Evaluate the improper integral.

[mathwizarddud]

\int_0^\infty \; \frac{ \ln\;(1+x^2)}{ x^2+2x\;\cos\;\theta + 1 }\;\;dx

\theta \in \mathbb{R}

 

[HallsofIvy]

Why?

 

[dirk_mec1]

Why?

Indeed why?

But If you don't know how to do this I think a contour integration is the best approach.

 

[nicksauce]

From Maple:

(1/2)*(ln(cos(theta)-sqrt(cos(theta)^2-1))*ln(I/(cos(theta)-sqrt(cos(theta)^2-1)+I))+ln(cos(theta)-sqrt(cos(theta)^2-1))*ln(-I/(cos(theta)-sqrt(cos(theta)^2-1)-I))+dilog(I/(cos(theta)-sqrt(cos(theta)^2-1)+I))+dilog(-I/(cos(theta)-sqrt(cos(theta)^2-1)-I)))/sqrt(cos(theta)^2-1)-(1/2)*(ln(cos(theta)+sqrt(cos(theta)^2-1))*ln((-2*sqrt(cos(theta)^2-1)+I)/(cos(theta)-sqrt(cos(theta)^2-1)+I))+ln(cos(theta)+sqrt(cos(theta)^2-1))*ln(-(2*sqrt(cos(theta)^2-1)+I)/(cos(theta)-sqrt(cos(theta)^2-1)-I))+dilog((-2*sqrt(cos(theta)^2-1)+I)/(cos(theta)-sqrt(cos(theta)^2-1)+I))+dilog(-(2*sqrt(cos(theta)^2-1)+I)/(cos(theta)-sqrt(cos(theta)^2-1)-I)))/sqrt(cos(theta)^2-1)

 

[Santa1]

Oh wow, that's a very insightful answer.

 

[Gib Z]

Differentiation under the integral sign looks like it'll work here.

 

[dirk_mec1]

Differentiation under the integral sign looks like it'll work here.

What function should you take: \ln(ax^2 +1)?