Expected value of the sech of a normal random variable 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.
 A: 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 $k=1$
$$\left(
\begin{array}{cc}
p & \sum_{n=0}^p (-1)^n\,a_n \\
 1 & 0.280129 \\
 10 & 0.388692 \\
 100 & 0.372607 \\
 1000 & 0.370831 \\
 10000 & 0.370652 \\
\cdots & \cdots  \\
\infty &0.370632
\end{array}
\right)$$
