Sophomore's dream

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

: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" = "x"^"n" + "y"^"n". (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.

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:

: $int\; x^m\; (ln\; x)^n;\; dx\; =\; frac\{x^\{m+1\}(ln\; x)^n\}\{m+1\}\; -\; frac\{n\}\{m+1\}int\; x^\{m+1\}\; frac\{(ln\; x)^\{n-1\{x\}\; dx\; qquadmbox\{(for\; \}m\; eq\; -1mbox\{)\}$

(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 $scriptstylelim\_\{x\; o\; 0^+\}\; x^m\; (ln\; x)^n\; ,\; =\; ,\; 0$ by l'Hôpital's rule, and all but the last term vanish at 1since 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" = 1instead of "n" = 0) yields the formula.

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,OEIS|id=A083648 and OEIS|id=A073009
* Pólya and Gábor Szegö, "Problems and Theorems in Analysis" (part I, problem 160).
* Weisstein, Eric W. [http://mathworld.wolfram.com/SophomoresDream.html "Sophomore's Dream."] From MathWorld--A Wolfram Web Resource.