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Homework Help: Stuck on an integral

  1. Mar 2, 2006 #1
    i have a double integral to reverse the order of and then integrate, i have reversed the order fine, however i am VERY stuck on the integration of the function in the first integral

    integrate sin(y)/(x+y) dx between 0 and y

    any pointers greately appreciated
     
  2. jcsd
  3. Mar 2, 2006 #2

    Galileo

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    You're integrating wrt x, right? So sin(y) is a contant as far as the integration is concerned.
     
  4. Mar 2, 2006 #3
    so it would just be sin(y) * integral 1/(x+y)

    therefore sin(y) * ln(x+y)

    if so im going to feel like a fool :P
     
  5. Mar 2, 2006 #4
    if this is the case i have now integrated it and subbed for (y) (from the limits) to get sin(y)*ln(2y) but i must now integrate that function wrt y, which is even harder than the first function >.< and i am very stuck, maybe integration by parts here?
     
  6. Mar 3, 2006 #5

    benorin

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    Don't feel like a fool; the thing is so. Post the bounds so we can help...
     
  7. Mar 3, 2006 #6
    ok the original integral was between y and 0 hence getting sin(y)*ln(2y)

    now the integral is that function ive just typed between pi/2 and 0 dy

    im muchos stuck
     
  8. Mar 3, 2006 #7

    dextercioby

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    [tex] -\int_{0}^{\frac{\pi}{2}} \sin y \ \ln 2y \ dy =-\left[ \mbox{Ci}\left( \frac{1}{2}\pi \right) +\ln 2-\gamma\right][/tex].

    Daniel.
     
  9. Mar 3, 2006 #8

    Galileo

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    The bounds on the first integral are y and 0 right? So just enter the bounds correctly. Don't skip too many steps or you might miss a simplification:
    [tex]\ln(x+y)|^{x=y}_{x=0}=\ln(2y)-\ln(y)=\ln(\frac{2y}{y})=\ln2[/tex]

    Which doesn't even depend on y, so the integral is easy as pie.
     
  10. Mar 3, 2006 #9
    thanks :)

    i just forgot to seperate the variables and made life hard on myself
     
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