2
$\begingroup$

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.

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ It's about $n/(\log n)^2$. The integral is dominated by the region near $1$, where you can use $-1/\Gamma(x-1)=(1-x) + O((1-x)^2)$ ... $\endgroup$
    – Lucia
    Jul 9, 2015 at 15:46
  • $\begingroup$ @Lucia You most likely wanted to write $1/(\log n)^{2}$. It seems reasonable for the leading term. $\endgroup$
    – Twi
    Jul 9, 2015 at 20:05
  • $\begingroup$ Yes, that's right -- $1/(\log n)^2$. $\endgroup$
    – Lucia
    Jul 10, 2015 at 3:37
  • 1
    $\begingroup$ 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. $\endgroup$
    – Brendan McKay
    Jul 10, 2015 at 4:42
  • $\begingroup$ @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. $\endgroup$
    – Twi
    Jul 10, 2015 at 7:22

1 Answer 1

Reset to default
1
$\begingroup$

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 $a > b.$

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.