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Problem with this integral

  1. Jun 24, 2006 #1
    i need help with this integral


    when i let u=ln(x) and v=e^x, i get

    [tex]\{e^x}{lnx}\ - \int{e^x/{x}}dx[/tex]

    and then it keeps looping
    Last edited: Jun 24, 2006
  2. jcsd
  3. Jun 24, 2006 #2


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    What do you mean by "it keeps looping?"
  4. Jun 24, 2006 #3


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    If by "looping", you mean that you let [itex]dv= \frac{1}{x}[/itex] and [itex]u= e^x[/itex] and everything canceled out leaving you with
    [tex]\int e^x lnx dx= \int e^x ln x dx[/itex]
    Of course!

    If you do integration by parts twice and the second time "swap" u and dv, that will always happen.

    Perhaps what you mean is that you again let [itex]dv= e^x[/itex] and [itex]u= \frac{1}{x}[/itex] so that you would up with
    [tex]e^x lnx dx= e^x lnx- \frac{1}{x}+ \int\frac{e^x}{x^2}dx[/itex]
    and continuing just gives you higher and higher powers in the denominator: Okay, that's good. What eventually happens is that, in the limit, you get ln x plus an infinite series. The integral cannot be done in terms of "elementary" functions.
  5. Jun 28, 2006 #4


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    If, when solving by parts, you ever end up with your original integral, you can substitute a algebraically (like substitute the letter I for the origingal integral anywhere you see it) then just solve for I and that's your solution.

    My professor broke solving by parts into three different types, and he called that one "around the world"
  6. Jun 28, 2006 #5
    Except for the case [tex] \int_0^{\infty}dxe^{-ax}lnx=-\frac{\gamma+lna}{a} [/tex] which is exact.
  7. Jun 28, 2006 #6

    matt grime

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    Your integral is a definite integral, the other is not, moreover, what is gamma?
  8. Jun 28, 2006 #7
  9. Jun 29, 2006 #8

    [tex]\text{ExpIntegralEi}[x] = - \int_{-x}^\infty \frac{e^{-t}}{t} dt[/tex]

    Mathematica solution:
    [tex]\int e^x ln{x} dx = \int_{-x}^\infty \frac{e^{-t}}{t} dt + e^x ln x[/tex]


    [tex]\gamma[/tex] - EulerGamma (Euler-Mascheroni Constant)

    [tex]\text{ExpIntegralEi}[z] = \sum_{k=1}^\infty \frac{z^k}{kk!} + \frac{1}{2} \left[ ln z - ln \left( \frac{1}{z} \right) \right] + \gamma[/tex]

    [tex]\int_{-z}^\infty \frac{e^{-t}}{t} dt = \sum_{k=1}^\infty \frac{z^k}{kk!} + \frac{1}{2} \left[ ln z - ln \left( \frac{1}{z} \right) \right] + \gamma[/tex]

    Orion1 solution:
    [tex]\int e^x ln{x} dx = \sum_{k=1}^\infty \frac{x^k}{kk!} + \frac{1}{2} \left[ ln x - ln \left( \frac{1}{x} \right) \right] + e^x ln x + \gamma[/tex]


    Last edited: Jun 29, 2006
  10. Jun 29, 2006 #9
    if "lnx" is the anle of any trignometric function. like intergration of "sinln dx "
  11. Jun 29, 2006 #10
    What? That made absolutely no sense whatsoever? ln(x) is the natural logarithm of x.
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