Help in proving, that $\int_{0}^{\infty} \frac{1}{\Gamma(x)} d x=e+\int_{0}^{\infty} \frac{e^{-x}}{\pi^{2}+(\ln x)^{2}} d x$ using real methods only The above identity is the difference formula for the Fransén-Robinson Constant. Proving this statement gave me severe headaches those last days, since everytime I try to calculate the RHS I either miss the $+e$ term or nothing converges. In the picture you can find one of my attempts. As it seems as of now, the integral in the last line on the left does not converge. I'm completely at a loss. I would appreciate some help alot!!
 
 A: I found the following proof in the book Ramanujan by G. H. Hardy, which Hardy attributes to Ramanujan.
Define
$$f(x,y) = \int_{-y}^\infty\frac{x^t}{\Gamma(1+t)}\,dt + \int_0^\infty t^{y-1}\frac{e^{-xt}}{\pi^2+\log^2t}(\cos(\pi y)-\frac{\sin(\pi y)}{\pi}\log t)\,dt,\ x\geq 0,\, y\geq 0$$
and first differentiate it wrt to $y$ to get
\begin{align}
\frac{\partial f}{\partial y}(x,y) 
&= \frac{x^{-y}}{\Gamma(1-y)}+\int_0^\infty\frac{e^{-xt}}{\pi^2+\log^2t}\frac{\partial}{\partial y}\left[t^{y-1}(\cos(\pi y)-\frac{\sin(\pi y)}{\pi}\log t)\right]\,dt \\
&= \frac{x^{-y}}{\Gamma(1-y)}-\int_0^\infty\frac{e^{-xt}}{\pi^2+\log^2t}t^{y-1}(\pi^2+\log^2t)\frac{\sin(\pi y)}{\pi}\,dt \\
&= \frac{x^{-y}}{\Gamma(1-y)}-\frac{\sin(\pi y)}{\pi}\int_0^\infty e^{-xt}t^{y-1}\,dt \\
&= \frac{x^{-y}}{\Gamma(1-y)}-\frac{\sin(\pi y)}{\pi} x^{-y}\Gamma(y) = 0
\end{align}
where the last equality is due to Euler's reflection formula.
Thus, $f$ doesn't depend on $y$.
Now, differentiate $f$ wrt to $x$ to get
\begin{align}
\frac{\partial f}{\partial x}(x,y) 
&= \int_{-y}^\infty\frac{x^{t-1}}{\Gamma(t)}\,dt - \int_0^\infty t^{y}\frac{e^{-xt}}{\pi^2+\log^2t}(\cos(\pi y)-\frac{\sin(\pi y)}{\pi}\log t)\,dt \\
&= \int_{-(y+1)}^\infty\frac{x^{t}}{\Gamma(t+1)}\,dt - \int_0^\infty t^{(y+1)-1}\frac{e^{-xt}}{\pi^2+\log^2t}(-\cos(\pi (y+1))+\frac{\sin(\pi (y+1))}{\pi}\log t)\,dt \\
&= f(x,y+1) = f(x,y).
\end{align}
It follows that $f(x,y) = Ce^x$, and by plugging in $x = y = 0$, we get $C = 1$. Thus, letting $y = 0$ we get integral formula
$$\int_{0}^\infty\frac{x^t}{\Gamma(1+t)}\,dt = e^x - \int_0^\infty \frac{e^{-xt}}{t(\pi^2+\log^2t)}\, dt$$ which was known to Ramanujan and used (and generalized) by Hardy in some of his papers.
All you have to do now is differentiate the last integral formula wrt $x$ and let $x = 1$ to obtain the desired result.
