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I ask for an inequality which is a follow up of this question: Prove that $\int_0^\infty\frac1{x^x}\, dx<2$ :

$$\int_{0}^{\infty}x^{-x}dx<\pi-\ln\pi$$

You can find a nice proof @RiverLi among others and also an attempt of mine.

My try

We have:

$$\int_{1}^{\phi}\phi^{\left(a-ax^{2}\right)}dx+\int_{0}^{1}x^{-x}dx+\int_{b}^{\infty}x^{-x}dx>\int_{0}^{\infty}x^{-x}dx$$

Where $\phi=\frac{1+\sqrt{5}}{2},a=1$ because we have the inequality for $x\in[1,b]$:

$$b^{a-ax^{2}}\geq x^{-x}$$

Unfortunetaly it's already bigger...

We also have:

$$\int_{1}^{\frac{1}{2}+\sqrt{\frac{5}{4}}}\left(\frac{1+\sqrt{5}}{2}\right)^{1-x^{2}}dx+\int_{0}^{1}x^{-x}dx+\int_{\frac{1}{2}+\sqrt{\frac{5}{4}}}^{1.8}f\left(x+\frac{1}{9}\right)dx+\int_{1.8}^{\infty}x^{-x}dx<2$$

Where:

$$f\left(x\right)=2^{\frac{1+\sqrt{5}}{2}}\left(x^{\frac{1}{3}}\right)e^{-x^{\frac{4}{3}}}$$

$$f(x+1/9)>x^{-x},x\in[\phi,\infty)$$

Until now all the derivatives are easy and there is one trick .

How to show it analytically?

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  • $\begingroup$ can you show your efforts? $\endgroup$
    – TShiong
    Dec 28, 2022 at 13:10
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    $\begingroup$ Please do not include "thanks" and other chit chat in posts. $\endgroup$ Dec 28, 2022 at 13:11
  • $\begingroup$ @ShivaVenkata See the link there is a try of mine . Do you read carefully ? $\endgroup$ Dec 28, 2022 at 13:11
  • $\begingroup$ I think you should show your own thoughts to get an answer or find if the existing asnwers help you $\endgroup$
    – TShiong
    Dec 28, 2022 at 13:12
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    $\begingroup$ Is there some significant difference between the linked question and this one that makes it not a duplicate? $\endgroup$ Dec 28, 2022 at 13:14

2 Answers 2

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Firstly, $$I_1=\int\limits_0^e x^{-x}\,\text dx = e\int\limits_0^1 (ex)^{-ex}\,\text dx = e\int\limits_0^1 e^{-ex(1+\ln x)}\,\text dx = e\int\limits_0^1\,\sum_{k=0}^\infty \dfrac{(-ex(1+\ln x))^k}{k!}\,\text dx$$ $$= \sum_{k=0}^\infty \dfrac{(-1)^k e^{k +1}}{k!} \int\limits_0^1\, x^k(1+\ln x)^k\,\text dx = e+\sum_{k=1}^\infty \dfrac{(-1)^k e^{k+1}}{k!} \sum_{j=0}^k\dbinom kj \int\limits_0^1\, x^k(\ln x)^j\,\text dx$$ $$= e+\sum_{k=1}^\infty \dfrac{(-1)^k e^{k+1}}{k!} \sum_{j=0}^k\dbinom kj \left(\dfrac2{k+1}\right)^{j+1} \int\limits_0^1\, \dfrac{k+1}2\,x^k \ln^j\left(x^{\large\frac{k+1}2}\right)\,\text dx$$ $$= e + \sum_{k=1}^\infty \dfrac{(-1)^k e^{k+1}}{k!} \sum_{j=0}^k\dbinom kj \left(\dfrac2{k+1}\right)^{j+1} \int\limits_0^1\, y\ln^j\left(y\right)\,\text dy$$ $$=e+\sum_{k=1}^\infty \dfrac{(-1)^k e^{k+1}}{k!} \sum_{j=0}^k\dbinom kj \left(\dfrac2{k+1}\right)^{j+1} (-1)^j \dfrac{j!}{2^{j+1}}$$ $$=e+\sum_{k=1}^\infty (-1)^k e^{k+1} \sum_{j=0}^k\dfrac{(-1)^j}{ (k-j)!(k+1)^{j+1}}$$ $$=e + \sum_{k=1}^\infty (-1)^k e^{k+1} e^{-k-1}\dfrac{(-1)^k}{(k+1)^{k+1} k!}\Gamma(k+1,-k-1),$$ $$I_1=e + \sum_{k=2}^\infty \dfrac{\Gamma(k,-k)}{k^k \Gamma(k)}$$ received an alternating series, wherein $$e + \sum_{k=2}^{22}\dfrac{\Gamma(k,-k)}{k^k \Gamma(k)} \le 1.96466\,14544\,475,\quad \dfrac{\Gamma(23,-23)}{\Gamma(23)}{23^{-23}}\le 2\cdot 10^{-13}.$$ Therefore, $$I_1\ge 1.96466\,14544\,477.$$

$\color{green}{\textbf{Edit of 21.02.23}}$

Obtained estimation can be significantly improved.

