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$\frac{\int_{0}^{\infty}x^{m+1}e^{x}dx}{\int_{0}^{\infty}x^{m}e^{x}dx}, m\in \mathbb{N}.$

I have checked numerically the value of this ratio of integrals considering some known value for $m$ and the taking some finite value (same for numerator and denomerator) in place of $\infty$ in upper limit of the integration. I noticed that the value of the ratio increases as the upper limit of the integration increases. But, I can't explain, what will be the value of the ration when the upper limit of the integration will be $\infty.$

mojammel
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  • https://math.stackexchange.com/q/46469/1012971 ..Maybe this helps. –  Jan 17 '22 at 12:20
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    None of the integrals, as currently written, converge. Maybe the integration is over a fine interval, or or the weight $e^x$ is meant to be $e^{-x}$. – Mittens Jan 17 '22 at 16:01

1 Answers1

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If $p$ is very large

$$I_{m,p}=\frac{\int_{0}^{p}x^{m+1}e^{x}dx}{\int_{0}^{p}x^{m}e^{x}dx}=-\frac{\Gamma (m+2)-\Gamma (m+2,-p)}{\Gamma(m+1)-\Gamma (m+1,-p)}\sim-\frac{\Gamma (m+2,-p)}{\Gamma (m+1,-p)}$$

Composing series

$$I_{m,p} \sim p-1+\frac{m}{p}-\frac{m(m-2)}{p^2}+\frac{m(m^2-6 m+6 )}{p^3}+O\left(\frac{1}{p^4}\right)$$

Trying for $m=10$ and $p=10^3$, the above truncated expansion gives $$\frac{49950496023}{50000000}=\color{red}{999.0099204}600$$ while the exact value is $\color{red} {999.0099204588}$.

Edit

This could be generalized for $$J_{m,n}=\frac{\int_{0}^{p}x^{m}e^{x}dx}{\int_{0}^{p}x^{n}e^{x}dx}\sim (-1)^{n-m}\,\,\frac{ \Gamma (m+1,-p)}{\Gamma (n+1,-p)}$$ $$J_{m,n}\sim p^{m-n} \Bigg[1+\frac{n-m}{p}+\frac{(m-1) (m-n)}{p^2}+O\left(\frac{1}{p^3}\right)\Bigg]$$

Claude Leibovici
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