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For context, I was reading this Math.SE answer which is a solution to the integral $I=\int_0^{\infty} \frac{\ln \cos^2 x}{x^2} \,dx=-\pi$ and was confused when reading this part:

$$ 2I = \int_{-\infty}^\infty\frac{\ln(\cos^2x)}{x^2} \,dx = \color{red}{\sum_{n=-\infty}^{+\infty}} \left(\color{red}{\int_{n\pi}^{(n+1)\pi}} \frac{\ln(\cos^2x)}{x^2} \,dx\right). $$

My question is, when does the following hold? $$\int_{-\infty}^\infty f(x) \,dx = \sum_{n=-\infty}^\infty \int_{n\pi}^{(n+1)\pi} f(x) \,dx$$

Thank you.

Amanuensis Frances
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    What is your definition of the improper integral? – Severin Schraven Nov 16 '22 at 16:59
  • @SeverinSchraven The improper integral is a definite integral evaluated using a limit because one or both endpoints of integration are either discontinuous or $\pm\infty$, I think? The one in the post is an example but I don't know how to justify changing it into an infinite series (Can one just simply do that?). I'm sorry if the question in the post should be trivial, I'm still self-studying integral calculus. – Amanuensis Frances Nov 16 '22 at 17:08
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    It follows from your definition of indefinite integral (write it as a limit, split it for finite endpoints) $$ \int_{-\infty}^\infty f(x) dx = \lim_{N\rightarrow \infty} \int_{-N}^N f(x) dx = \lim_{N\rightarrow \infty} \sum_{n=-N}^{N-1} \int_{n\pi}^{(n+1)\pi} f(x)dx = \sum_{n=-\infty}^\infty \int_{n\pi}^{(n+1)\pi} f(x)dx.$$ – Severin Schraven Nov 16 '22 at 18:37

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