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Homework Help: Evaluate the improper integral.

  1. Jun 28, 2008 #1
    [tex]\int_0^\infty \; \frac{ \ln\;(1+x^2)}{ x^2+2x\;\cos\;\theta + 1 }\;\;dx[/tex]

    [tex]\theta \in \mathbb{R}[/tex]
     
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  3. Jun 29, 2008 #2

    HallsofIvy

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  4. Jun 29, 2008 #3
    Indeed why?

    But If you don't know how to do this I think a contour integration is the best approach.
     
  5. Jun 29, 2008 #4

    nicksauce

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    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)
     
  6. Jun 29, 2008 #5
    Oh wow, thats a very insightful answer.
     
  7. Jun 30, 2008 #6

    Gib Z

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    Differentiation under the integral sign looks like it'll work here.
     
  8. Jun 30, 2008 #7
    What function should you take: [tex]\ln(ax^2 +1) [/tex]?
     
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