# Interchanging the summation and integral sign

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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 \[\int_0^1 \dfrac{\ln(1-x)}{x} dx.\]

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 \[\sum_{n=1}^\infty\int_0^1 \dfrac{x^{n-1}}{n} dx=\sum_{n=1}^\infty \dfrac{1}{n^2} = \dfrac{\pi^2}{6}.\ _\square\]

**Cite as:**Interchanging the summation and integral sign.

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