Sophomore's dream

The original proof, given in Bernoulli,[2] and presented in modernized form in Dunham,[3] differs from the one above in how the termwise integral ${\textstyle \int _{0}^{1}x^{n}(\log x)^{n}\,dx}$ is computed, but is otherwise the same, omitting technical details to justify steps (such as termwise integration). Rather than integrating by substitution, yielding the Gamma function (which was not yet known), Bernoulli used integration by parts to iteratively compute these terms.

The integration by parts proceeds as follows, varying the two exponents independently to obtain a recursion. An indefinite integral is computed initially, omitting the constant of integration ${\displaystyle +C}$ both because this was done historically, and because it drops out when computing the definite integral.

Integrating ${\textstyle \int x^{m}(\log x)^{n}\,dx}$ by substituting ${\textstyle u=(\log x)^{n}}$ and ${\textstyle dv=x^{m}\,dx}$ yields:

$${\displaystyle {\begin{aligned}\int x^{m}(\log x)^{n}\,dx&={\frac {x^{m+1}(\log x)^{n}}{m+1}}-{\frac {n}{m+1}}\int x^{m+1}{\frac {(\log x)^{n-1}}{x}}\,dx\qquad {\text{(for }}m\neq -1{\text{)}}\\&={\frac {x^{m+1}}{m+1}}(\log x)^{n}-{\frac {n}{m+1}}\int x^{m}(\log x)^{n-1}\,dx\qquad {\text{(for }}m\neq -1{\text{)}}\end{aligned}}}$$

(also in the list of integrals of logarithmic functions). This reduces the power on the logarithm in the integrand by 1 (from ${\displaystyle n}$ to ${\displaystyle n-1}$) and thus one can compute the integral inductively, as

$${\displaystyle \int x^{m}(\log x)^{n}\,dx={\frac {x^{m+1}}{m+1}}\cdot \sum _{i=0}^{n}(-1)^{i}{\frac {(n)_{i}}{(m+1)^{i}}}(\log x)^{n-i}}$$

where ${\textstyle (n)_{i}}$ denotes the falling factorial; there is a finite sum because the induction stops at 0, since n is an integer.

In this case ${\textstyle m=n}$, and they are integers, so

$${\displaystyle \int x^{n}(\log x)^{n}\,dx={\frac {x^{n+1}}{n+1}}\cdot \sum _{i=0}^{n}(-1)^{i}{\frac {(n)_{i}}{(n+1)^{i}}}(\log x)^{n-i}.}$$

Integrating from 0 to 1, all the terms vanish except the last term at 1,[note 2] which yields:

$${\displaystyle \int _{0}^{1}{\frac {x^{n}(\log x)^{n}}{n!}}\,dx={\frac {1}{n!}}{\frac {1^{n+1}}{n+1}}(-1)^{n}{\frac {(n)_{n}}{(n+1)^{n}}}=(-1)^{n}(n+1)^{-(n+1)}.}$$

This is equivalent to computing Euler's integral identity ${\displaystyle \Gamma (n+1)=n!}$ for the Gamma function on a different domain (corresponding to changing variables by substitution), as Euler's identity itself can also be computed via an analogous integration by parts.

Footnotes