# Evaluate the improper integral.

1. Jun 28, 2008

### mathwizarddud

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

$$\theta \in \mathbb{R}$$

2. Jun 29, 2008

### HallsofIvy

Staff Emeritus
Why?

3. Jun 29, 2008

### dirk_mec1

Indeed why?

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

4. Jun 29, 2008

### 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)

5. Jun 29, 2008

### Santa1

Oh wow, thats a very insightful answer.

6. Jun 30, 2008

### Gib Z

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

7. Jun 30, 2008

### dirk_mec1

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