# Asymptotic Behaviour of $\Gamma^{(k)}(1)$

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?

• 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... – metamorphy Apr 23 '20 at 16:07

$$\Gamma(1+z)$$ 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 $$|z|, and the last series converges at $$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$$.)

• Kind of weird, that despite the weak growth of $(\log t)^k$ compared to $t^k$, the asymptotic behaviour is pretty much the same... – Diger Apr 22 '20 at 21:43
• @Diger: it is $t\to 0$ (not $t\to\infty$) that causes such a growth. – metamorphy Apr 22 '20 at 22:24
• 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 $(1,\infty)$. – Diger Apr 22 '20 at 22:41
• Needed to modify my question... – Diger Apr 22 '20 at 23:37

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 $$\left| \Gamma^{k}(1) - (-1)^k k! \right| = O\left(r^k k!\right)\ \text{for all } r \in (0,1/2)$$