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I have to compute the following limit and Lebesgue integral:

$$\lim\limits_{n \rightarrow +\infty} \int_{[0,+\infty)} \ln \left(2 + \frac{x^2}{n} \right)\frac{1}{x^2 +1}d\mu$$

I already looked up into some questions about a practical computation of a Lebesgue integral, but none was clear to me. So far, what I've done in my attempt to solve the problem consists in, basically, compute the limit:

$$\lim\limits_{n \rightarrow +\infty} \ln \left(2 + \frac{x^2}{n} \right)\frac{1}{x^2 +1} = \frac{\ln \left(2 \right)}{x^2 +1}$$

So that:

$$\lim\limits_{n \rightarrow +\infty} \int_{[0,+\infty)} \ln \left(2 + \frac{x^2}{n} \right)\frac{1}{x^2 +1}d\mu = \int_{[0,+\infty)} \frac{\ln \left(2 \right)}{x^2 +1} d \mu$$

And then, I evaluate the RHS as a Riemann integral:

$$\int_{[0,+\infty)} \frac{\ln \left(2 \right)}{x^2 +1} d \mu = \int_0^{+\infty} \frac{\ln \left(2 \right)}{x^2 +1} d x = \ln(2) \tan^{-1}(x) \bigg|_{0}^{+ \infty} = \ln(2) \frac{\pi}{2}$$

But I have two problems, or doubts, with this procedure:

  1. Am I even allowed to do those two last steps or are they plenty wrong?
  2. If the last step is incorrect then how do we integrate with respect to the measure $d \mu$? I think that there must be a way to reduce this integral to a Riemann integral if I compute some explicit expression for $d\mu(x)$ for the set of integration, namely $[0, +\infty)$, or something of the sort, but I'm not so sure if this is really the case or how to do this in practice.

Any help would be appreciated.

Nicolas Agote
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Gustavo Silva
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  • Welcome. Please use more descriptive titles – FShrike Oct 09 '22 at 21:58
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    You're not allowed to do the steps unless you can justify the limit interchange, e.g. with dominated convergence – FShrike Oct 09 '22 at 22:00
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    It depends on your measure $\mu$. – copper.hat Oct 09 '22 at 22:02
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    See here and here for references to compute this integral exactly, for every $n$! – FShrike Oct 09 '22 at 22:03
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    $$I_n=\frac{\pi}{2}\ln(2)+\pi\cdot\ln\left(1+\frac{1}{\sqrt{2n}}\right)\overset{n\to\infty}{\longrightarrow}\frac{\pi}{2}\ln(2)$$ – FShrike Oct 09 '22 at 22:05
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    You can use the dominated convergence theorem https://en.wikipedia.org/wiki/Dominated_convergence_theorem to justify taking limit under integral sign. In your case $f_n=\ln \left(2 + \frac{x^2}{n} \right)\frac{1}{x^2 +1}\leqslant g(x)=\frac{\ln(2+x^2)}{x^2 +1}$ for all $n\geqslant1$ – Svyatoslav Oct 09 '22 at 22:22

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