How to integrate a function using a power series

Sometimes the integral is hard to recognize, but the function inside it looks like a series we already know. If we can write that function as a power series, we can integrate each power separately. Let’s see how that works with a geometric series.

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Rewrite the integrand as a geometric series

For $\left|u\right|<1$, the series ${\sum }_{n=0}^{\infty }{u}^n$ adds up to $\frac{1}{1-u}$. The denominator $1-r^6$ tells us to use $u=r^6$, then multiply by $r$.

$\begin{array}{l}\frac{r}{1-r^6}=r{\sum }_{n=0}^{\infty }{r}^{6n}\\ ={\sum }_{n=0}^{\infty }{r}^{6n+1}\end{array}$

Integrate one power at a time

Increase each exponent by one and divide by the new exponent. The resulting series gives an antiderivative on $-1<r<1$.

$\begin{array}{l}\int \frac{r}{1-r^6}dr\\ =C+{\sum }_{n=0}^{\infty }\frac{r^{6n+2}}{6n+2}\\ =C+\frac{r^2}{2}+\frac{r^8}{8}+\frac{r^{14}}{14}+\cdots \end{array}$

Find the radius and test the endpoints

The ratio of successive terms in absolute value tends to $|r{|}^6$, so the radius is $1$. At $r=1$ and $r=-1$, all the powers in the numerator are even. Both endpoint sums become ${\sum }_{n=0}^{\infty }\frac{1}{6n+2}$, which diverges by comparison with the harmonic series. The interval of convergence is $(-1,1)$.

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