# generalization of sophomore’s dream integral/series identities

This post grew out of one of my MO questions and Sivaram Ambikasaran’s questions on M.SE. We’re currently poking at $M(z)=1+g(z)$ to see how deformed exponential and trigonometric functions would operate, but the following piece of reasoning occurred to me tonight and is too good not to share:
Define $g(z)=\sum_{n=1}^{\infty} \frac{z^{n}}{n^n}$

$g(z)$ can be used to evaluate integrals of the “Sophomore’s dream” form:

$\int_{0}^{1} x^{ax} dx$

So:

$\int_{0}^{1} x^{ax} dx = \sum_{n=0}^{\infty} \int_{0}^{1} \frac{(ax)^{n}(\log x)^{n}}{n!} dx$

We can easily factor that $a^{n}$ out:

$\int_{0}^{1} x^{ax} dx = \sum_{n=0}^{\infty} a^{n} \int_{0}^{1} \frac{x^{n}(\log x)^{n}}{n!} dx$

Wikipedia’s page on the Sophomore’s dream gives an explicit evaluation
of the integral inside as:
$\int_{0}^{1} \frac{x^{n}(\log x)^{n}}{n!} dx = \frac{(-1)^{n}}{(n+1)^{n+1}}$

Therefore, making the appropriate substitution:
$\int_{0}^{1} x^{ax} dx = \sum_{n=0}^{\infty} a^{n} \frac{(-1)^{n}}{(n+1)^{n+1}} = -\frac{1}{a}\sum_{n+1=1}^{\infty} \frac{(-a)^{n+1}}{(n+1)^{n+1}} = -\frac{1}{a}\sum_{m=1}^{\infty} \frac{(-a)^{m}}{(m)^{m}} = -\frac{1}{a}g(-a)$

(Additional: This document discusses another direction to go in terms of generalizing the sophomore’s dream integrals)