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I would like to find $$\int_{0}^{1}\frac{\ln x}{1-x^2} dx$$ using power series approach by first finding the power series of $$\frac{\ln x}{1-x^2} $$ and then showing that it is uniformly convergent and thus we can interchange summation and integration. So the integral is equal to $$-\sum_{n=0}^{\infty } \frac{1}{(2n+1)^2} $$May I ask how can I approach the question? I could find the power series expansion but upon integration, it does not seem to get me anywhere near the form that I would like to get...

Giorgos Giapitzakis
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dilemma
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    See https://math.stackexchange.com/questions/1913260/find-integral-int-01-frac-ln-x1-x2-mathrmdx? – Robert Z Nov 16 '21 at 13:41
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    Hint:- Note $$\sum_{k\ge 0}\frac{1}{2k+1}=\sum_{k\ge 1}\frac{1}{k}-\sum_{k\ge 1}\frac{1}{2k}$$. – RAHUL Nov 16 '21 at 13:50

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Since you already proved that, $$\int_{0}^{1}\frac{\ln x}{1-x^{2}} dx=-\sum_{k\ge 0}\frac{1}{(2n+1)^{2}}$$ To evaluate just note that, $$-\sum_{k\ge 0}\frac{1}{(2n+1)^{2}}=-\sum_{k\ge 1}\frac{1}{k^{2}} +\sum_{k\ge 1}\frac{1}{4k^2}$$ So it's the series of zeta function i.e $\zeta(s)$ evaluated at $s=2$ which was proved to be $\frac{\pi^{2}}{6}$. So the required sum will evaluate to, $$-\frac{\pi^{2}}{6}+\frac{\pi^{2}}{24}$$.

2 is even prime
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