Prove Integral Evaluates to Fransén–Robinson constant

Question

How can I prove the following identity, $$\int_{1}^{\infty} \frac{1}{\pi^2x^2 + x^2\log^2\log(x)} \mathrm dx = F-e$$ where $F$ is the Fransén–Robinson constant.

I have verified the result numerically using Wolfram alpha. This integral was given as a representation for $F$ in the paper Fransén-Robinson Constant (Edgar Valdebenito), equation $(33)$. Little context/proof is given.

The Fransén–Robinson, $F \approx 2.8077702420285\cdots$ represents the area between the graph of the reciprocal Gamma function $\Gamma(x)^{-1}$, and the positive x axis. It is no coincidence that $F$ is somewhat close to $e$; by considering the series expansion of the integral-definition of $F$ gives (wikipedia)

$$F=\int _{0}^{\infty }{\frac {1}{\Gamma (x)}}\,dx\approx \sum _{n=1}^{\infty }{\frac {1}{\Gamma (n)}}=\sum _{n=0}^{\infty }{\frac {1}{n!}} = e$$

The integral somehow captures the difference between the two constants, $F-e$. I am unsure how to relate the original integral to the closed form above (performing substitution on $\log\log(x)$ leads to another tricky integral). Similar integrals are given in the same paper, yet how can they be shown to clearly evaluate to $F-e$?

@Tyma Gaidash, how can I transform this integral to the one you linked? — Maxime Jaccon, Jan 21 '25 at 00:21
Don't plagiarize from Wikipedia, of all places. Remove the first sentence from your second paragraph since you already linked to the Wikipedia article, or at the very least add "From Wikipedia, '...'" to your post. — teadawg1337, Jan 22 '25 at 01:27

score 2 · Accepted Answer · answered Jan 22 '25 at 02:48

Let $I$ be the integral given. Applying the substitution $x= e^t$ (provided by Tyma in the comments) gives the following integral

\begin{align*} I &= \int_{0}^{\infty} \frac{e^t}{\pi^2 e^{2t} + \log^2(t)e^{2t}}\mathrm dt = \int_{0}^{\infty} \frac{e^{-t}}{\pi^2 + \log^2(t)} \mathrm dt= \end{align*}

Which can be evaluated using the following identity (by Ramanujan). See the very nice answer given in this post, which gives a proof of the identity.

$$\boxed{\int_{0}^\infty\frac{x^t}{\Gamma(1+t)}\,dt = e^x - \int_0^\infty \frac{e^{-xt}}{t(\pi^2+\log^2t)}\, dt}$$

$$\int_{0}^\infty\frac{tx^{t-1}}{\Gamma(1+t)}\,dt = e^x + \int_0^\infty \frac{e^{-xt}}{(\pi^2+\log^2t)}\, dt$$

$$\int_{0}^\infty\frac{x^{t-1}}{\Gamma(t)}\,dt = e^x + \int_0^\infty \frac{e^{-xt}}{(\pi^2+\log^2t)}\, dt$$

Where I differentiated the equation with respect to $x$ and simplified using properties of $\Gamma(x)$. Substituting $x=1$ (and recalling the definition of the Fransén–Robinson constant), we see that

$$\int_{0}^\infty\frac{1}{\Gamma(t)}\,dt = e + \int_0^\infty \frac{e^{-t}}{(\pi^2+\log^2t)}\, dt$$

$$F = e + \int_0^\infty \frac{e^{-t}}{(\pi^2+\log^2t)}\, dt$$

$$F = e + I$$

as desired.

Prove Integral Evaluates to Fransén–Robinson constant

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