# Finding an integral using a series

## Homework Statement

$\displaystyle \int_0^1 \frac{\arctan x}{x}dx$

## The Attempt at a Solution

I converted the integral to the following; $\displaystyle \int_0^1 \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{2n+1}dx$. In this case am I allowed to swap the summation and integral signs?

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Mark44
Mentor

## Homework Statement

$\displaystyle \int_0^1 \frac{\arctan x}{x}dx$

## The Attempt at a Solution

I converted the integral to the following; $\displaystyle \int_0^1 \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{2n+1}dx$. In this case am I allowed to swap the summation and integral signs?
Since the series is absolutely convergent on the interval in question, yes, you can exchange the operations. Here's a link to another thread that discusses this -- https://www.physicsforums.com/threads/interchanging-summation-with-integral-differentiation-with-integral.564620/.

Ray Vickson
Homework Helper
Dearly Missed
Since the series is absolutely convergent on the interval in question, yes, you can exchange the operations. Here's a link to another thread that discusses this -- https://www.physicsforums.com/threads/interchanging-summation-with-integral-differentiation-with-integral.564620/.
This argument does not quite work: the series is absolutely convergent on the open interval $0 \leq x < 1$ but is only conditionally convergent at $x=1$. However, the OP can try another argument instead: he can check for uniform convergence on the closed interval, and that will be enough. See, eg.,
https://www.dpmms.cam.ac.uk/~agk22/uniform.pdf .

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Mark44
Mentor
This argument does not quite work: the series is absolutely convergent on the open interval $0 \leq x < 1$ but is only conditionally convergent at $x=1$. However, the OP can try another argument instead: he can check for uniform convergence on the closed interval, and that will be enough. See, eg.,
https://www.dpmms.cam.ac.uk/~agk22/uniform.pdf .
I was a bit worried about that endpoint at x = 1, but didn't include my concern in what I wrote.