2n} \sim (-1)^n 8 \sqrt{\frac{n}{\pi}} \Bigg[\frac {4n}{e \pi} \,\frac{480 n^2+9}{480 n^2-1}\Bigg]^{2n}$$ gives $$\left|\frac{a_{n+1}}{a_n}\right|=\frac{8 n}{\pi ^2}\left(1+\frac{1}{2 n}+ O\left(\frac{1}{n^6}\right)\right)$$ probably implies that the series is of no use.

Answer 2

Another approach.

Write $$\text{sech}(t)=2 \sum_{n=0}^\infty (-1)^n e^{-(2 n+1) t}$$ $$H(k)=\sum_{n=0}^\infty (-1)^n\, e^{\frac{(2 n+1)^2}{2 k^2}}\, \text{erfc}\left(\frac{2 n+1}{k\sqrt{2} }\right)$$ which, as a summation, is much more pleasant since; if

$$a_n= e^{\frac{(2 n+1)^2}{2 k^2}}\, \text{erfc}\left(\frac{2 n+1}{k\sqrt{2} }\right)\qquad \implies \qquad \frac{a_{n+1}}{a_n}=1-\frac{1}{n}+\frac{3}{2 n^2}+O\left(\frac{1}{n^2}\right)$$

where we could even use $$e^{\frac{m^2}{2 k^2}} \text{erfc}\left(\frac{m}{\sqrt{2} k}\right)=\sqrt{\frac{2}{\pi }}\sum_{p=0}^\infty (-1)^p \, (2 p-1)!! \, \left(\frac{k}{m}\right)^{2 p+1}$$

If we have to compute the sum as $$H(k)=\sum_{n=0}^p (-1)^n\,a_n+\sum_{n=p+1}^\infty (-1)^n\,a_n$$ and we search for $p$ such that $$a_{p+1} \le \epsilon$$ we need to solve for $t$ $$e^{t^2} \text{erfc}(t)\le \epsilon \qquad \text{where} \qquad t=\frac{2 p+3}{k\sqrt{2} }$$ Using $$e^{t^2} \text{erfc}(t)\sim \frac{1}{t\sqrt{\pi } } \implies p=\frac{k}{\epsilon \sqrt{2 \pi } }-\frac{3}{2}$$ and then a lot of terms will need to be added.

Computing the partial sums for $k=1$

$$\left( \begin{array}{cc} p & \sum_{n=0}^p (-1)^n\,a_n \\ 1 & 0.280129 \\ 10 & 0.388692 \\ 100 & 0.372607 \\ 1000 & 0.370831 \\ 10000 & 0.370652 \\ \cdots & \cdots \\ \infty &0.370632 \end{array} \right)$$