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Pretty simple question to which I haven't found much in the literature. Are there proper results for the asymptotic behaviour of the derivatives of the Gamma-function? That is for $$\Gamma^{(k)}(1) = \int_0^\infty (\log t)^k e^{-t} \, {\rm d}t$$ as $k$ gets large?


The range of integration over the interval $(0,1)$ is responsible, for the strong factorial growth which I oversaw. Hence, I'd be somewhat more interested in the asymptotics of $$\int_1^\infty (\log t)^k e^{-t} \, {\rm d}t \, .$$ The integrand has a maximum at approximately $t\approx \frac{k}{\log k}$, so it will blow up somehow. Laplace's method is what comes to my mind, which gives $$\int_1^\infty (\log t)^k e^{-t} \, {\rm d}t \sim \sqrt{\frac{2\pi k}{W(k)+1}} \, W(k)^k \, e^{-\frac{k}{W(k)}}$$ where $W(k)$ is the principal branch of LambertW.

The remainder for the asymptotic expansion of the starting expression (given by metamorphy) is then just $$\frac{\int_1^\infty (\log t)^k e^{-t} \, {\rm d}t}{k!} \sim \frac{1}{\sqrt{W(k)+1}} \left( \frac{e \, W(k)}{k \, e^{\frac{1}{W(k)}}} \right)^k \, ,$$ vanishing faster than any power.

Is it possible to obtain an asymptotic expansion, instead of just the asymptotics?

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  • $\begingroup$ For the new part of the question - yes, I think Laplace's mtehod is the way to go, too. And for higher-order asymptotics, I see nothing more than just to continue the computations of the method... $\endgroup$ – metamorphy Apr 23 '20 at 16:07
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$\Gamma(1+z)$ has simple poles at $z=-n$ with the residue $(-1)^{n-1}/(n-1)!$ where $n$ takes positive integer values. Thus, for any positive integer $m$, the function $$\Gamma(1+z)+\sum_{n=1}^{m}\frac{(-1)^n}{(n-1)!}\frac{1}{n+z}=\sum_{k=0}^{\infty}\left(\frac{\Gamma^{(k)}(1)}{k!}+(-1)^k\sum_{n=1}^{m}\frac{(-1)^n}{n^k\cdot n!}\right)z^k$$ is regular in $|z|<m+1$, and the last series converges at $z=m$ (at least). This gives the asymptotics $$\Gamma^{(k)}(1)\asymp(-1)^k k!\sum_{n=1}^{(\infty)}\frac{(-1)^{n-1}}{n^k\cdot n!},\qquad k\to\infty.$$ Despite the convergence, this is only an asymptotic equality; there is a remainder that is not captured. (It is coming from $\int_1^\infty t^z e^{-t}\,dt$ which is an entire function of $z$.)

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  • $\begingroup$ Kind of weird, that despite the weak growth of $(\log t)^k$ compared to $t^k$, the asymptotic behaviour is pretty much the same... $\endgroup$ – Diger Apr 22 '20 at 21:43
  • $\begingroup$ @Diger: it is $t\to 0$ (not $t\to\infty$) that causes such a growth. $\endgroup$ – metamorphy Apr 22 '20 at 22:24
  • $\begingroup$ Sorry, you are right...That's precisely the definition of the Gamma function $$\int_0^1 (-\log t)^k \, {\rm d}t$$ upon change of variables. So the asymptotics pretty much follow by splitting the integral in two parts: $(0,1)$ and $(1,\infty)$. $\endgroup$ – Diger Apr 22 '20 at 22:41
  • $\begingroup$ Needed to modify my question... $\endgroup$ – Diger Apr 22 '20 at 23:37
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Note that $$\sum_{k=0}^\infty \Gamma^{(k)}(1) \frac{z^k}{k!} = \Gamma(1+z)$$ The closest singularity of $\Gamma(\zeta)$ to $\zeta=1$ is at $\zeta=0$, where $\Gamma$ has a simple pole with residue $1$. The next closest singularity is at $\zeta=-1$. Thus $$\Gamma(1+z) - \frac{1}{1+z} = \sum_{k=0}^\infty \left(\frac{\Gamma^{(k)}(1)}{k!} - (-1)^{k}\right) z^k $$ has radius of convergence $2$. Thus $\Gamma^{(k)}(1) \sim (-1)^k k!$, with $$ \left| \Gamma^{k}(1) - (-1)^k k! \right| = O\left(r^k k!\right)\ \text{for all } r \in (0,1/2)$$

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