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Homework Help: Double Integration problem

  1. Nov 3, 2013 #1
    1. The problem statement, all variables and given/known data

    “Fun Homework Problem” assigned for extra credit...
    ∫ 0 to ∞ [(e^-x)-(e^-3x)]/x dx

    2. Relevant equations (supplied hints)

    I can get an abstract answer from wolfram but it’s not how the professor wants us to do it.
    He gave us a couple “hints”; try to introduce a second variable and reverse the boundries
    for example: ∫∫ f(y) f(x) dy dx? I’m assuming he means integrate with respect to y first.
    A second possible hint was to try to get to ∫ 1/x dx by getting -[(e^-x)-(e^-3x)] on top to cancel the other.

    3. The attempt at a solution

    The biggest problem I’m running into is the x on the bottom.
    Can I set new boundries by introducing 1 to y+1? I feel like the only way to do this problem is by getting +n in the denominator in order to prevent division by 0 when evaluating 0 to infinity.
    I’ve also tried multipling by x/x in order to get an x^2 and move it to the top but the following integration by parts seems to get me no where.

    Is there a trig sub I’m not seeing?

    Even my professor admits he’s having trouble with it, but “knows” it’s possible.

    Thanks for any help...
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Nov 3, 2013 #2


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    Scrutinize the integral [itex]\int_{a}^{b}e^{-yx}dy[/itex]
  4. Nov 3, 2013 #3


    Yeah, that x on the bottom. How about this: I haven't worked this problem yet ok but that's not the point I wish to make. Rather, the essential quality in successful mathematics of trying things and cultivating a tolerance of not getting discouraged when they don't work but rather, motivating yourself into finding other things to try.

    Ok, back to that x on the bottom. Well, that reminds me of another problem recently in here:


    Now, I'm not sure that's even relevant to solving this problem but again, that's not the point I'm trying to make. :)
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