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I'm studying for a qualifying exam and I'm stuck on this problem from Bass's "Real Analysis for Graduate Students" (Exercise 7.14). It asks us to prove that $$\sum_{k=1}^\infty\frac{1}{(p+k)^2}=-\int_0^1\frac{x^p}{1-x}\log(x)dx$$ for $p>0$.

Note that this exercise comes from the chapter that introduces the monotone convergence theorem, Fatou's lemma, and the dominated convergence theorem. My issue is probably that I don't immediately see how to apply any of these theorems to this particular problem. I have tried playing around with Feynman's trick and log series, but haven't made any notable progress. I've been stuck here for awhile so any help is appreciated.

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  • $\begingroup$ Maybe you could write the RHS into an integral of an infinite sum, then interchange the $\int$ and $\sum$ according to the convergence theorems. $\endgroup$ – xbh Jun 2 at 3:32
  • $\begingroup$ @xbh Right, so one of the things I mentioned playing around with was writing the log(x) term in series form. I tried doing exactly what you mentioned, but I wasn't able to get a convenient integrand. $\endgroup$ – zbrads2 Jun 2 at 3:41
  • $\begingroup$ Try the Gamma function. $\endgroup$ – xbh Jun 2 at 3:41
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First, expand $1/(1-x)$ into its geometric series to get $$-\int_0^1\sum_{k\ge 0} x^mx^p\log x\,dx$$ Now, consider the partial sums $$S_N=\sum_{k=0}^Nx^k=\frac{1-x^{N+1}}{1-x}$$ We can easily show that $S_N\le S_{N+1}$, as $S_{N+1}-S_N=x^{N+1}\ge 0$ as $x\in[0,1]$. By the monotone convergence theorem, \begin{align} -\int_0^1\sum_{k\ge 0} x^k x^p\log x\,dx&=-\int_0^1\lim_{N\to\infty}\sum_{k= 0}^N x^k x^p\log x\,dx \\ &=-\lim_{N\to\infty}\int_0^1\sum_{k= 0}^N x^k x^p\log x\,dx \tag{1} \\ &=-\lim_{N\to\infty}\sum_{k= 0}^N\int_0^1 x^k x^p\log x\,dx \\ &=-\sum_{k\ge 0}\int_0^1 x^k x^p\log x\,dx \\ &=\sum_{k\ge 0}\frac{1}{(k+p+1)^2} \tag{2}\\ &=\sum_{k\ge 1}\frac{1}{(k+p)^2} \\ \end{align} Where the monotone convergence theorem was used in $(1)$ and integration by parts in $(2)$.

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  • $\begingroup$ Thanks for pointing out where the monotone convergence theorem was used. This is exactly how I just did it myself. $\endgroup$ – zbrads2 Jun 2 at 4:05
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Proceeding naively,

$\begin{array}\\ \int_0^1\frac{x^p}{1-x}\log(x)dx &=\int_0^1x^p\log(x)\sum_{n=0}^{\infty} x^ndx\\ &=\sum_{n=0}^{\infty} \int_0^1x^{p+n}\log(x)dx\\ &=\sum_{n=0}^{\infty} \dfrac{x^{n+p+1}((n+p+1)\ln(x)-1}{(n+p+1)^2}|_0^1 \qquad\text{(according to Wolfy)}\\ &=\sum_{n=0}^{\infty}\dfrac1{(n+p+1)^2} (x^{n+p+1}((n+p+1)\ln(x)-1)|_0^1\\ &=\sum_{n=0}^{\infty}\dfrac{-1}{(n+p+1)^2}\\ &=\sum_{n=1}^{\infty}\dfrac{-1}{(n+p)^2}\\ \end{array} $

Looks OK to me.

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  • $\begingroup$ Let $t = -\log x$, then the integral is actually $-\Gamma (2)/(n+p+1)^2$. $\endgroup$ – xbh Jun 2 at 3:43
  • $\begingroup$ Ofcourse! The geometric series because $x$ is in $(0,1)$. I was so settled on writing the log as a series, I didn't see this. Thanks. $\endgroup$ – zbrads2 Jun 2 at 3:44

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