I hope this does not seem like a too easy question for Overflow. I would like to find an easier method than mine to prove the above statement for the FransénRobinson Constant. My first method was to use the reflection formula, and then a laplace transform for the sin($\pi$x). Due to convergence problems with the integral definition of the gamma function, I was always missing the factor of $+e$. After days of work, I was actually able to prove and derive a way stronger argument, where the above equality is just a special case (see pictures). I would appreciate some help in simplifying the process of proving it!

5$\begingroup$ for a helpful answer you will really want to invest the effort to write out what you have done in the question box, in a compact way like you would for a research publication; a link to hand written notes is not that helpful. $\endgroup$ – Carlo Beenakker Jul 9 at 6:01
This is Exercise 100 of Chapter 9 of my GTM 240 (Exercise 99 gives the complex analytic method): set $$I(a,t)=\int_{t}^\infty\dfrac{a^x}{\Gamma(x+1)}\,dx+\int_0^\infty\dfrac{e^{ax}x^{t1}}{\log^2(x)+\pi^2}\Bigl(\cos(\pi t)\dfrac{\sin(\pi t)}{\pi}\log(x)\Bigl)\,dx$$ 1) After proving absolute convergence, prove that its derivative with respect to $t$ vanishes, so that $I(a,t)=I(a)$.
2) Prove that $I'(a)=I(a)$.
3) Make $a\to0^+$, set $x=\exp(t)$, and deduce the value of $I(a)$.
4) Deduce for instance the value of $$\int_0^\infty \dfrac{x(x1)\cdots(xk+1)a^x}{\Gamma(x+1)}\,dx$$ I must confess that I forgot where I found this proof (perhaps in one of the Borwein's books ?)

$\begingroup$ The proof is due to Ramanujan (according to Hardy), can be found in Ramanujan: Twelve Lectures on Subjects Suggested by His Life and Work. $\endgroup$ – Ennar Jul 12 at 2:11