Prove that $\int_0^\infty\frac1{x^x}\, dx<2$

Solution 1:

$$\int_{0}^{+\infty}e^{-x\log x}\,dx = \underbrace{\int_{0}^{1}e^{-x\log x}\,dx}_{I_1}+\underbrace{\int_{1}^{+\infty}e^{-x\log x}\,dx}_{I_2} $$

$$ I_1=\sum_{n\geq 0}\frac{(-1)^n}{n!}\int_{0}^{1}x^n\left(\log x\right)^n\,dx = \sum_{n\geq 0}\frac{1}{(n+1)^{n+1}}=\sum_{n\geq 1}\frac{1}{n^n}\tag{A}$$

$$ I_2 = \int_{0}^{+\infty}e^{-(x+1)\log(x+1)}\,dx=\int_{0}^{+\infty}\frac{e^{-x}}{W(x)+1}\,dx\tag{B} $$ where $(A)$ gives $I_1\leq 1.292$ and high-order Padé approximants give $I_2\leq 0.705$.
It is a very tight inequality: I wonder if it can be proved in a more elementary way, maybe by writing the whole integral as $\int_{1}^{+\infty}e^{-x}g(W(x))\,dx$.

Solution 2:

This approach is based on the intermediate inequalities, which follow from the series representations.

Using inequality $$x^{-x}\le1+(\sqrt[e]e -1)ex(2-ex)+ \frac{116-ex}{635} \sqrt{e x} (ex-1)^2,\quad\text{if}\quad x\in[0,1],\tag1$$

one can get $$I_1 \le \dfrac1e\int\limits_0^e (1 + (\sqrt[e]e - 1) (2 y - y^2) + \dfrac{116-y}{635} \sqrt y\, (y - 1)^2)\,\text dy,$$ $${\small I_1 \le 1 - e + \dfrac{e^2}3 + \sqrt e\left(\dfrac{232}{1905} - \dfrac{466 e}{3175} + \dfrac{236 e^2}{4445} - \dfrac{2 e^3}{5715}\right) + \dfrac{\sqrt[e]e (3e-e^2)}3 < 1.291733}\tag2$$ (see also Wolfram Alpha integration).

If $\;x\in[1,3],\;$ then $$\begin{align} &(2-x)\left(1+\dfrac1{120}(x-1)^6\right) + \dfrac1{12}(x-1)^3 (x^2-6x+11)\\[4pt] &+ \dfrac1{602910}(x-1)^8(470 +381(x-e) -270 (x-e)^2 +115(x-e)^3) -x^{-x} \le 0, \end{align}\tag3$$ $$\begin{align} &I_2 = \int\limits_1^3 x^{-x}\,\text dx \le \int\limits_0^2\bigg(1-y)\left(1+\dfrac1{120}y^6\right) + \dfrac1{12}y^3 (y^2-4y+6) + \dfrac1{602910}y^8\\[4pt] &\times\bigg(470 +381(y+1-e) -270 (y+1-e)^2 +115(y+1-e)^3\bigg)\bigg)\,\text dy = \bigg(\dfrac{y^9}{5426190}\\[4pt] &\times\left(\dfrac{345}4 y^3 - \dfrac{135(23e - 5)}{11} y^2 + \dfrac{27(62 - 50 e + 115 e^2)}{10} y + 696 - 186 e + 75 e^2 - 115 e^3\right) \\[4pt] & - y^8/960 + y^7/840 + y^6/72 - y^5/15 + y^4/8 - y^2/2 + y\bigg)\bigg|_0^2, \end{align}$$ $$I_2 \le \dfrac{2 (61687647 - 11159040 e + 4899840 e^2 - 809600 e^3)}{149220225} < 0.687550.\tag4$$

At last, $$I_3 = \int\limits_3^\infty x^{-x}\,\text dx = 3\int\limits_0^\infty (3+3y)^{-3-3y}\,\text dy = 3\int\limits_0^\infty3^{-3-3y}(1+y)^{-3-3y}\,\text dy$$ $$I_3 \le \dfrac19\int\limits_0^\infty \dfrac{27^{-y}}{(1+y)^3}\,\text dy = \dfrac{1-3\ln3\,(1 + 81\ln3\, \text{Ei}(-3\ln3))}{18} < 0.018865\tag5$$ (see also WA integration)

