Sophomore's dream

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Graph of a function and a region bounded by it in an interval [0,1].
Graph of a function y=1/x^{x}\!\, and a region bounded by it in an interval [0,1].
Graph of a function and a region bounded by it in an interval [0,1].
Graph of a function y=x^{x}\!\, and a region bounded by it in an interval [0,1].

In mathematics, sophomore's dream is a name occasionally used for the identities

\begin{align} \int_0^1 x^{-x}\,dx &= \sum_{n=1}^\infty n^{-n}&&(= 1.291285997\dots)\\ \int_0^1 x^x \,dx &= \sum_{n=1}^\infty (-1)^{n+1}n^{-n} &&(= 0.783430510712\dots) \end{align}

discovered in 1697 by Johann Bernoulli (especially the first).

The name is in contrast to the "freshman's dream" which is given to the mistake (x + y)n = xn + yn. (The correct result is given by the binomial theorem.) The sophomore's dream names a result with a similarly too-good-to-be-true feel, that

\int_0^1 \frac{1}{x^x}\,dx = \sum_{n=1}^\infty \frac{1}{n^n}.

However, this result is in fact true.

[edit] Proof

We prove the second identity; the first is completely analogous.

The key ingredients of the proof are:

  • Write xx = exp(x ln x).
  • Expand exp(x ln x) using the power series for exp.
  • Integrate termwise.
  • Integrate by parts.

Expand xx as

x^x = \exp(x \ln x) = \sum_{n=0}^\infty \frac{x^n(\ln x)^n}{n!}.

Thus by termwise integration,

\int_0^1 x^xdx = \sum_{n=0}^\infty \int_0^1 \frac{x^n(\ln x)^n}{n!}dx.

Evaluate the terms by integration by parts; integrate \int x^m (\ln x)^n\; dx by taking u = (lnx)n and dv = x^m\; dx, which yields:

\int x^m (\ln x)^n\; dx = \frac{x^{m+1}(\ln x)^n}{m+1} - \frac{n}{m+1}\int x^{m+1} \frac{(\ln x)^{n-1}}{x} dx \qquad\mbox{(for }m\neq -1\mbox{)}

(also in the list of integrals of logarithmic functions).

Thus inductively,

\int x^m (\ln x)^n\; dx = \frac{x^{m+1}}{m+1} \cdot \sum_{i=0}^n (-1)^i \frac{(n)_i}{(m+1)^i} (\ln x)^{n-i}

where (n)i denotes the falling factorial.

In this case m=n, and they are an integer, so

\int x^n (\ln x)^n\; dx = \frac{x^{n+1}}{n+1} \cdot \sum_{i=0}^n (-1)^i \frac{(n)_i}{(n+1)^i} (\ln x)^{n-i}

Integrating from 0 to 1, all the terms vanish except the last term at 1 (all the terms vanish at 0 because \lim_{x \to 0^+} x^m (\ln x)^n = 0 by l'Hôpital's rule, and all but the last term vanish at 1 since ln(1) = 0), which yields:

\int \frac{x^n (\ln x)^n}{n!}\; dx = \frac{1}{n!}\frac{1^{n+1}}{n+1} (-1)^n \frac{(n)_n}{(n+1)^n} = (-1)^n (n+1)^{-(n+1)}

Summing these (and changing indexing so it starts at n = 1 instead of n = 0) yields the formula.

[edit] References