Thursday, 19 November 2015

calculus - How to integrate inti0nftyfracarctan(x)dxx(1+x2)?



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



10duu20dx(11u2+x211+x2)=10du1u20dx(11+x211u2+x2)=10du1u2π2(1u)=π2log2


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