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Integrating a natural log in double integral

  1. May 12, 2013 #1
    1. The problem statement, all variables and given/known data

    [itex]\int^{1}_{0}[/itex][itex]\int[/itex][itex]^{e^x}_{e^-x}[/itex][itex]\frac{lny}{y}[/itex]dydx

    The attempt at a solution

    So I am integrating ln(y)/y and I tried it by parts, first with u = ln(y), dv = 1/y, and therefore du = 1/y, and v = ln y

    but if I use that I get

    (ln(y))2-[itex]\int[/itex][itex]\frac{lny}{y}[/itex] again.

    So I tried switching u and v around. I got: u = 1/y, dv = ln(y) dy, du = 1/y2 and v = 1/y.

    On that basis I get [itex]\frac{1}{y^2}[/itex]-[itex]\int[/itex][itex]\frac{1}{y^3}[/itex]|[itex]^{e^x}_{e^-x}[/itex]dx

    and from there I get

    [itex]\int^{1}_{0}[/itex] [itex]\frac{1}{y^2}[/itex]-[itex]\frac{1}{4y^4}[/itex]|[itex]^{e^x}_{e^-x}[/itex]dx

    But I have a sneaking suspicion I have done something horribly wrong. I see the integral I want as (ln(y))2/2. But that doesn't make sense to me.

    So, I did something messed up. If someone could tell me where it is, that would be much appreciated.
     
  2. jcsd
  3. May 12, 2013 #2

    SammyS

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    Try substitution.

    Let u = ln(y).

     
  4. May 12, 2013 #3
    I did that the first time, and showed what I got. u=ln y so du = 1/y, correct? Or not? If that's so and I et dv = 1/y then v has to be ln y, no? "try substitution" is what I just did.
     
  5. May 12, 2013 #4

    SammyS

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    No.

    You tried integration by parts.

    Do integration by substitution, with the substitution u = ln(y). No v is needed.
     
  6. May 12, 2013 #5
    ah, i see (I was thinking u-substitutions in integration by parts) -- NOW It is a lot clearer. Thanks much.
     
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