This integral came up while I was trying to evaluate $$\sum_{n=1}^\infty \frac {1}{n^2}.$$ The value should be $$\frac {\pi^2}{6},$$ but how do I solve the integral and evaluate it?
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@AnginaSeng He's trying to use that to evaluate the sum – Gareth Ma Mar 29 '20 at 06:13
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Does this answer your question? https://math.stackexchange.com/questions/1023832/evaluate-lim-n-rightarrow-infty1-frac1n1-frac2n-frac12 – jeremy909 Mar 29 '20 at 06:19
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I see, sorry haha – Gareth Ma Mar 29 '20 at 06:21
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Look at this answer. – metamorphy Mar 29 '20 at 07:53
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Does this answer your question? Different methods to compute $\sum\limits_{k=1}^\infty \frac{1}{k^2}$ (Basel problem) – mrtaurho Mar 29 '20 at 13:44
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Substitute $t=1-x$ to reexpress the integral as
$$\int_0^1 -\frac {\ln(1-x)}{x}dx = \int_0^1 \frac {\ln t}{t-1}dt = \frac {\pi^2}6$$
where $\int_0^1 \frac {\ln x}{x-1}dx = \frac {\pi^2}6$ is used in the last step
Quanto
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$$I=\int_{0}^{1} \frac{\ln(1-x)}{x}dx= \int_{0}^{1} \sum_{k=1}^{\infty}\frac{x^{k-1}}{k}=\sum_{k=1}^{\infty} \frac{1}{k^2}=\zeta(2)=\frac{\pi^2}{6}.$$
Z Ahmed
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Using the definition of polylogarithms $$I(a)=-\int_0^a \frac {\ln(1-x)}{x}dx=\text{Li}_2(a)$$ Expandinga series around $a=1$ gives $$I(a)=\frac {\pi^2}6-\frac{1}{4} (1-a) (5-a-2 (3-a) \log (1-a))+\cdots$$
If you use this approximation for $a=0.75$, it will give $0.989$ while the exact value is $0.978$.
Claude Leibovici
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