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I am trying to show that the following interchange of summation and integration is valid: $$\int_0^{+\infty} \sum_{n=1}^{+\infty} \frac{e^{-nx}}{n}\,dx=\sum_{n=1}^{+\infty} \int_0^{+\infty} \frac{e^{-nx}}{n}\,dx$$

I first tried using the Weierstrass $M$-test in order to show that the series summing the function $f_n(x)=\frac{e^{-nx}}{n}$ converges uniformly on $[0,+\infty)$, but I was unsuccessful. Indeed, I could not find a sequence $M_n$ such that $$\left|\frac{e^{-nx}}{n}\right|\le M_n \qquad x\in[0,+\infty)$$ and $$\sum_{n=1}^{+\infty}M_n\le +\infty$$ I tried $M_n=1/n$ but its series does not converge.

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  • $\begingroup$ I think the relevant interval for the Weierstrass M-Test should be $(0,\infty)$, otherwise clearly it doesn't converge. Also, the summation should go from $n=1$ instead of $0$. $\endgroup$ – Václav Mordvinov May 13 '18 at 20:57
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    $\begingroup$ The summand/integrand is non-negative, the easiest to use tool is Tonelli's theorem (for non-negative functions) $\endgroup$ – achille hui May 13 '18 at 21:02
  • $\begingroup$ @Mordvinov But would that interval help me since the integral is on $[0,+\infty)$? $\endgroup$ – Zachary May 13 '18 at 21:08
  • $\begingroup$ @achille hui Would I use Tonelli's theorem with the counting measure for my sum? $\endgroup$ – Zachary May 13 '18 at 21:08
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    $\begingroup$ @Zachary yes, Tornelli work for $\sigma$-finite measures and both the Lebesgue measure on $\mathbb{R}$ and counting measure on $\mathbb{N}$ are $\sigma$-finite. $\endgroup$ – achille hui May 13 '18 at 21:15
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The series does not converge uniformly on $(0,\infty)$ since for $x_n = 1/n$,

$$\sup_{x \in (0,\infty)}\sum_{k = n+1}^{\infty} \frac{e^{-kx}}{k}\geqslant \sum_{k = n+1}^{2n} \frac{e^{-kx_n}}{k} > n \frac{e^{-2nx_n}}{2n} = \frac{e^{-2}}{2}.$$

So you will not have any luck with the Weierstrass M-test.

However, note that as $n \to \infty$ we have monotone convergence:

$$\sum_{k=1}^n \frac{e^{-kx}}{k} \uparrow \begin{cases}+\infty, \quad x = 0\\ -\log(1 - e^{-x}), \quad x > 0 \end{cases} $$

We can apply the monotone convergence theorem, and since the limit function is Lebesgue integrable on $(0,\infty),$ we have

$$-\int_{(0,\infty)} \log(1 - e^{-x}) = \frac{\pi^2}{6} = \sum_{n=1}^\infty\frac{1}{n^2} = \sum_{n=1}^\infty \int_{(0,\infty)} \frac{e^{-nx}}{n}$$

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