How was the difference of the Fransén–Robinson constant and Euler's number found?

I recently ran across the following integral:

$$ \int_{0}^{\infty}\frac{1}{\Gamma(x)}dx $$

Which I learned is equal to the Fransén-Robinson constant. On the linked wikipedia page for the Fransén-Robinson constant, it states that the difference between Fransén-Robinson constant and Euler's number can be expressed by this:

$$ F = e + \int_{0}^{\infty}\frac{e^{-x}}{\pi^2+\ln^2(x)}dx $$

Where on earth did the difference come from? How do we know this?

Solution 1:

It is a consequence of the $\Gamma$ reflection formula: $$ \Gamma(z)\,\Gamma(1-z) = \frac{\pi}{\sin(\pi z)} \tag{1} $$ and the Cantarini's trick (aka the Laplace transform of the sine function): $$ \int_{0}^{+\infty} \sin(a t)\,e^{-bt} = \frac{a}{a^2+b^2}\tag{2} $$ from which:

$$ \frac{1}{\pi^2+\log^2(x)} = \int_{0}^{+\infty} \frac{\sin(\pi t)}{\pi} x^{-t}\,dt \qquad \left(\log(x)>0\right)\\\frac{1}{\pi^2+\log^2(x)} = \int_{0}^{+\infty} \frac{\sin(\pi t)}{\pi} x^{t}\,dt \qquad \left(\log(x)<0\right)\tag{3}$$ so $ \int_{0}^{+\infty}\frac{e^{-x}}{\pi^2+\log^2(x)}\,dx$ is related with: $$ \int_{0}^{+\infty}\frac{\sin(\pi t)}{\pi}\,\Gamma(1-t)\,dt = \int_{0}^{+\infty}\frac{dt}{\Gamma(t)}\tag{4}$$