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I want to show that $$\int_{0}^{\sqrt{n}} \exp(-s^4)s^{n+1} ds - \int_{\sqrt{n}}^{\infty} \exp(-s^4)s^{n+1} ds \geq 0\, \quad \text{for all }n\in\mathbb{N}.$$

These integrals can in fact be written as upper and lower incomplete gamma functions: $$\int_{0}^{\sqrt{n}} \exp(-s^4)s^{n+1} ds = \frac{1}{4} \gamma\left(\frac{n+2}{4},n^2\right)$$ and $$\int_{\sqrt{n}}^{\infty} \exp(-s^4)s^{n+1} ds = \frac{1}{4} \Gamma\left(\frac{n+2}{4},n^2\right)$$

So essentially, I want to prove $$\gamma\left(\frac{n+2}{4},n^2\right)\geq\Gamma\left(\frac{n+2}{4},n^2\right)\quad \text{for all }n\in\mathbb{N},$$ but I could not find anything useful on this. Any suggestions?

Thanks already,
Nils

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  • $\begingroup$ This is equivalent to $\Gamma\left(\frac{n+2}{4}\right)\ge 2\Gamma\left(\frac{n+2}{4},n^2\right)$ $\endgroup$
    – Zima
    Commented Jun 13 at 10:11
  • $\begingroup$ dlmf.nist.gov/8.11 could be useful $\endgroup$ Commented Jun 13 at 10:59

2 Answers 2

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Let $n\ge 1$ be a real number. We split into two cases.

Case 1. $1 \le n < 2$

We have \begin{align*} \int_0^{\sqrt{n}} \exp(-s^4)s^{n+1} \,\mathrm{d}s &\ge \int_0^1 \exp(-s^4)s^{n+1} \,\mathrm{d}s\\ &\ge \int_0^1 \exp(-s^4)s^{3} \,\mathrm{d}s\\ &= \frac14 - \frac14 \mathrm{e}^{-1}. \end{align*}

Also, we have \begin{align*} \int_{\sqrt{n}}^\infty \exp(-s^4)s^{n+1} \,\mathrm{d}s &\le \int_1^\infty \exp(-s^4)s^{n+1} \,\mathrm{d}s\\ &\le \int_1^\infty \exp(-s^4)s^{3} \,\mathrm{d}s\\ &= \frac14 \mathrm{e}^{-1}. \end{align*}

Since $\frac14 \mathrm{e}^{-1} < \frac14 - \frac14 \mathrm{e}^{-1}$, the desired inequality is true.

Case 2. $n\ge 2$

We have \begin{align*} \int_{\sqrt{n}}^{\infty} \exp(-s^4)s^{n+1} \,\mathrm{d}s &= \int_{n^2}^\infty \frac{\mathrm{e}^{-x}}{4}x^{n/4 - 1/2}\, \mathrm{d} x\\ &= \int_{n^2}^\infty \frac{\mathrm{e}^{-x/2}}{4}\mathrm{e}^{-x/2}x^{n/4 - 1/2}\, \mathrm{d} x\\ &\le \int_{n^2}^\infty \frac{\mathrm{e}^{-x/2}}{4}\, \mathrm{d} x\\ &= \frac12\mathrm{e}^{-n^2/2}\\ &\le \frac12\mathrm{e}^{-2} \end{align*} where we use $\mathrm{e}^{-x/2}x^{n/4 - 1/2} \le 1$ for all $x \ge n^2$ (use derivative).

Also, we have \begin{align*} \int_0^{\sqrt{n}} \exp(-s^4)s^{n+1} \,\mathrm{d}s &= \int_0^{n^2} \frac{\mathrm{e}^{-x}}{4}x^{n/4 - 1/2}\, \mathrm{d} x\\ &\ge \int_1^{n^2} \frac{\mathrm{e}^{-x}}{4}x^{n/4 - 1/2}\, \mathrm{d} x\\ &\ge \int_1^{n^2} \frac{\mathrm{e}^{-x}}{4}\, \mathrm{d} x\\ &\ge \int_1^{4} \frac{\mathrm{e}^{-x}}{4}\, \mathrm{d} x\\ &= \frac14 \mathrm{e}^{-1} - \frac14 \mathrm{e}^{-4}. \end{align*}

Since $\frac12\mathrm{e}^{-2} < \frac14 \mathrm{e}^{-1} - \frac14 \mathrm{e}^{-4}$, the desired inequality is true.

We are done.

