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I've been trying to prove this inequality for a while.

$$\int_{0}^{\infty}{e^{-2(m-2)s}(1+s)^m ds}\leq\frac{1}{m-3}\text{ for }m>3.$$ It can be rewritten in terms of the incomplete gamma function $$\Gamma(m+1,2(m-2))\leq\frac{(2(m-2))^{m+1}e^{-2(m-2)}}{m-3}$$ or the generalized exponential integral $$E_{-m}(2(m-2))\leq\frac{e^{-2(m-2)}}{m-3}.$$

For any version, integrating by parts gives a recurring expression, but it quickly becomes intractable.

The inequality can be directly checked for low values of $m$. It can also be proved that $(m-3)\int_{0}^{\infty}{e^{-2(m-2)s}(1+s)^m ds}$ converges to one as $m$ grows to infinity. But I have not been able to prove that the latter expression is always increasing in $m$.

Using the fact that $1+s\leq e^s$ holds for every $s\geq 0$, we can see that the left hand side is smaller than $\frac{1}{m-4}$, but I am interested in the tighter bound.

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  • $\begingroup$ Is $m$ actually an integer? $\endgroup$ – Diger Aug 19 '18 at 0:14
  • $\begingroup$ You may assume that if that helps. Then the inequality can be written as $m!\sum_{k=0}^{m}{\frac{(2(m-1))^{k}}{k!}}\leq\frac{(2(m-1))^{m+1}}{m-3}$. $\endgroup$ – userq3125 Aug 19 '18 at 1:08
  • $\begingroup$ So you still need the proof? $\endgroup$ – Diger Aug 21 '18 at 5:48
  • $\begingroup$ Yes, I do. Any ideas? $\endgroup$ – userq3125 Aug 21 '18 at 6:52
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On $(3,\infty)$ we have \begin{align} 0 \leq &(m-3)\int_0^\infty e^{-2(m-2)s} (1+s)^m \, {\rm d}s \\ \stackrel{s=e^{t}-1}{=} &(m-3)\int_{0}^\infty e^{-2(m-2)\left(e^{t}-1\right)+(m+1)t} \, {\rm d}t \\ \leq &(m-3) \int_{0}^\infty e^{-2(m-2)\left(t+\frac{t^2}{2}+\frac{t^3}{6}\right)+(m+1)t} \, {\rm d}t \\ \stackrel{u=(t+1)^3}{=} &(m-3) \, \underbrace{e^{-3}\int_1^\infty \frac{e^{-\frac{m-2}{3}(u-1)+3u^{{1}/{3}}}}{3u^{{2}/{3}}} \, {\rm d}u}_{\equiv I_m} \, . \end{align} Now \begin{align} \lim_{m\rightarrow \infty} (m-3)I_m &= \lim_{m\rightarrow \infty} \frac{m-3}{m-2} \, e^{-3} \sum_{k=0}^\infty \frac{3^k}{k!} \left(\frac{m-2}{3}\right)^{-\frac{k+1}{3}+1} e^{\frac{m-2}{3}} \, \Gamma\left(\frac{k+1}{3} , \frac{m-2}{3}\right) \\ &=1 \end{align} as can be seen by expanding $e^{3u^{1/3}}$.

It therefore remains to show that $(m-3)I_m$ is an increasing function. For that the idea is to remove the $m$-dependent pre-factor before taking the derivative to keep it tractable. In fact we have the identity $$ (m-2)I_m = 1 + e^{-3} \int_1^\infty \left\{ u^{-4/3} - \frac{2 u^{-5/3}}{3} \right\} \, e^{-\frac{m-2}{3}(u-1)+3u^{{1}/{3}}} \, {\rm d}u $$ by partial integration. Hence \begin{align} \frac{\rm d}{{\rm d}m} (m-3)I_m &= \frac{\rm d}{{\rm d}m} (m-2)I_m - \frac{\rm d}{{\rm d}m} I_m \\ &=e^{-3} \int_1^\infty \underbrace{\left\{ \frac{u^{1/3}}{9} - \frac{u^{-1/3}}{3} + \frac{u^{-2/3}}{9} + \frac{u^{-4/3}}{3} - \frac{2 u^{-5/3}}{9} \right\}}_{\equiv f(u)} \, e^{-\frac{m-2}{3}(u-1)+3u^{{1}/{3}}} \, {\rm d}u \\ &\geq 0 \end{align} because $f(u)\geq 0$ which can be seen as follows:

