# Asymptotic expansion of a sequence given by an integral with reciprocal Gamma function

I would like to know the asymptotic expansion of the sequence of positive numbers given by $$I_{n}:=-\int_{0}^{1}\frac{n^{x-1}}{\Gamma(x-1)}dx,$$ for $n\rightarrow\infty$.

One can easily derive an upper estimation since $$I_{n}\leq\max_{-1<x<0}\frac{1}{|\Gamma(x)|}\int_{0}^{1}n^{x-1}dx\approx0.282\,\frac{n-1}{n\ln n}\leq0.282\,\frac{1}{\ln n}.$$ However, by some numerical experiments, it seems that the sequence $I_{n}$ decreases to zero slightly faster than $1/\ln n$. So far, however, I do not know how to obtain at least the leading term of the asymptotic expansion. Any idea? Many thanks.

• It's about $n/(\log n)^2$. The integral is dominated by the region near $1$, where you can use $-1/\Gamma(x-1)=(1-x) + O((1-x)^2)$ ... – Lucia Jul 9 '15 at 15:46
• @Lucia You most likely wanted to write $1/(\log n)^{2}$. It seems reasonable for the leading term. – Twi Jul 9 '15 at 20:05
• Yes, that's right -- $1/(\log n)^2$. – Lucia Jul 10 '15 at 3:37
• By Lucia's method, I get $1/(\ln n)^2 - 2\gamma/(\ln n)^3 + O(1/(\ln n)^4)$, where $\gamma$ is Euler's constant. I didn't prove it rigorously, but I'm sure it wouldn't be hard. In general, the coefficients of the negative powers of $\ln n$ are a mad mix of $\gamma$, $\pi$ and $\zeta$ function values. – Brendan McKay Jul 10 '15 at 4:42
• @Brendan McKay I arrived at the same result. It can be proved rigorously, indeed (though it's a bit tedious). You are right, the general expansion seems complicated due to the complicated form of coefficients in the power series expansion of $1/\Gamma(x)$. Nevertheless, this is sufficient to me. Thanks. – Twi Jul 10 '15 at 7:22

## 1 Answer

I am not sure about the asymptotic expansion, but approximating $1/\Gamma(x-1)$ by $x$ near $0$ (and $1/(1-x)$ near $1$) does show that the decrease is faster than $1/\log n$ - the integral of $x n^{x-1}$ is completely explicit, and equals $$\frac{n^{x-1} (x \log (n)-1)}{\log ^2(n)},$$ while $$\int_a^b n^{x-1} = \frac{n^b-n^a}{n \log (n)} \sim - n^{a-1}/\log n,$$ for $a > b.$