∫((ln(x^2+1))/x^2,x,-(∞),∞) -> 4*pi

[ftneek]

Hello, I noticed this improper integral ∫((ln(x^2+1))/x^2,x,-(∞),∞) returns 4*π but I believe the answer to be 2*π.

We can split the interval at x=0 and integrate from (-inf,0) + (0,+inf), taking the limit at the endpoints.

The indefinite integral ∫((ln(x^2+1))/x^2,x) returns 2*atan(x)-ln(x^2+1)/x

limit(2*atan(x)-ln(x^2+1)/x,x,0,-1) -> 0
limit(2*atan(x)-ln(x^2+1)/x,x,0,1) -> 0
limit(2*atan(x)-ln(x^2+1)/x,x,inf) -> π
limit(2*atan(x)-ln(x^2+1)/x,x,-inf) -> -π

Putting it all together:

∫((ln(x^2+1))/x^2,x,-(∞),∞) = (π - 0) + (0 - (-π)) = 2*π

What's the reason for the difference by a factor of 2?

[parisse]

I did not check that 1/x^2 roots were real, and was using incorrectly the residue formula.