First $f(1)=0$ and $$f'(u) = \frac{u^{-2/3}}{27} + \frac{u^{-4/3}}{9} - \frac{2u^{-5/3}}{27} - \frac{4u^{-7/3}}{9} + \frac{10u^{-8/3}}{27} \geq 0 \, .$$

The latter because $f'(1)=0$ and $f'(\infty)=0$ while $$ f''(u)=-\frac{2u^{-5/3}}{81} - \frac{4u^{-7/3}}{27} + \frac{10u^{-8/3}}{81} + \frac{28u^{-10/3}}{27} - \frac{80u^{-11/3}}{81} = 0 $$ has only $1$ real-valued solution $u_0=z^3>1$ where $z$ is the only real solution of $$ z^5 + z^4 + 7z^3 + 2z^2 + 2z - 40 = 0 $$ which corresponds to a maximum of $f'(u)$, since \begin{align} f'''(u_0)&=\frac{10u_0^{-8/3}}{243} + \frac{28u_0^{-10/3}}{81} - \frac{80u_0^{-11/3}}{243} - \frac{280u_0^{-13/3}}{81} + \frac{880u_0^{-14/3}}{243} \\ &\approx -0.002446492623 < 0 \end{align} with $u_0 \approx 3.101517308$.

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  • $\begingroup$ Really like the flow of the proof. Could you add details on how you found $\lim_{m->\infty}(m-3)I_m=1$? Thanks. $\endgroup$ – skbmoore Aug 21 '18 at 19:34
  • $\begingroup$ Do you mean how I arrived at the series expansion, or why the limit using the series is $1$? $\endgroup$ – Diger Aug 21 '18 at 20:04
  • $\begingroup$ Thanks a lot, that was extremely helpful! Perhaps the limiting result can be more easily seen using the expression that you found for $(m-2)I_m$. $\endgroup$ – userq3125 Aug 22 '18 at 8:00
  • $\begingroup$ Ah indeed ;) That's much simpler. $\endgroup$ – Diger Aug 22 '18 at 8:09
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    $\begingroup$ It also seems to be the best bound, since any value smaller $3$ violates the inequality for large enough $m$. In fact it seems that $$(m-2) \int_{0}^{\infty}{e^{-2(m-2)s}(1+s)^m \, {\rm d}s} \geq 1$$ for any $m>2$. $\endgroup$ – Diger Aug 22 '18 at 9:08
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$$\int_{0}^{\infty}{e^{-2(m-2)s}(1+s)^m ds}\leq\frac{1}{m-3}\text{ for }m>3$$

put $1+s=t$ thus $ds=dt$

also as $s$ changes from 0 to $\infty$ thus $t$ changes from 1 to $\infty$.

$$I=\int_{1}^{\infty}{e^{-2(m-2)(t-1)}t^m dt}=e^{2(m-2)}\int_{1}^{\infty}{e^{-2(m-2)t}t^m dt}$$

put $2(m-2)t=x$ thus $2(m-2)dt=dx$

$x$ changes from $2(m-2)$ to $\infty$

$$I=\frac{e^{2(m-2)}}{2^{m+1}(m-2)^{m+1}}\int_{2(m-2)}^{\infty} e^{-x}x^m dx$$ $$\Gamma(m+1)=\int_{0}^{\infty} e^{-t} t^m dt$$

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  • $\begingroup$ As $s$ changes from $0$ to infinity, $t$ changes from $1$ to infinity. $\endgroup$ – Ahmad Bazzi Aug 18 '18 at 1:46
  • $\begingroup$ yes it was a mistake I corrected it $\endgroup$ – Deepesh Meena Aug 18 '18 at 1:48
  • $\begingroup$ I added the version of the inequality for the incomplete gamma function that can be obtained using your derivation. $\endgroup$ – userq3125 Aug 18 '18 at 5:46

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