From $(2),(4),(5)$ should $$I=I_1+I_2+I_3 < 1.9982 <2.$$

On the other hand, $$I_1=\sum\limits_{n=1}^\infty \dfrac1{n^n}\le 1+\dfrac14+\dfrac1{27} +\dfrac1{256} + \dfrac1{3125}+\sum\limits_{n=6}^\infty\dfrac1{n^6},$$ $$I_1 \le 1+\dfrac14+\dfrac1{27} +\dfrac1{256} + \dfrac1{3125}+\dfrac{\pi^6}{945} < 1.291302,$$

$$\color{brown}{\mathbf{I< 1.291302 + 0.687550 + 0.018865 < 1.99772}}.$$

Solution 3:

This is a well-known result but it is hard to Google it if you do not know what it is called. This, and a similar identity are known as the sophomore's dream and the proof is given here.

Important Note: This Proof is also present in Nahin’s Inside Interesting Integrals. However, this is how we learnt it in class as well.

We start with the identity $$x^{cx^a} = e^{cx^a\ln x} \tag 1$$ where $a$ and $c$ are constants. Using the power series expansion of the exponential $$e^y = 1 + y + \frac{y^2}{2!} + … $$ with $y = cx^a \ln x$ gives us: $$x^{cx^a} = 1+ cx^a\ln x + \frac{1}{2!}c^2x^{2a}\ln^2 x + \ldots $$ and so: $$\int_{0}^{1} x^{cx^a} \, dx = \int_{0}^{1}\, dx + c\int_{0}^1{} x^a \ln x\, dx + \frac{c^2}{2!} \int_{0}^{1} x^{2a}\ln^2 x\, dx + \ldots \tag 2$$

Now, to solve integrals of the form: $$I(m,n) = \int_{0}^{1} x^m\ln^n x \, dx $$ we simply make the substitution $u=-\ln x$, followed by another substitution $\chi = (m+1)u$ giving us: $$I(m,n) = \frac{(-1)^nn!}{(m+1)^{n+1}}$$

Hence, $(2)$ becomes $$\int_{0}^{1} x^{cx^a}\, dx = 1- \frac{c}{(a+1)^2} + \frac{c^2}{(2a+1)^3} - \frac{c}{(3a+1)^4}+\ldots$$

Now, if $c=-1$ and if $a=1$, it gives us: $$\int_{0}^{1} x^{-x}\, dx = 1 + \frac{1}{2^2}+\frac{1}{3^3}+\frac{1}{4^4}+\ldots$$

See also here.

Solution 4:

This comment is not the proof, just wanted to share a quick idea. Lots of assumptions, because no proofs are provided.

Assume we know that $a(x)=\int_0^\infty\frac1{x^x}\, dx = a< \infty$. Then $a(x)/a$ is the valid probability density of some random variable, say $B$. Due to the Markov inequality we have $$ P(B>1)={1 \over a} \int_1^\infty\frac{1}{x^x}dx < {1 \over a} \int_0^\infty\frac{x}{x^x}dx, $$ wherefrom we get $$ \int_1^\infty\frac{1}{x^x}dx < \int_0^\infty\frac{1}{x^{x-1}}dx. $$

Assume the integral $\int_0^\infty\frac{1}{x^{x-1}}dx= \int_0^\infty a_1(x) dx=a_1<\infty$. Then $a_1(x)/a_1$ is the valid probability density of some random variable, say $B_1$. Again due to the Markov inequality we have $$ P(B_1>1)={1 \over a_1} \int_1^\infty\frac{1}{x^{x-1}}dx < {1 \over a_1} \int_0^\infty\frac{1}{x^{x-2}}dx. $$ Proceeding in the same way we get $$ \int_0^\infty\frac{1}{x^x}dx < \int_0^1\frac{1}{x^{x}}dx + \int_0^1\frac{1}{x^{x-1}}dx + \int_0^1\frac{1}{x^{x-2}}dx + \int_0^1\frac{1}{x^{x-3}}dx +\dots = \sum_{n=0}^\infty \int_0^1 {x^{n-x}}dx $$

Since Markov bound is very crude, not sure if this converges.