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

Integration by parts is out of the question. If we let $f(x)=\dfrac1{x^x}$ and $g'(x)=1$ then $f'(x)=-x^x(\ln x + 1)$ by implicit differentiation and $g(x)=x$. The integral $\int f'(x)g(x)\, dx$ looks even harder to evaluate.

I tried to use the formula $$(b-a)\inf_{x\in [a,b]}f(x)\le\int_a^b f(x)\,dx\le(b-a)\sup_{x\in [a,b]}f(x)$$ with $f(x)=\dfrac1{x^x}$ but since in this case we have $b=\infty$, this method is not possible. The Cauchy-Schwarz inequality wouldn't work either.

I then tried Frulliani's integral $$\int_0^\infty\frac{f(ax)-f(bx)}x\, dx=(f(0)-f(\infty))\ln\frac ba$$ with $f(ax)-f(bx)=x^{1-x}$. However, I can't seem to find a continuous function $f$ such that it holds. Is there such a function?

Also, I've seen in a maths formula handbook the identity $$\int_0^1\frac1{x^x}\, dx=\sum_{k=1}^{\infty}\frac1{k^k}$$ Is there a way to prove this as well?

  • $\begingroup$ just an idea (I dont know if this can work): try to compare your integral with $\sum_{k=1}^\infty\frac1{k^2}<2$ $\endgroup$ – Masacroso Dec 30 '17 at 16:47

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

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

$$ I_2 = \int_{0}^{+\infty}e^{-(x+1)\log(x+1)}\,dx=\int_{0}^{+\infty}\frac{e^{-x}}{W(x)+1}\,dx\tag{B} $$ 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 $\int_{1}^{+\infty}e^{-x}g(W(x))\,dx$.


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.

  • $\begingroup$ but the integral is from zero to infinity $\endgroup$ – Masacroso Dec 30 '17 at 16:40

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