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Improper integration change of variables

  1. Jun 12, 2010 #1
    1. The problem statement, all variables and given/known data


    [tex]\int _1^{\infty }\frac{1}{x^2}\text{Log}[x]dx=-\int_0^1 \text{Log}[x] \, dx [/tex]

    similarly show

    [tex] \int _0^{\infty }\frac{1}{x^2+1}\text{Log}[x]dx = 0 [/tex]

    3. The attempt at a solution

    For the first part a substitution 1/x works.

    The second part I cannot do, I thought about

    [tex] \int _0^{\infty }\frac{1}{x^2+1}\text{Log}[x]dx=\int _1^{\infty }\frac{1}{x^2+1}\text{Log}[x]dx+\int _0^1\frac{1}{x^2+1}\text{Log}[x]dx [/tex]

    and then trying to maybe show

    [tex] \int _1^{\infty }\frac{1}{x^2+1}\text{Log}[x]dx=-\int _0^1\frac{1}{x^2+1}\text{Log}[x]dx [/tex]

    but for now I am not sure what to do.
  2. jcsd
  3. Jun 12, 2010 #2


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    Science Advisor
    Homework Helper

    Try substituting u=1/x in one of your final two integrals.
  4. Jun 12, 2010 #3
    [tex] \int _1^{\infty }\frac{1}{x^2+1}\text{Log}[x]dx=\int _1^0\frac{1}{u^2+1}\text{Log}du [/tex]

    [tex] \int _1^{\infty }\frac{1}{x^2+1}\text{Log}[x]dx+\int _0^1\frac{1}{x^2+1}\text{Log}[x]dx=\int _0^{\infty }\frac{1}{x^2+1}\text{Log}[x]dx=0 [/tex]

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