JEEOlympiad
Evaluate I = \int_0^{1} \frac{(\ln x)^2}{1+x^2}\,dx. It equals…
Question
Evaluate I = \int_0^{1} \frac{(\ln x)^2}{1+x^2}\,dx. It equals \frac{m}{n}\pi^3 with m/n in lowest terms. Find m+n.
✓ Verified answer: 17checked by our engine — not a guess
Step-by-step solution
Expand \frac{1}{1+x^2}=\sum_{k\ge0}(-1)^k x^{2k} for 0<x<1, then I = \sum_{k\ge0}(-1)^k \int_0^1 x^{2k}\ln^2 x\,dx.
The standard integral \int_0^1 x^{m}\ln^2 x\,dx = \frac{2}{(m+1)^3}.
With m=2k: I = \sum_{k\ge0}(-1)^k \frac{2}{(2k+1)^3} = 2\sum_{k\ge0}\frac{(-1)^k}{(2k+1)^3} = 2\beta(3), where \beta is the Dirichlet beta function.
Since \beta(3)=\frac{\pi^3}{32}, we get I = \frac{\pi^3}{16}.
So m=1, n=16 and m+n = 17.
Final answer17
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