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  • $\begingroup$ Thank you very much! This works well. $\endgroup$
    – Nils
    Commented Jun 18 at 10:01
  • $\begingroup$ @Nils You are welcome. $\endgroup$
    – River Li
    Commented Jun 18 at 12:08
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Substitute $x=s^4$. Then $s=x^{1/4}, \text{d}s = \frac14x^{-3/4}\text{ d}x$. The integrand then takes the form $$ \exp(-s^4)s^{n+1}\text{ d}s=\exp(-x)x^{\frac{n+1}4-\frac34}\text{ d}x=x^{\frac{n+2}4-1}e^{-x}\text{ d}x $$ which is the integrand for $\Gamma\left(\frac{n+2}4\right)$. By adding $\int_0^{n^2}x^{\frac{n+2}4-1}e^{-x}\text{ d}x$ to both sides, the inequality becomes $$ \int_0^{\infty}x^{\frac{n-2}4}e^{-x}\text{ d}x\ge2\int_{n^2}^\infty x^{\frac{n-2}4}e^{-x}\text{ d}x $$ or $\Gamma\left(\frac{n+2}4\right)\ge2\Gamma\left(\frac{n+2}4,n^2\right)$ as @Zima pointed out.

If you test this inequality for a couple of $n$, you'll see that the right-hand side is often much much smaller than the left. Furthermore, $\Gamma\left(\frac{n+2}4, n^2\right)$ actually decreases with $n$ (extremely rapidly!) while $\Gamma\left(\frac{n+2}4\right)$ increases:

$n$ $\Gamma(\frac{n+2}4)$ $2\Gamma\left(\frac{n+2}4,n^2\right)$
1 1.22542 0.637266
2 1 0.0366313
3 0.906402 0.00043854
4 0.886227 9.27608*10^-7
5 0.919063 3.19771*10^-10
6 1 1.71645*10^-14
7 1.133 1.39425*10^-19
8 1.32934 1.68109*10^-25
9 1.60836 2.96752*10^-32
10 2 7.59044*10^-40

This means that the inequality is not tight at all, and we can afford to be very lax with our bounds.

Here's one approach: I'll try to simplify the integration by bounding the right-hand side with an exponential: $$ x^{\frac{n-2}4} \le e^{ax} $$ for where $a < 1$ is some constant that might change with $n$. Then, the right-hand side can be bounded as $$\begin{aligned} 2\int_{n^2}^\infty x^{\frac{n-2}4}e^{-x}\text{ d}x &\le 2\int_{n^2}^\infty e^{-(1-a)x}\text{ d}x = \frac{2e^{-(1-a)n^2}}{1-a} \end{aligned}$$ The defining inequality for $a$ can be rearranged as $$ax \ge \ln\left(x^{\frac{n-2}4}\right) = \frac{n-2}4\ln x$$ Thus, $a$ should satisfy $$1\ge a\ge \frac{n-2}4\frac{\ln x}x$$ for all $x\ge n^2$. Since $\frac{\ln x}x$ is decreasing for $x>e$, this is equivalent to requiring $$1\ge a\ge \frac{n-2}4\frac{\ln\left(n^2\right)}{n^2} = \frac{(n-2)\ln n}{2n^2}$$ At this stage, it's fairly clear that there are many choices for $a$, since the last expression doesn't even exceed $0.1$: it's global maximum for $n\ge1$ is $0.0997059$ near $n=6.3$. To be concrete, let's pick $a=1/2$. This clearly satisfies the required bound since $$\ln n \le n-1 \implies \ln n < n \implies \frac12 > \frac{n\ln n}{2n^2}>\frac{(n-2)\ln n}{2n^2}$$ Thus, $$\begin{aligned} 2\Gamma\left(\frac{n+2}4,n^2\right) = 2\int_{n^2}^\infty x^{\frac{n-2}4}e^{-x}\text{ d}x \le \frac{2e^{-(1-a)n^2}}{1-a} = 4e^{-n^2/2} \end{aligned}$$ is a really small number which decreases faster-than-exponentially! Since $\Gamma(\frac{n+2}4)$ increases for large $n$ and never dips below $\Gamma\left(\frac32\right)=\frac{\sqrt{\pi}}2\approx0.886$, we have $$\Gamma\left(\frac{n+2}4\right)\ge\Gamma\left(\frac32\right)=\frac{\sqrt{\pi}}2\approx0.886\ge4e^{-2}\approx0.541\ge4e^{-n^2/2}\ge2\Gamma\left(\frac{n+2}4,n^2\right)$$ for $n\ge2$. The inequality is a little too weak for the $n=1$ case, but you can check that case directly: $\Gamma\left(\frac34\right)\approx1.225>2\Gamma\left(\frac34,1\right)\approx0.637$.

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