There was a friend asking me how to prove

$$\int_0^\infty x^{-x}\,dx<2$$

Mathematica showed that its approximate value is 1.99546, so I think it isn't easy to solve it, can you provide me some ideas about this question?


You may write $$ \begin{align} \int_0^\infty x^{-x}\,dx& =\int_0^6 x^{-x}\,dx+\int_6^\infty x^{-x}\,dx\\\\ &<\int_0^6 x^{-x}\,dx+\int_6^\infty 6^{-x}\,dx\\\\ &=\int_0^6 x^{-x}\,dx+\frac{1}{466565\log6}\\\\ &=\int_0^6 x^{-x}\,dx+0.000011962...\\\\ &=1.99544...+0.000011962...\\\\ &<2.\\\\ \end{align} $$ We are left to approximate $\displaystyle \int_0^6 x^{-x}\mathrm d x$ by a numerical integration rule. The point is that the second derivative of $f(x)=x^{-x}$ is not bounded on $(0,6]$, in fact not bounded near $0^+$.

So let's write $$ \int_0^6 x^{-x} \mathrm d x =\int_0^1 x^{-x} \mathrm d x +\int_1^6 x^{-x} \mathrm d x. $$ Observe that $$ x^{-x}=e^{-x\ln x}=1-x\ln x+\frac{(x\ln x)^2}{2!}-\frac{(x\ln x)^3}{3!}+..., \quad 0<x\leq 1, $$ and that $$ \int_0^1 (x\ln x)^n\,dx = (-1)^n \int_0^\infty u^ne^{-nu}\,du = (-1)^n \frac{n!}{(1+n)^{n+1}}, \, (u=-\ln x), $$ leading to the series $$ \int_0^1 x^{-x}\,dx =\sum_{k=1}^{\infty}\frac{1}{n^{n}}. $$ Consequently $$ \left|\int_0^1 x^{-x}\,dx -\sum_{k=1}^{7} \frac{1}{n^n} \right|=\left|\sum_{k=8}^{\infty} \frac{1}{n^n} \right| < \left| \frac{1}{8^8} \cdot\frac{1}{1-\frac{1}{8}}\right|<10^{-7} $$ and we are done, since we may now use the trapezoidal rule to approximate $\displaystyle \int_1^6 x^{-x}\,dx$ taking into account that $$|f''(x)|=\left|\frac{d^2}{dx^2}\left(x^{-x}\right)\right|=\left|\frac{1}{x}+(1+\ln x)^2 \right| x^{-x} <1, \quad 1<x\leq6.$$

  • 1
    $\begingroup$ To call this a proof, I would probably ask for precisions on the numerical integration rule, and a careful evaluation of the error. If I need to rely on a black box, I'm not perfectly happy (and I could show how to misuse a blackbox like Mathematica to get numerical b*t). $\endgroup$ – Jean-Claude Arbaut Aug 26 '14 at 11:33
  • $\begingroup$ @Jean-ClaudeArbaut You are right... $\endgroup$ – Olivier Oloa Aug 26 '14 at 11:35
  • $\begingroup$ Why did you choose $6$ as splitting point of the integration domain? $\endgroup$ – Ruslan Aug 26 '14 at 14:36
  • 4
    $\begingroup$ @Ruslan My guess would be that he went backwards. Start from $\int_0^\infty x^{-x} dx = \int_0^y x^{-x} dx + \int_y^\infty x^{-x} dx$. Bound the second term by $\int_y^\infty y^{-x} dx = y^{-y}/(\ln(y))$. Then take $y$ large enough that $y^{-y}/\ln(y) \ll 0.004$, since you need to estimate the overall integral to within about $0.0045$. Dealing with integers, $4$ is almost surely too small: $4^{-4}/\ln(y) \approx 0.0028$. $5$ isn't great either: $5^{-5}/\ln(y) \approx 0.0002$, which absorbs about 5% of your room to work. $6$ leaves you with over 99% of your error to use in the bounded part. $\endgroup$ – Ian Aug 26 '14 at 17:00
  • $\begingroup$ Typo, should be $4^{-4}/\ln(4) \approx 0.0028$ and $5^{-5}/\ln(5) \approx 0.0002$. $\endgroup$ – Ian Aug 26 '14 at 17:08

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