I am stuck on the following question from Folland's Real Analysis book (chapter 2, problem 31b).

Derive the following formula by expanding part of the integrand into an infinite series and justifying the term-by-term integration. For p > -1, $$ \int_0^1 x^p (x-1)^{-1} dx = - \sum_{k=1}^{\infty} \frac{1}{(p+k)^2} $$

I can expand the second factor as a power series, but I don't see how that helps me. Worse yet, I don't even see how this can be correct. Taking for example $p=0$, I get

$$ \int_0^1 (x-1)^{-1} dx = \int_{-1}^0 \frac{du}{u} = - \int_{0}^1 \frac{dy}{y} $$

which as far as I know doesn't converge. I'm not sure then if Folland meant for $p \in (-1,0)$, or for $p > -1, p \neq 0$, or if the problem is just incorrect. Or maybe I am just missing something obvious. Any help would be appreciated!

  • $\begingroup$ Graphing $x^p/(x-1)$, it would appear to be meant $p\in (-1,0)$ $\endgroup$
    – Yankl
    Jun 20, 2018 at 17:40
  • $\begingroup$ @JacobMazor Can you provide more reasoning? The problem when integrating is near $x=1$, and near 1, $x^p$ approaches 1 for $p \in (-1,0)$, so it still doesn't seem likely that that $x^p/(x-1)$ is integrable over $(0,1)$. $\endgroup$ Jun 20, 2018 at 17:59
  • 3
    $\begingroup$ You are missing a factor of $\log x$: The integral as it appears in (my copy of ) Folland is $\int_0^1 x^p(1-x)^{-1}\log x\,dx$. (Actually, Folland uses the letter $a$ rather than $p$.) $\endgroup$ Jun 20, 2018 at 18:10
  • $\begingroup$ @JohnDawkins I must have a different version of Folland. I will try your version and see if that works out. $\endgroup$ Jun 20, 2018 at 19:20
  • 2
    $\begingroup$ Clearly a typo since $\int_{0}^{1}\frac{x^p}{x-1}\,dx$ is blatantly divergent due to the simple pole at $x=1$ of the integrand function. $\endgroup$ Jun 20, 2018 at 19:24

1 Answer 1


As John Dawkins stated in a comment following the OP post, the OP is missing a factor of $\log(x)$ in the integrand. Without that factor, the integrand has a non-removable singularity at $x=1$ and the integral diverges. We now proceed to evaluate the "corrected integral" that has the factor $\log(x)$ in the integrand.

For $p>-1$ we have

$$\int_0^1 x^{p+n}\log(x)\,dx=-\frac{1}{(p+n+1)^2}\tag1$$

Next, summing $(1)$ over $n$ reveals

$$\begin{align} -\sum_{n=1}^\infty \frac{1}{(p+n)^2}&=\sum_{n=0}^\infty \int_0^1 x^{p+n}\log(x)\,dx\\\\ &=\lim_{N\to \infty}\int_0^1 \frac{x^p(1-x^{N+1})}{1-x}\,\log(x)\,dx\tag 2 \end{align}$$

Noting that for $x\in [0,1]$, have $\left|\frac{x^p(1-x^{N+1})}{1-x}\,\log(x)\right|\le -2\frac{x^p}{1-x}\log(x)$ and that $0\le \int_0^1 -2\frac{x^p}{1-x}\log(x)\,dx<\infty$, the Dominated Convergence Theorem guarantees that we can interchange the limit with the integral on the right-hand side of $(2)$ yields

$$-\sum_{n=1}^\infty \frac{1}{(p+n)^2}=\int_0^1 \frac{x^p}{1-x}\,\log(x)\,dx$$

as was to be shown!

  • $\begingroup$ Was this answer useful? Please let me know how I can improve my answer. I really want to give you the best answer I can. $\endgroup$
    – Mark Viola
    Jul 21, 2018 at 18:20

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