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Prove that

$$\int_0^1 \frac{\log^2 x-2}{x^x}dx<0$$

One way to do this is use the idea in the proof of Sophomore's dream. We have $$x^{-x}=\exp(-x\log x)=\sum_{n=0}^\infty\frac{(-1)^nx^n\log^n x}{n!}$$ Therefore, using the change of variable $x=\exp(-t/(n+1))$ we have

$$\begin{aligned}\int_0^1 \frac{\log^2 x}{x^x}dx=&\sum_{n=0}^\infty\frac{(-1)^n}{n!}\int_0^1x^n\log^{n+2}xdx\\ =&\sum_{n=0}^\infty\frac{(n+1)^{-(n+3)}}{n!}\int_0^\infty t^{n+2}e^{-t}dt\\ =&\sum_{n=0}^\infty\frac{(n+1)^{-(n+3)}(n+2)!}{n!}\\ =&\sum_{n=0}^\infty(n+1)^{-(n+2)}(n+2)\\ =&\sum_{n=1}^\infty n^{-(n+1)}(n+1)\\ =&\sum_{n=1}^\infty n^{-n}+n^{-(n+1)}\\ <&2\sum_{n=1}^\infty n^{-n}\\ =&\int_0^1\frac{2}{x^x}dx \end{aligned}$$

Hence the result follows.

I am curious if there are any other methods to prove this, especially I am interested in easier approaches.

P.S. This was a bonus problem in an assignment from a multivariable calculus class.

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    $\begingroup$ Its numerical calculation results $-.1528960021. $ $\endgroup$ – user64494 Jun 2 '15 at 18:55
  • $\begingroup$ I gave a completely wrong answer by disregarding very small variations of the positive $x^x$. Calculating like user64494 I realized that the integral becomes negative just near enough of 1. For example, from 0 to 0.92 it is still positive and only from 0 to 0.93 the integral becomes negative (just for approximations of 1/100). I see now as very difficult an "easier way" as you ask. $\endgroup$ – Piquito Jun 5 '15 at 12:52
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Note that $f(x)=x^x$ is convex in $[0,1]$ with its minimum happening at $1/e$. Now since $2^e>e$ ($e>1$) it holds that $f(1/e)>1/2$. Therefore \begin{align} \int_0^1 \frac{\log^2 x-2}{x^x}dx<2\int_0^1 (\log^2 x-2)dx=0 \end{align}

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Using the inequality $\log(x)<x-1$ for all $x>0$ we have \begin{align} \int_0^1 {\frac{{\log ^2 \left( x \right) - 2}}{{x^x }}dx} < \int_0^1 {\frac{{\left( {x - 1} \right)^2 - 2}}{{x^x }}dx} \end{align} Also we use the inequality $e^{x} \ge 1+x$ for all $x>0$ to express the denumeantor $$x^x=\exp(x\ln(x))\ge 1+x\ln(x).$$

Therefore, \begin{align} \int_0^1 {\frac{{\log ^2 \left( x \right) - 2}}{{x^x }}dx} < \int_0^1 {\frac{{\left( {x - 1} \right)^2 - 2}}{{x^x }}dx} <\int_0^1 {\frac{{\left( {x - 1} \right)^2 - 2}}{{1+x\ln(x)}}dx}. \end{align} Now, using the inequality $\log(x) > \frac{x-1}{x}$ for all $x > 0$, we egt \begin{align} \int_0^1 {\frac{{\log ^2 \left( x \right) - 2}}{{x^x }}dx} < \int_0^1 {\frac{{\left( {x - 1} \right)^2 - 2}}{{x^x }}dx} &<\int_0^1 {\frac{{\left( {x - 1} \right)^2 - 2}}{{1+x\ln(x)}}dx} \\ &<\int_0^1 {\frac{{\left( {x - 1} \right)^2 - 2}}{{1+x\cdot \frac{x-1}{x}}}dx} \\ &<\int_0^1 {e\cdot x\cdot \frac{{\left( {x - 1} \right)^2 - 2}}{{1+x\cdot \frac{x-1}{x}}}dx} \,\,\,\,\,\,\,\,\,\,\,\,(\text{since $0<x<1$}) \\ &<\int_0^1 {e\cdot \left({\left( {x - 1} \right)^2 - 2}\right)dx} = -\frac{5}{3} e <0 \end{align}

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    $\begingroup$ You have $\log x<x-1<0$ and hence $\log^2x>(x-1)^2$ for $0<x<1$. Hence your first inequality is wrong. $\endgroup$ – mickep Jul 22 '15 at 8:30
  • $\begingroup$ Quick question: Am I missing something important here, or is it possible to just state that as $\int\limits_0^1 \frac{\log^2 x-2}{x^x}< \int\limits_0^1 \frac{(x-1)^2 -2}{x^x}$, and the integrand on the right side is negative for all $x\in (0,1]$, the original integral (if it exists) must be less than zero? Or do we need to show that it doesn't diverge also? $\endgroup$ – Scounged Jul 22 '15 at 8:34
  • $\begingroup$ No we have $x>0$ so for any value of $x>0$ the inequality valid. We do not change the inequality sign if we squared the both sides $\endgroup$ – Mohammad W. Alomari Jul 22 '15 at 8:35
  • $\begingroup$ $-2<-1$ but $(-2)^2>(-1)^2$. $\endgroup$ – mickep Jul 22 '15 at 8:36

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