Existence of an Analytic Formula for a Definite Integral It would be very helpful if the following definite integral or a similar one had an analytic solution:
$$\int\nolimits_{-\infty}^{\infty}\mathrm{sech}^2(x)\exp(-\alpha x^2)\,\mathrm dx,\qquad \alpha \geq 0$$
I have attacked this problem from several directions now, including contour integration (the Gaussian blows up along Re=0), differentiation under the sign (no luck), and interpreting it as the expected value when sampling with a certain distribution (because both of these functions may be interpreted that way easily).
To me, it seems like this could have a nice formula because both functions separately do, there is so much symmetry, and the integrand seems like just a bunch of exponentials to me.  I just wanted to know if anyone had a compelling reason why this won't have a 'nice' analytic solution - or even what the intuition of some more experienced people is concerning my probability of success or direction of my efforts.
 A: Let 
$$\begin{eqnarray*}
I(\alpha ) &:&=\int_{-\infty }^{+\infty }\left( \text{sech }\left( x\right)
\right) ^{2}\exp (-\alpha x^{2})\ \mathrm{d}x \\
&=&\int_{-\infty }^{+\infty }\left( \cosh x\right) ^{-2}\exp (-\alpha x^{2})\ \mathrm{d}x\qquad (1)
\\
&=&\int_{-\infty }^{+\infty }\frac{4}{\left( e^{x}+e^{-x}\right)
^{2}}e^{-\alpha x^{2}} \ \mathrm{d}x.
\end{eqnarray*}$$
Wolfram Alpha cannot evaluate $I(\alpha )$ in terms of standard mathematical functions, so most likely there is no closed form for it. In SWP I got the following expansion:

which I rewrote as
$$I(\alpha )=\sum_{k=0}^{\infty }4\left( -1\right) ^{k+1}e^{\frac{\left( k+1\right) ^{2}%
}{\alpha }}\sqrt{\pi }\left( -1+\mathrm{erf }\left( \frac{k+1}{\sqrt{\alpha }}%
\right) \right) \frac{k+1}{\sqrt{\alpha }},\qquad (2)$$
where $\mathrm{erf }(x)$ is the error function
$$\mathrm{erf }\left( x\right) =\frac{2}{\sqrt{\pi }}\int_{0}^{x}e^{-t^{2}}dt.\qquad(3)$$
In terms of the complementary error function
$$\mathrm{erfc }(x)=1-\mathrm{erf }(x)=\frac{2}{\sqrt{\pi }}\int_{x}^{\infty }e^{-t^{2}}\mathrm{d}t\qquad (4)$$  
it may be written as
$$I(\alpha )=\frac{4\sqrt{\pi }}{\sqrt{\alpha }}\sum_{k=0}^{\infty }
\left( -1\right) ^{k}e^{\left( k+1\right) ^{2}/\alpha } \left(
k+1\right)\ \mathrm{erfc }\left( \frac{1}{\sqrt{\alpha }}\left( k+1\right)
\right) . \qquad (5)$$
For $\alpha =1/2$, both $(1)$ and $(5)$ give $I(1/2)\approx 1.5183$; for $\alpha =1$, $I(1)\approx 1.2874$; and for $\alpha =10$, $I(10)\approx 0.53494.$
