Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Problem with an Integral

  1. Apr 21, 2005 #1

    Pyrrhus

    User Avatar
    Homework Helper

    [tex] \int \frac{1 - x \exp{xy}}{\frac{1}{y} + x \ln {y}} dx [/tex]

    Here's what i have done

    [tex] \int \frac{1 - x \exp{xy}}{\frac{1}{y} + x \ln {y}} dx = \int \frac{1}{\frac{1}{y} + x \ln {y}} dx + \int \frac{- x \exp{xy}}{\frac{1}{y} + x \ln {y}} dx [/tex]

    so

    [tex] I_{1} = I_{2} + I_{3} [/tex]

    where

    [tex] I_{1} = \int \frac{1 - x \exp{xy}}{\frac{1}{y} + x \ln {y}} dx [/tex]

    [tex] I_{2} = \int \frac{1}{\frac{1}{y} + x \ln {y}} dx [/tex]

    [tex] I_{3} = \int \frac{- x \exp{xy}}{\frac{1}{y} + x \ln {y}} dx [/tex]

    Now,

    [tex] I_{2} = \int \frac{1}{\frac{1}{y} + x \ln {y}} dx [/tex]

    multiplying by [itex] \frac{\ln{y}}{\ln{y}} [/itex]

    [tex] I_{2} = \frac{\ln{y}}{\ln{y}} \int \frac{1}{\frac{1}{y} + x \ln {y}} dx [/tex]

    [tex] I_{2} = \frac{1}{\ln{y}} \int \frac{\ln{y}}{\frac{1}{y} + x \ln {y}} dx [/tex]

    Using substitution [itex] u = \frac{1}{y} + x \ln {y} [/itex] and therefore [itex] du = \ln{y} dx [/itex]

    Now,

    [tex] I_{2} = \frac{1}{\ln{y}} \int \frac{1}{u} du [/tex]

    [tex] I_{2} = \frac{1}{\ln{y}} \ln{|\frac{1}{y} + x \ln {y}|} [/tex]

    Any ideas about [itex] I_{3} [/itex] or the whole integral?
     
  2. jcsd
  3. Apr 21, 2005 #2

    I dont think thats fair game.
     
  4. Apr 21, 2005 #3

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    There you go

    [tex] \int \frac{1-xe^{xy}}{\frac {1}{y}+x\ln y}dx=\allowbreak \frac{\ln \left( \frac {1}{y}+x\ln y\right) }{\ln y}-\frac {1}{y\ln y}e^{xy}-\frac {1}{y\ln^{2}y}e^{-\frac 1{\ln y}}\func{Ei}\left( 1,-xy-\frac 1{\ln y}\right) +C[/tex]


    Daniel.
     
  5. Apr 22, 2005 #4

    Pyrrhus

    User Avatar
    Homework Helper

    Thanks Dexter :smile:, i guess my first integral was correct ([itex] I_{2} [/itex]) but the second one ([itex] I_{3} [/itex]) , whoa what a mess...

    By the way, any idea for solving the [itex] I_{3} [/itex] ?, i haven't been able to see anything.... maybe i can rewrite it someway... i'll let you all know if i can.

    Whozum? what do you mean? y is being kept as a constant...
     
    Last edited: Apr 22, 2005
  6. Apr 22, 2005 #5

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    You'll have to dig it somehow (make algebraic calculations and prossibly changes of integration variable) as to get it in the standard form of the Ei function.



    Daniel.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Problem with an Integral
  1. Integration problem (Replies: 3)

  2. Integration problem (Replies: 4)

  3. Integration problems (Replies: 3)

  4. Integral problem (Replies: 3)

Loading...