Let $x\in\big(a,(1+\delta a\big),\;\delta\ll1,\;$ then $$x=a(1+\delta z)=a +\delta a z,\qquad \big(z\in(0,1),\;a\in(1,\infty)\big),$$ $$x^{-x}=a^{\large -a-\delta a z}\;\left(1+\delta z \right) ^{\large -a(1+\delta z)},$$

Taking in account the inequality $$(1+t)^{\large-a(1+t)} -e^{-at} \le \dfrac a2 t^2\left(1-\dfrac{1+3a}3 t\right)\tag1$$ and identity $$\int\limits_0^d t^k a^{-a t}\,\text dt =\dfrac{Γ(k+1)-Γ(k+1,ad\ln a)}{(a\ln a)^{k+1}},\tag2$$ one can get

$$I(a, \delta)=\int\limits_a^{a+\delta a} x^{-x}\,\text dx = \delta a\cdot a^{-a} \int\limits_0^1 a^{-\delta az}\left(1+\delta z\right)^{\large- a(1+\delta z)}\,\text dz$$ $$= a^{2-a}\int\limits_0^\delta a^{-at}\left(1+t\right)^{\large-a(1+t)}\,\text dt \le a^{2-a} \int\limits_0^\delta a^{\large-at} \left(e^{-at}+\frac a2 t^2 - \frac {a(3a+1)}6 t^3\right)\,\text dt$$ $$=a^{1-a} \int\limits_0^\delta (ea)^{\large-a t}\,\text dt + \dfrac12a^{2-a} \int\limits_0^\delta t^2 a^{\large-a t}\,\text dt -\dfrac {3a+1}6 a^{2-a} \int\limits_0^\delta t^3 a^{\large-a t}\,\text dt,$$ $$\color{brown}{\mathbf{I(a, \delta)=a^{-a}\left(\dfrac{1-(ea)^{-\large \delta a}}{\ln a} +\dfrac{2-\Gamma(3,a\delta\ln a)}{2a\ln^3a} -(3a+1)\dfrac{6-\Gamma(4,a\delta\ln a)}{6a\ln^4a}\right)}},\tag3$$ $$\int\limits_e^\infty x^{-x} dx = \lim\limits_{\delta \to 0} \sum\limits_{k=0}^\infty I(e\delta^k, \delta).\tag4$$

Comparing of this expression (for $\delta=\sqrt[\large 6000]2-1,\; n=3000$) with the numeric value of the integral is shown in the table below $$\left[\begin{matrix} [a,b] & I(a,\delta) & \text{numeric}\\ \left[e, e\sqrt2\right] & 0.02848\,08301\,58887 & 0.02848\,08301\,58764\\ \left[e\sqrt2, 2e\right] & 0.00227\,71750\,39440 & 0.00227\,71750\,39412\\ \left[2e, 2e\sqrt2\right] & 0.00003\,64476\,51537 & 0.00003\,64476\,51535\\ \left[2e\sqrt2,4e\right] & 0.00000\,00502\,01224 & 0.00000\,00502\,01224\\ [e,4e] & 0.03079\,45030\,51082 & 0.03079\,45030\,50937 \end{matrix}\right]$$

Besides, $$(1+t)^{-a(1+t)}\le e^{-at},$$ $$I(a,\infty)=\int\limits_a^\infty x^{-x}\,\text dx =a^{-a}\int_0^\infty a^{-at}\left(1+t\right)^{\large- a(1+t)}\,\text dt$$ $$\le a^{-a} \int\limits_0^\infty a^{-at} e^{-at}\text dz = \dfrac{a^{-a}}{\ln(ae)} (ae)^{-at}\bigg|_0^\infty = \dfrac{a^{-a}}{\ln(ae)},$$ $$I(4e,\infty) \le \dfrac{(4e)^{-4e}}{\ln(4e^2)} \le 1.592\cdot 10^{-12}.$$

Finally, $$I\le 1.96466\,14544\,477 + 0.03079\,45030\,51082 + 1.5917\cdot 10^{-12} < \color{brown}{\mathbf{1.99545\,59576}}.$$ Numeric calculations give $$I\approx 1.99545\,59575.$$

