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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?

user85798
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    $\pi,\Gamma,e,\ln,\infty,0$ together. I am not sure if this is elegant or an abomination. – Guy Apr 05 '14 at 16:31
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    This is the Fransen-Robinson constant, the continuous equivalent of e, and whose approximate value is about $2.708880^+$. – Lucian Apr 05 '14 at 16:36
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    IMHO, in a whole year I joined MSE, this is probably the most beautiful identity I have seen. – achille hui Apr 05 '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 05 '14 at 16:45
  • @SanathDevalapurkar First, in your $u$-sub, it should be $$\int_{-1}^\infty \frac1{u},{\rm d}u$$ and second, to integrate you need a continuous function, so back to $\Gamma$. – Guy Apr 05 '14 at 16:48
  • As far as I know(from the book "Mathmatical Constant" of S.R.Finch, p263-p264), this formula was discovered by Ramanujan. Hardy proved a generalized version of this formula in his paper "Another formula of Ramanujan", J.London Math.Soc. (1937) s1-12 (4): 314-318. – zy_ Apr 05 '14 at 17:16
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    @yzhao Why does that surprise me? – user85798 Apr 05 '14 at 17:24

1 Answers1

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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.

zy_
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