8
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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

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

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

$\endgroup$
4
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.