Definite integral

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#1 h Mar 28, 2025, 11:58 PM

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If $n$ is a nonnegative integer, evaluate $\int_0^1 \frac{x^n}{1 + x^2}\,dx$ .

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#2 h Mar 29, 2025, 12:25 AM

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We have $a_0=\pi/4, a_1={{\ln 2}\over 2}$ and the recursion $a_n+a_{n+2}=\int_0^1 x^n dx={1\over {n+1}}$ .

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#3 h Mar 31, 2025, 1:03 PM • 1 Y

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As far as I know, when using the word "evaluate", recursion is not really enough, because the questioner is curious about the final result, right? Based on the previous recursion, these are the following, and this is the only way to complete the task:
${{a}_{2k}}={{\left( -1 \right)}^{k}}\left( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+...-\frac{1}{2k+1}-\frac{\pi }{4} \right)$ or
${{a}_{2k+1}}={{\left( -1 \right)}^{k-1}}\left( \frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+...-\frac{1}{2k}-\frac{\ln 2}{2} \right)$

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#4 h Apr 3, 2025, 7:45 PM

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You're on the right track, but the pattern is more subtle. Check your formulas against these four values calculated by hand and verified by $\textit{Maple}$ :
$a_0 = \frac{\pi}{4}, \quad \quad a_1 = \frac{\ln 2}{2}, \quad \quad a_2 = 1 - \frac{\pi}{4}, \quad \quad a_3 = \frac{1}{2} - \frac{\ln 2}{2}.$

This post has been edited 1 time. Last edited by PolyaPal, Apr 3, 2025, 7:46 PM
Reason: I forgot the dollar signs to indicate LaTeX text.

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#5 h Apr 9, 2025, 4:18 PM

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One form of the solution is
$I_n = (-1)^{n/2}\cdot \frac{\pi}{4} + \sum_{k = 0}^{(n - 2)/2} \frac{(-1)^k}{n - 2k -1}$ for $n$ even and not zero. Directly, $I_0 = \pi/4$ . For $n$ odd and not equal to 1, we have
$I_n = (-1)^{(n - 1)/2}\cdot \frac{\ln 2}{2} + \sum_{k = 0}^{(n - 3)/2} \frac{(-1)^k}{n - 2k - 1}$ and $I_1 = \ln 2/2$ .

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