Prove $\int_{0}^\infty \frac{1}{\Gamma(x)}\, \mathrm{d}x = e + \int_0^\infty \frac{e^{-x}}{\pi^2 + \ln^2 x}\, \mathrm{d}x$

I came across this nice identity: $$\int_{0}^\infty \frac{1}{\Gamma(x)}\, \mathrm{d}x = e + \int_0^\infty \frac{e^{-x}}{\pi^2 + \ln^2 x}\, \mathrm{d}x$$

Is there an elementary proof?

• $\pi,\Gamma,e,\ln,\infty,0$ together. I am not sure if this is elegant or an abomination. – Guy Apr 5 '14 at 16:31
• This is the Fransen-Robinson constant, the continuous equivalent of e, and whose approximate value is about $2.708880^+$. – Lucian Apr 5 '14 at 16:36
• IMHO, in a whole year I joined MSE, this is probably the most beautiful identity I have seen. – achille hui Apr 5 '14 at 16:37
• @Lucian (+1 for the link), what do you mean "continuous equivalent" of $e$ , are you referring to $$e=\sum_{n=0}^\infty \frac{1}{n!}?$$ – Guy Apr 5 '14 at 16:45
• @yzhao Why does that surprise me? – user85798 Apr 5 '14 at 17:24

Hardy recorded in his 1937 papers that the formula was discovered by Ramanujan, who did not profess to prove such an identity. Hardy published his proof, which was based on the "Plana's formula"(A proof of Plana's formula can be found here). Here is an outline of Hardy's original proof. I doubt whether an "elementary" proof exists(without complex analysis).

The Plana's formula asserts that

$$\sum_{n=0}^{\infty}f(n)-\int_{0}^{\infty}f(x)dx=\frac{1}{2}f(0)+i\int_{0}^{\infty}\frac{f(it)-f(-it)}{e^{2\pi t}-1}dt$$

Let $f(u)=1/\Gamma(u)$, then $$\sum_{n=0}^{\infty}\frac{1}{n!}-\int_{0}^{\infty}\frac{1}{\Gamma(x)}dx=i\int_{0}^{\infty}\frac{1/\Gamma(it)-1/\Gamma(-it)}{e^{2\pi t}-1}dt$$

An elementary transform of RHS gives

$$i\int_{0}^{\infty}\frac{1/\Gamma(it)-1/\Gamma(-it)}{e^{2\pi t}-1}dt=-\frac{1}{2}\int_{-\infty}^{\infty}\frac{1/\Gamma(it)}{\sin \pi i t}e^{-\pi\vert t\vert}dt$$

Also $$e^{-\pi\vert t\vert}=\int_{0}^{\infty}\frac{u^{it}}{u(\pi^2+\log^2u)}du$$

Then RHS is equal to $$\int_{-\infty}^{\infty}\int_{0}^{\infty}\frac{u^{it}}{u(\pi^2+\log^2u)}\frac{-1/\Gamma(it)}{2\sin \pi i t}dudt$$

Mellin transform gives $$-\frac{1}{2\pi i}\int_{C}\frac{\pi u^{z-1}}{\Gamma(z)\sin \pi z}dz=-e^{-u}$$

where C is the imaginary axis.

And we are done.