6
$\begingroup$

How to prove the following formula, providing a generalization of the Fransén-Robinson constant?

$$\boxed{ \int_0^\infty \frac{t^k}{\Gamma(t)} dt = \sum_{n=1}^\infty \frac{n^{k+1}}{n!} + (2k)!!\sum_{r=0}^k \frac{(-1)^r}{2^{2r}} \binom{k-r}{r} \int_0^\infty \frac{e^{-x}\log^{k-2r}(x)}{[\pi^2+\log^2(x)]^{k-r+1}}dx =: F_k, }$$ where $k\in\mathbb Z_{\geq 0}$. Here's a run-through for the first few values of $k$ (the sums $S_k$ have been calculated using Dobiński's formula): setting $\varphi_k:=F_k-S_k$, $$ \begin{array} FF_0 = &e+\displaystyle\int_0^\infty \frac{e^{-x}}{\pi^2+\log^2(x)}dx = 2.8077702420285\dots \equiv F&\to \quad \varphi_0\approx 0.089489,\\ F_1 = &2e+2\displaystyle\int_0^\infty \frac{e^{-x}\log(x)}{[\pi^2+\log^2(x)]^2}dx = 5.43181977215\dots &\to \quad \varphi_1 \approx -0.004744, \\ F_2 = &5e +2 \displaystyle\int_0^\infty \frac{e^{-x}(\log^2(x)-\pi^2)}{[\pi^2+\log^2(x)]^3}dx= 13.5797413956\dots &\to \quad \varphi_2 \approx -0.011668, \\ F_3 = &15e + 24 \displaystyle\int_0^\infty \frac{e^{-x}(\log^3(x)-\pi^2 \log(x))}{[\pi^2+\log^2(x)]^4}dx = 40.7762149851\dots &\to \quad \varphi_3\approx 0.001988,\ \dots \end{array} $$

One could also add $$ F_{-1} = (e-1)+\frac 1 2 -\frac 1 \pi \int_0^\infty e^{-x} \arctan\left(\frac{\log(x)}\pi \right) dx = 2.2665345077\dots \quad \to \quad \varphi_{-1} \approx 0.548253. $$


I have arrived at the formula above by observing that the integrand in $\varphi_0$ is such that $$\frac{e^{-x}}{\pi^2 +\log^2(x)} = \frac {e^{-x}} \pi \mathcal L\{\sin(\pi t)\}(\log x),$$ where $\mathcal L$ is the Laplace transform, and by guessing that the integrand for $k>0$ should similarly involve the Laplace transform of $t^k \sin(\pi t)$. Same for $k=-1$.

My conjecture checks out numerically, but I'd love to know what a rigorous proof of this would look like!

$\endgroup$

1 Answer 1

3
$\begingroup$

Your argument can be justified and developed. Another one (which is easier for me) is Hankel's formula $$\frac{1}{\Gamma(s)}=\frac{1}{2\pi i}\int_\lambda z^{-s}e^z\,dz$$ (valid for all $s\in\mathbb{C}$), where the contour $\lambda$ encircles the negative real axis (here and below, the principal value of any complex exponentiation is taken). Now, for any $k\in\mathbb{C}$ with $\Re k>-1$, we can write $$\int_0^\infty\frac{t^k\,dt}{\Gamma(t)}=\frac{1}{2\pi i}\int_0^\infty t^k\int_\lambda z^{-t}e^z\,dz\,dt$$ and, if we deform $\lambda$ so that it encircles the disc $|z|\leqslant 1$ (and not only the negative real axis), we get an absolutely convergent double integral, so that the integrations may be interchanged, and we obtain $$\int_0^\infty\frac{t^k\,dt}{\Gamma(t)}=\frac{\Gamma(k+1)}{2\pi i}\int_\lambda\frac{e^z\,dz}{(\log z)^{k+1}}$$ (a side note: the integral on the LHS converges for $\Re k>-2$, so the formula above holds analytically continued, if additionally $k\neq-1$ of course). One may continue this evaluation even for these "general" $k$, by "squeezing" $\lambda$ close to the real axis.

But let's get back to the case of integer $k\geqslant 0$ (see the end for $k=-1$). Then we have $$\int_0^\infty\frac{t^k\,dt}{\Gamma(t)}=k!\ \underbrace{\operatorname*{Res}_{z=1}\frac{e^z}{(\log z)^{k+1}}}_{=A_k}+\frac{k!}{2\pi i}\underbrace{\int_{\lambda'}\frac{e^z}{(\log z)^{k+1}}}_{=J_k},$$ where $\lambda'$ encircles the negative real axis closely. $A_k$ can be computed using the generating function: $$\sum_{k=0}^\infty A_k t^k=\frac{1}{2\pi i}\sum_{k=0}^\infty\int_C\frac{e^z}{\log z}\left(\frac{t}{\log z}\right)^k dz=\frac{1}{2\pi i}\int_C\frac{e^z\,dz}{\log z-t}=\operatorname*{Res}_{z=e^t}\frac{e^z}{\log z-t}=e^{e^t+t}$$ (where the contour $C$ is chosen with $|t/\log z|<1$ uniformly on it).

Hence $k! A_k=eB_{k+1}$ express in terms of the Bell numbers. As for $J_k$, we take the limit of "closely": $$J_k=\int_0^\infty\big((\log x-\pi i)^{-k-1}-(\log x+\pi i)^{-k-1}\big)e^{-x}\,dx,$$ which can be "simplified" using the binomial formula, and giving the result (for $k\geqslant 0$).

For $k=-1$, one may (better) write the integral as $\int_0^\infty\frac{dt}{\Gamma(1+t)}$ and repeat all the steps.

$\endgroup$
6
  • $\begingroup$ Thank you! Is there a way to see this without resorting to complex methods? $\endgroup$
    – giobrach
    Sep 1, 2020 at 10:26
  • $\begingroup$ Ok, in my head I'd categorized the Laplace transform as a complex method, at least in spirit. I was thinking more of differentiation under the integral sign, etc. I tried making my Laplace transform idea-argument more rigorous but in the end, I didn't quite manage to connect the two sides of the formula. Is it really that hard to make it work? (If needed, I can update my post with the relevant steps and doubts) $\endgroup$
    – giobrach
    Sep 1, 2020 at 12:54
  • 1
    $\begingroup$ @giobrach: I'm sorry but it looks like I don't actually see a generalization. The equality $$\Gamma(1-t)=\sum_{n=0}^\infty\frac{(-1)^n}{n!}\frac{2t}{(n+1)^2-t^2}+\int_0^1 x^t e^{-x}\,dx+\int_1^\infty x^{-t}e^{-x}\,dx$$ is a straightforward way to handle $F_0$ (multiply by $(\sin\pi t)/\pi$ and integrate), but not $F_k$. You may want to try it yourself. $\endgroup$
    – metamorphy
    Sep 3, 2020 at 13:36
  • $\begingroup$ (A line of attack seems visible, but it quickly becomes boring, in comparison to the above.) $\endgroup$
    – metamorphy
    Sep 3, 2020 at 13:44
  • $\begingroup$ That formula is really impressive and might just be what I need! Where/how did you find it? (I'm working on a solution, which I will post as a new answer) $\endgroup$
    – giobrach
    Sep 3, 2020 at 14:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .