# Integration by parts and infinite series

I was teaching tabular integration yesterday and as I was preparing, I was playing around with using it on integrands that don’t ‘disappear’ after repeated differentiation. In particular, the problem I was doing was this:

$\int x^2\ln{x}dx$

Now this is done pretty quickly with only one integration by parts:

Let $u=\ln{x}$ and $dv=x^2dx$. Then $du=\frac{1}{x}dx$ and $v=\frac{x^3}{3}$. Rewriting the integral and evaluating, we find

$\int x^2\ln{x}dx = \frac{1}{3}x^3\ln{x}-\int \left(\frac{x^3}{3}\cdot\frac{1}{x}\right)dx$

$=\frac{1}{3}x^3\ln{x}-\int \frac{x^2}{3}dx$

$= \frac{1}{3} x^3 \ln{x} - \frac{1}{9} x^3 + c$.

But I decided to try tabular integration on it anyway and see what happened. Tabular integration requires us to pick a function $f(x)$ and compute all its derivatives and pick a function $g(x)$ and compute all its antiderivatives. Multiply, then insert alternating signs and voila! In this case, we choose $f(x)=\ln{x}$ and $g(x)=x^2$. The result is shown below.

$\int x^2\ln{x}dx = \frac{1}{3}x^3\ln{x}-\frac{1}{12}x^3-\frac{1}{60}x^3-\frac{1}{180}x^3-\cdots$

$= \frac{1}{3}x^3\ln{x} - x^3 \sum_{n=0}^\infty \frac{2}{(n+4)(n+3)(n+2)(n+1)} +c$

If I did everything right, then the infinite series that appears in the formula must be equal to $\frac{1}{9}$. Checking with wolframalpha, we see that indeed,

$\sum_{n=0}^\infty \frac{2}{(n+4)(n+3)(n+2)(n+1)} = \frac{1}{9}$.

Wow!! That’s pretty wild. It seemed like any number of infinite series could pop up from this kind of approach (Taylor series, Fourier series even). In fact, they do. Here are just three nice resources I came across which highlight this very point. I guess my discovery is not so new.