How would one evaluate the integral ∫∞0arctan(x)dxx(1+x2)
?
I was told it had a nice closed form and could have been solved with differentiation under the integral sign; however, I tried to set I(α)=∫∞0arctan(αx)dxx(1+x2)
and got nowhere (the resulting integral was very messy). Is there a much more clever substitution that can be used to tackle the integral?
Answer
Rewrite
arctanxx=∫10du1+x2u2
Then plugging this in and reversing the order of integration, we get the integral; value as
∫10duu2∫∞0dx(11u2+x211+x2)=∫10du1−u2∫∞0dx(11+x2−11u2+x2)=∫10du1−u2π2(1−u)=π2log2
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