Can someone please help me in finding the solution to the following integral:
$$\int_0^\infty \frac{e^{a u}}{2\mathrm{cosh}\left(\frac{u^2}{2}-b\right)+1} du,$$ where $a>0$ and $b>0$.
I have looked at all the table integrals, tried with Mathematica, without any success. So far I have a "good'' solution only when $b$ is large enough.
This integral is a result of a research problem that I am currently working on, and it represents a major blockade in moving forward.
$\mathbf{{Update\; 1:}}$
I good approximation to this integral, in the domain that I am interested in, would be the following:
$$\int_{\sqrt{2b}}^\infty \frac{e^{a u}}{e^{\frac{u^2}{2}-b}+1} du,$$ which again I struggle to solve. Note that if +1 is removed from the denominator, this integral is easy to be solved. But removing the +1 makes the solution not quite accurate.
I suppose that you could use $$\int_{\sqrt{2b}}^\infty \frac{e^{a u}}{e^{\frac{u^2}{2}-b}+1} du=\sum_{n=0}^\infty(-1)^n \int_{\sqrt{2b}}^\infty e^{a u} \left(e^{b-\frac{u^2}{2}}\right)^{n+1}\,du$$ $$I_n=\int_{\sqrt{2b}}^\infty e^{a u} \left(e^{b-\frac{u^2}{2}}\right)^{n+1}\,du$$ $$I_n=\sqrt{\frac{\pi }{2(n+1)}}\exp\left(\frac{a^2}{2 (n+1)}+b(n+1) \right) \text{erfc}\left(\sqrt{b (n+1)}-\frac{a}{ \sqrt{2(n+1)}}\right)$$
The problem will be how fast the sum converges.
Just to try, I used $a=\pi$ and $10$ terms for the summation $$\left( \begin{array}{ccc} b &\sum_{n=0}^{10} (-1)^n\, I_n &\text{exact value} \\ 0 & 332.862 & 332.396 \\ 1 & 844.917 & 841.868 \\ 2 & 2044.39 & 2031.30 \\ 3 & 4712.69 & 4669.65 \\ 4 & 10385.3 & 10264.2 \\ 5 & 21975.0 & 21668.4 \\ 6 & 44842.8 & 44124.8 \\ 7 & 88600.9 & 87019.1 \\ \end{array} \right)$$