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Graph of the functions y = x^x and y = x^−x on the interval x ∈ [0, 1].

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

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

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 has a similarly too-good-to-be-true feel, but is in fact true.

[edit] Proof

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

The key ingredients of the proof are:

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

Expand x^x 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 $\scriptstyle \int x^m (\ln x)^n\; dx$ by taking u = (ln x)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 integers, 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 $\scriptstyle\lim_{x \to 0^+} x^m (\ln x)^n \, = \, 0$ by l'Hôpital's rule, and all but the last term vanish at 1 since ln(1) = 0, which yields:

$\int_0^1 \frac{x^n (\ln x)^n}{n!}\; dx = \frac{1}{n!}\frac{1^{n+1}}{n+1} (-1)^n \frac{(n)_n}{(n+1)^n} = (-1)^n (n+1)^{-(n+1)}.$

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

[edit] References

Jonathan Borwein, David H. Bailey, Roland Girgensohn Experimentation in Mathematics: Computational Paths to Discovery Page 44.
William Dunham, The Calculus Gallery, Masterpieces from Newton to Lebesgue, Princeton University Press, Princeton, NJ 2005, p. 46-51.
N. J. A. Sloane, (sequence A083648 in OEIS) and (sequence A073009 in OEIS)
Pólya and Gábor Szegő, Problems and Theorems in Analysis (part I, problem 160).
Weisstein, Eric W. Sophomore's Dream. From MathWorld--A Wolfram Web Resource.