Sophomore's Dream

#Sophomore'sDream

Problem Statement

Prove that

$$ \int_{0}^{1} x^{-x} dx = \sum_{n=1}^{\infty} n^{-n} $$

Solution: This statement is called Sophomore's Dream for a reason. The two sides are almost equivalent, whilst one is the discrete form, the other is its continuous form. To prove that both sides are equal, we begin by writing $x^x = e^{x \log x}. $

$$ \implies \int_{0}^{1} x^{-x} dx = \int_{0}^{1} e^{-x \log x} dx $$

$$ = \int_{0}^{1} \sum_{n=0}^{\infty} \frac{(-1)^n x^n (\log x)^n}{n!} $$

By uniform convergence of power series, we can flip the order of integral and sum

$$ = \sum_{n=0}^{\infty} \int_{0}^{1} \frac{(-1)^n x^n (\log x)^n}{n!} $$

For $n\geq 0$, let

$$ I_n=\int_{0}^{1}\frac{(-1)^n x^n(\log x)^n}{n!},dx. $$

To evaluate $I_n$, make the substitution

$$ x=e^{-\frac{u}{n+1}}. $$

Then

$$ \log x=-\frac{u}{n+1}, \qquad dx=-\frac{1}{n+1}e^{-\frac{u}{n+1}},du. $$

As $x$ goes from $0$ to $1$, $u$ goes from $\infty$ to $0$. Thus,

$$ \begin{aligned} I_n &=\frac{(-1)^n}{n!} \int_{\infty}^{0} e^{-\frac{nu}{n+1}} \left(-\frac{u}{n+1}\right)^n \left(-\frac{1}{n+1}e^{-\frac{u}{n+1}}\right),du\ &=\frac{1}{n!(n+1)^{n+1}} \int_{0}^{\infty}u^n e^{-u},du. \end{aligned} $$

Now, invoking the Gamma function,

$$ \Gamma(n+1) =\int_{0}^{\infty}u^n e^{-u},du =n!. $$

Hence,

$$ I_n=\frac{1}{(n+1)^{n+1}}. $$

Therefore,

$$ \begin{aligned} \int_{0}^{1}x^{-x},dx &=\sum_{n=0}^{\infty}I_n\ &=\sum_{n=0}^{\infty}\frac{1}{(n+1)^{n+1}}\ &=\boxed{\sum_{n=1}^{\infty}n^{-n}}. \end{aligned} $$

For the Gamma function, you can refer to my book on calculus published by the World Scientific.