The Sophomore’s Dream

I get back to continue this journey into the world of curious integrals by telling you about the ‘Sophomore’s Dream’. The Sophomore’s Dream is the identity

$\displaystyle \int_0^1 x^{-x} \: \mathrm{d}x = \sum_{n=1}^{\infty} n^{-n}.$

Discovered in 1697 by the Swiss mathematician Johann Bernoulli, this identity (or some minor variant of it) is today called the Sophomore’s Dream because it has a feel that it is too-good-to-be-true and it is true. Bernoulli himself called it his “series mirabili” or “marvelous series”! In other words, really? We can swap the integral out for a sum and the result doesn’t change? It miraculously doesn’t and we’ll see how!

We expand the integrand $x^{-x}$ using the exponential series as

$\displaystyle x^{-x} = \exp (-x \ln x) = \sum_{n=0}^{\infty} \frac{(-x)^n (\ln x)^n}{n!},$

and, therefore,

$\displaystyle \int_0^1 x^{-x} \: \mathrm{d}x = \int_0^1 \sum_{n=0}^{\infty} \frac{(-x)^n (\ln x)^n}{n!} \: \mathrm{d}x.$

The power series on the right hand side is uniformly convergent and we can interchange the order of summation and integration to yield

$\displaystyle \displaystyle \int_0^1 x^{-x} \: \mathrm{d}x = \sum_{n=0}^{\infty} \int_0^1 \frac{(-x)^n (\ln x)^n}{n!} \: \mathrm{d}x = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \underbrace{\int_0^1 x^n (\ln x)^n \: \mathrm{d}x}_{I}.$

It then remains to evaluate the integral

$\displaystyle I=\int_0^1 x^n (\ln x)^n \: \mathrm{d}x .$

Performing the substitution $x=\exp (-\frac{u}{n+1})$ , we have that

$\begin{aligned} \displaystyle I &= -\frac{1}{n+1} \int_{\infty}^{0} \exp \left(-\frac{u}{n+1}\right) \exp \left(-\frac{u n}{n+1}\right)\left(-\frac{u}{n+1}\right)^{n} \: \mathrm{d} u \\ &= \frac{(-1)^n}{(n+1)^{n+1}} \int_0^{\infty} e^{-u} u^{(n+1)-1} \: \mathrm{d}u \\ &= \frac{(-1)^n \Gamma(n+1)}{(n+1)^{n+1}} \\ &= \frac{(-1)^n n!}{ (n+1)^{n+1} }, \end{aligned}$

where the last two equalities are the result of the definition of the Gamma function and the Gamma function being an analytic continuation of the factorial for positive integers. We now put it all back to get

$\displaystyle \int_0^1 x^{-x} \: \mathrm{d}x = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \frac{(-1)^n n!}{(n+1)^{n+1}} =\sum_{n=0}^{\infty} (n+1)^{-(n+1)}.$

Shifting the index of the summation to begin from 1, we get

$\displaystyle \int_0^1 x^{-x} \: \mathrm{d}x = \sum_{n=1}^{\infty} n^{-n},$

and we’re done! For the curious among you, the summation on the right hand side of the Sophomore’s Dream does converge, and is given by

$\displaystyle \int_0^1 x^{-x} \: \mathrm{d}x = 1 + \frac{1}{2^2} + \frac{1}{3^3} + \frac{1}{4^4} + \ldots = 1.29128\ldots .$

Exercise:

Can you show, in a similar vein, that

$\displaystyle \int_0^1 x^{x^2} \: \mathrm{d}x = 1 - \frac{1}{3^2} + \frac{1}{5^3} - \frac{1}{7^4} + \ldots = 0.896488 \ldots ?$

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