$\textbf{Previous version.}$

On the other hand, $$I_2=\int\limits_e^\infty x^{-x}\,\text dy =e\int\limits_0^\infty (e(1+y))^{-e(1+y)}\,\text dy =\int\limits_0^\infty \dfrac{e^{1-e(1+y)}}{(1+y)^3} (1+y)^{3-e(1+y)}\,\text dy=I_{21}+I_{22},$$ where $$I_{21}=\int\limits_1^2 \dfrac{e^{1-ez}}{z^3} z^{3-ez}\,\text dz,\quad I_{22}=\int\limits_2^\infty \dfrac{e^{1-ez}}{z^3}\, z^{3-ez}\,\text dz.$$ Applying inequality $z^{3-ez}\le f(z),$ where $$f(z)=\begin{cases} e^{-0.7177ez}(-26.937+49.92z-15.94z^2),\quad\text{if}\quad z\in[1,2)\\[8pt] (z-0.48) e^{6-1.4868ez},\quad\text{if}\quad z\in[2,+\infty) \end{cases}$$ (see also WA plot), one can get $$I_{21}\leq \int\limits_1^2 \dfrac{e^{1-ez}}{z^3}\,(e^{0.7177ez}(-26.937+49.92z-15.94z^2))\,\text dz$$ $$=\int\limits_1^2 \dfrac1{z^3}\,(e^{1-1.7177ez}(-26.937+49.92z-15.94z^2))\,\text dz$$ $$=\left(\dfrac{36.6112 - 306.641z}{z^2}\,e^{-4.66919 z} - 1475.1\operatorname{ Ei}(-4.66919 z)\bigg|_1^2\right)\le 0.03105\,33417\,253,$$(see also WA checking) $$I_{22}\leq \int\limits_2^\infty \dfrac{e^{1-ez}}{z^3}\,((z-0.48) e^{6-1.4868ez})\,\text dz =\int\limits_2^\infty \dfrac{z-0.48}{z^3}\,e^{7-2.4868ez})\,\text dz$$ $$=\dfrac{263.192-2875.76 z}{z^2}e^{-6.75982 z} - 19439.7 Ei(-6.75982 z) \bigg|_2^\infty \le 0.00003\,70357\,268$$ (see also WA checking).

Finally, $$I=I_1+I_{21}+I_{22}\le 1.96466\,14544\,477+0.03105\,33417\,253+0.00003\,70357\,268,$$ $$\color{brown}{\mathbf{I\le1.99575\,18318\,998 \le \pi-\ln\pi.}}$$

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  • $\begingroup$ Impressive. I assume there is an analytic way to guess-timate the coefficients in, e.g., $z^{3-ez}<f(z)$? $\endgroup$
    – FShrike
    Feb 7, 2023 at 16:55
  • $\begingroup$ @FShrike The accuracy of the last integrals can be improved via increasing of the polynomial order, but this way is not trivial for the second integral. Looks ugly the further fragmentation of the intervals, but it can be reserved as brute force approach. $\endgroup$ Feb 7, 2023 at 18:02
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I put here my progress so far we have :

$$\int_{2}^{\pi}k\left(x\right)dx+\int_{1}^{2}f\left(x\right)dx+\int_{0}^{1}h\left(x\right)dx+\int_{\pi}^{4}\frac{3+\frac{3.5}{1000}}{\ln3}x^{\frac{12}{1000}+\frac{2}{5}}\left(\frac{3+\frac{3.5}{1000}}{\ln3}\right)^{-x^{1+\frac{2}{5}+\frac{12}{1000}}}dx+\int_{4}^{5}ex^{\frac{12}{1000}+\frac{2}{5}}e^{-x^{\left(1+\frac{2}{5}+\frac{12}{1000}\right)}}dx+\int_{5}^{\infty}5.25^{-x}e^{\left(-x+5+0.25\right)}dx\simeq 1.996912$$

Where :

$$h\left(x\right)=x^{-x}$$

$$f\left(x\right)=a\left(\left((1-c)^{2}+ac(2-c)-ac(1-c)\ln a\right)\right)$$

$$k\left(x\right)=a^{2}\left(\left((1-d)^{2}+ad(2-d)-ad(1-d)\ln a\right)\right)$$

$$a=1/x,c=x-1,d=x-2$$

For some references see my first answer on TheSimplifire's question .

I got it !!!

$$\int_{0}^{1}x^{-x}dx+\int_{1}^{2}f\left(x\right)dx+\int_{2}^{3-0.75}k\left(x\right)dx+\int_{3-0.75}^{4}\frac{3+\frac{3.5}{1000}}{\ln3}x^{\frac{12}{1000}+\frac{2}{5}}\left(\frac{3+\frac{3.5}{1000}}{\ln3}\right)^{-x^{1+\frac{2}{5}+\frac{12}{1000}}}dx+\int_{4}^{4.8}ex^{\frac{13}{1000}+\frac{2}{5}}e^{-x^{\left(1+\frac{2}{5}+\frac{13}{1000}\right)}}dx+\int_{4.8}^{\infty}5^{-x}e^{-x+5}dx<\pi-\ln \pi$$

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