Evaluating $\int_0^{\infty} \frac {e^{-x}}{a^2 + \log^2 x}\, \mathrm d x$ I am trying to evaluate this integral
$$I=\int_0^{\infty} \frac {e^{-x}}{a^2 + \log^2 x}\, \mathrm d x$$
for $a \in \mathbb R_{>0}$.
Any ideas?
In the case $a=\pi$ we have $I= F - e$ where $F$ is the Fransén–Robinson constant.
 A: This is not an answer but a complement to the obtention of the Fransén–Robinson constant by Ramanujan (i.e. the case $a=\pi$ in Hardy's book "Ramanujan" $11.10$). Not sure it will help much here but...
The Formula C  referenced is :
$$\phi(0)+\phi(1)+\phi(2)+\cdots=\int_0^\infty \phi(x)\,dx+\int_0^\infty\frac{\phi(0)-x\phi(1)+x^2\phi(2)-\cdots}{x\;(\pi^2+(\log\,x)^2)}dx$$
and the Abel-Plana formula is from Wikipedia.



A: This is more a comment/hint than a real answer, but it's too long for the comment section.
First do the substitution $e^u = x$, so that $dx = e^udu$ and your integral becomes
$$\color{red}{-}\int_{-\infty}^\infty\color{red}{(-}e^u\color{red}{)}e^{-e^u}\frac{1}{a^2+u^2}du.$$
Now you could try to substitute using the Lambert $W$ function to get rid of the first term:
$$W(z) = -e^u,\qquad du = \frac{1}{W(z)}\frac{1}{z+e^{W(z)}}dz$$
(if I worked that out correctly), and your integral takes the form
$$\int_{\color{red}{?}}^{\color{red}{?}}\frac{z~dz}{(a^2+\log(-W(z))^2)W(z)(z+e^{W(z)})}.$$
Now this is of course very ugly, but maybe it can be used somehow. Honestly, I have no idea. Where is @Cleo when you need her?
A: Sorry this answer is not a complete on but I have not enough reputation to put this as comment. If some admin cares to change this to a comment, please do so.
The standard approach to such a thing would be to look at the integral as part of a closed path integral. Let $R_0, R_1>0$, $R_0\rightarrow 0,\;\; R_1 \rightarrow \infty$ and integrate from $I_1=\int_{R_0}^{R_1}$ along the real axis, a quarter circle back at $r=R_1$ $I_2=\int_{0}^\frac\pi{2}$, down the imaginary line $I_3=-\int_{R_0}^{R_1}$ and on a quater circle with $r=R_0\;\;I_4=-\int_0^\frac\pi{2}$. Observe that for $a>\frac\pi 2$ there are no poles inside the closed integration path.
Hope this helps someone else to come up with a proper answer.
A: I don't know if it makes you go further, but with a couple of substitutions I arrived at this expression:
$$
\int_0^{\infty} \frac {e^{-x}}{a^2 + \log^2 x}\, \mathrm d x = K \int_0^{\infty} \frac {e^{-t}}{\pi^2 + \log^2 t}e^{-t^{-q}-1}t^{-q-1}\, \mathrm d x
$$
It is nice because it seems to mix the integral of $\Gamma$ with the integral of its inverse.
If you need I'll double-check the calculations to give you the correct values for $K$, while $q=\pi/a$. 
