Interchanging the summation and integral sign

Several integrals without a closed form can be solved by converting it into a summation. The same goes for summations. Let us see one example, for which we will need the knowledge of Taylor series.

Find $$\int_0^1 \dfrac{\ln(1-x)}{x} dx.$$

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First, use the taylor series of $\ln(1-x)$: $$\int_0^1 \dfrac{\ln(1-x)}{x} dx=\int_0^1 \sum_{n=1}^\infty \dfrac{x^{n-1}}{n} dx.$$ We now use a trick: since the summation is absolutely convergent (skip ahead to learn what this is), we can interchange the summation and integral. We then have $$\sum_{n=1}^\infty\int_0^1 \dfrac{x^{n-1}}{n} dx=\sum_{n=1}^\infty \dfrac{1}{n^2} = \dfrac{\pi^2}{6}.\ _\square$$