\( \int_{0}^{1}x^{-x}\, dx \)

[Tolaso J Kos]

Prove that \( \displaystyle \int_0^1 x^{-x} \, dx=\sum_{n=1}^{\infty }n^{-n} \).



1. The above function is known as Sophomore's Dream Function.

2. I hope we haven't seen it before.

[Papapetros Vaggelis]

Good evening .

\(\displaystyle{0<x\leq 1\implies x^{-x}=e^{-x\,\ln\,x}>0}\) .

We define \(\displaystyle{f:\left(0,1\right]\longrightarrow \mathbb{R}}\) by \(\displaystyle{f(x)=x^{-x}=e^{-x\,\ln\,x}}\) .

The function \(\displaystyle{f}\) is continuous and differentiable at \(\displaystyle{\left(0,1\right]}\) with

\(\displaystyle{f^\prime(x)=-\left(\ln\,x+1\right)\,e^{-x\,\ln\,x}}\) and we have that

\(\displaystyle{f^\prime(x)<0\iff x\in\left(\frac{1}{e},1\right]\,,f^\prime(x)=0\iff x=\frac{1}{e}\,,f^\prime(x)>0\iff x\in\left(0,\frac{1}{e}\right)}\) , so

\(\displaystyle{0<x\leq 1\implies f(x)\leq f\,\left(\frac{1}{e}\right)\implies f(x)\leq e^{1/e}}\) and for each \(\displaystyle{x\in\left(0,1\right]}\) holds :

\(\displaystyle{\left|\int_{x}^{1}f(t)\,dt\right|\leq \int_{x}^{1}e^{1/e}\,dt=\left(1-x\right)\,e^{1/e}<e^{1/e}}\) and that's why

\(\displaystyle{\int_{0}^{1}x^{-x}\,dx<\infty}\) . Futhermore, \(\displaystyle{\int_{0}^{1}x^{-x}\,dx\in\left(0,e^{1/e}\right)}\) .

Now, \(\displaystyle{\int_{0}^{1}x^{-x}\,dx=\int_{0}^{1}e^{-x\,\ln\,x}\,dx=\int_{0}^{1}\sum_{n=0}^{\infty}\dfrac{\left(-x\,\ln\,x\right)^{n}}{n!}\,dx}\) .

Let \(\displaystyle{\left(g_{n}\right)_{n\in\mathbb{N}\cup\left\{0\right\}}}\) be the sequence of the real functions \(\displaystyle{g_{n}:\left(0,1\right]\longrightarrow \mathbb{R}\,,g_{n}(x)=\dfrac{\left(-x\,\ln\,x\right)^{n}}{n!}}\)

We have that \(\displaystyle{f(x)\leq e^{1/e}\implies -x\,\ln\,x\leq \frac{1}{e}\stackrel{-x\,\ln\,x>0}{\implies} \left(-x\,\ln\,x\right)^{n}\leq \dfrac{1}{e^{n}}<1\,,\forall\,n\in\mathbb{N}}\) .

Therefore, for each \(\displaystyle{x\in\left(0,1\right]}\) and \(\displaystyle{n\in\mathbb{N}\cup\left\{0\right\}}\) holds :

\(\displaystyle{\left|g_{n}(x)\right|=\dfrac{\left|-x\,\ln\,x\right|^{n}}{n!}\leq \dfrac{1}{n!}=M_{n}}\) and

\(\displaystyle{\sum_{n=0}^{\infty}M_{n}=\sum_{n=0}^{\infty}\dfrac{1}{n!}=e<\infty}\) .

Thus, the sequence \(\displaystyle{\left(\sum_{k=1}^{n}g_{k}\right)_{n\in\mathbb{N}}}\) converges uniformly and we get :

\(\displaystyle{\int_{0}^{1}x^{-x}\,dx=\sum_{n=0}^{\infty}\dfrac{1}{n!}\,\int_{0}^{1}x^{n}\,\left(-\ln\,x\right)^{n}\,dx=\sum_{n=0}^{\infty}\dfrac{1}{n!}\,I_{n}\,(I)}\)

where \(\displaystyle{I_{n}=\int_{0}^{1}x^{n}\,\left(-\ln\,x\right)^{n}\,dx\,,n\in\mathbb{N}\cup\left\{0\right\}}\) .

We apply the substitution \(\displaystyle{t=-\ln\,x\,,t\in\left[0,+\infty\right)}\) and we have

\(\displaystyle{x=e^{-t}\implies \rm{dx}=-e^{-t}\,\rm{dt}}\) , so

\(\displaystyle{\begin{aligned}I_{n}&=\int_{0}^{+\infty}e^{-n\,t}\,t^{n}\,e^{-t}\,dt\\&=\int_{0}^{+\infty}t^{(n+1)-1}\,e^{-(n+1)\,t}\,dt\\&=\dfrac{1}{n+1}\,\dfrac{1}{\left(n+1\right)^{n}}\,\int_{0}^{+\infty}\left((n+1)t\right)^{(n+1)-1}\,e^{-(n+1)\,t}\,d\,\left[(n+1)\,t\right]\\&=\dfrac{\Gamma\,\left(n+1\right)}{\left(n+1\right)^{n+1}}\\&=\dfrac{n!}{\left(n+1\right)^{n+1}}\end{aligned}}\)

So, according to \(\displaystyle{(I)}\) we have :

\(\displaystyle{\begin{aligned} \int_{0}^{1}x^{-x}\,dx&=\sum_{n=0}^{\infty}\dfrac{n!}{n!\,\left(n+1\right)^{n+1}}\\&=\sum_{n=0}^{\infty}\dfrac{1}{\left(n+1\right)^{n+1}}\\&=\sum_{n=1}^{\infty}\dfrac{1}{n^{n}}\\&=\sum_{n=1}^{\infty}n^{-n}\end{aligned}}\)

Note : \(\displaystyle{n\in\mathbb{N}\implies n^{n}\geq n^2\implies \sum_{n=1}^{\infty}n^{-n}\leq \sum_{n=1}^{\infty}\dfrac{1}{n^2}=\dfrac{\pi^2}{6}<\infty}\) .

[Tsakanikas Nickos]

Nice Solution!

Let me complement something on the calculation of the integral \( \displaystyle I_{n} \). For those not enough acquainted with the Gamma Function, here is another way of computing the integral \( I_{n} \) using Laplace Transforms.

Let \( \displaystyle n \in \mathbb{N}\cup\{0\}. \) Then
\begin{align*}
I_{n}
&= \int_{0}^{\infty} {e}^{-nt}t^{n}{e}^{-t}\mathrm{d}t \\
&= \mathcal{L} \left\{ t^{n}{e}^{-t} \right\} \Big|_{s=n} \\
&= \mathcal{L} \left\{ t^{n} \right\} \Big|_{s\mapsto s+1 \; , \; s=n} \\
&= \frac{n!}{s^{n+1}} \Big|_{s\mapsto s+1 \; , \; s=n} \\
&= \frac{n!}{ (n+1)^{n+1} }
\end{align*}