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An idea for integration that has been bugging me.

  1. Apr 14, 2008 #1
    I've only recently started teaching myself Calculus, so you'll have to forgive me if I'm trying to do something silly or impossible. This has been bugging me for a few days, so I figured it's high time I ask someone.

    Since we can write [tex]\int_{a}^b f(x)[/tex] as [tex]\int_{a}^b\ e^\ln(f(x))[/tex] would it be possible to use substitution to yank [tex]ln(f(x))[/tex] out of the integral and leave us with

    [tex]g(\int_{ln(f(a))}^\ln(f(b))} e^x } )[/tex]

    With g(x) as what we pulled out of the integral. I'm very new to integration, but this idea has been puzzling me for a while. If this is possible, it would seem pretty dang handy when integrating. Then again, I'm playing with methods I've only recently taught myself (I came up with this while running on a tredmil, actually.) It's fully possible I misunderstood something or overlooking a rule that makes this impossible. Anyway, thanks for any information that flies my way.
  2. jcsd
  3. Apr 14, 2008 #2


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    I'm intrigued by what you're trying to do here, and am not too much a veteran at integration myself, but I don't understand where the g() came from. I can see your first step, as they are equivalent, and the second because they evaluate to the same thing.. but I don't see the g().
  4. Apr 14, 2008 #3


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    Well if as you say [tex]g(x) = \ln{f(x)}[/tex], i.e. g(x) is what we pulled out.
    Now try evaluating your specific g(x):
    [tex]g(\int_{ln(f(a))}^\ln(f(b))} e^x } ) = \ln{f(\int_{ln(f(a))}^\ln(f(b))} e^x })}[/tex]

    Now if you plug g(x) back into the original integral you get a whole mess.

    In reality your specific g(x) is the following:

    [tex]g(\int_{ln(f(a))}^\ln(f(b))} e^x } ) = g(f(b)-f(a))[/tex] by the fundamental theorem of calculus, which still makes no sense when you plug it back into the original integral.

    What you may be thinking of is u substitution. Suppose you had the following integral:
    then setting [tex]u = x^{2}[/tex] you get [tex]du = 2xdx[/tex] and the integral can be re-written as
    [tex]\int e^{u}du[/tex]
    and the rest of the integration is simple.
  5. Apr 14, 2008 #4
    Hmm, I don't know if I fully follow. I'm not trying to integrate e to the power of something, but rather define a function in terms of [tex]e^{ln(f(x))}[/tex], which then becomes e to the power of something, and then integrate in terms of that. And yeah, I got the idea from u-substitution (hence I even plugged my original ln(f(x)) into the bounds of integration.)

    I think I failed to convey myself. g( ) was a mistake, I was trying to put something 'outside' of the integral to represent what was removed. I'm not sure how to denote that though, so I just plugged the integral into a function -- I think that wasn't the best of ideas, and changed my intentions into something else.
  6. Apr 14, 2008 #5


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    Sorry it was confusing, I just tried to follow what you had ;p

    Suppose you wanted to do u substitution on your original then
    [tex] u = ln(f(x))[/tex] and [tex]du = \frac{f'(x)}{f(x)} [/tex]
    which doesn't help you with the particular integration.

    There may be an f(x) for which your idea is beneficial, but I can't think of one at the moment ;(
  7. Apr 14, 2008 #6
    I think it would have to be a function s.t. [itex]f(x) \propto \frac{f'(x)}{f(x)}[/itex] in order to play nicely with the u-sub, no?
  8. Apr 14, 2008 #7
    Nooooo! My dreams of becoming famous for finding a universal one step method of integration are dashed!

    Okay, I'll admit it, I had a dream like that one time. =P
  9. Apr 14, 2008 #8


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    Yes. I guess you could have some horrendous integral like this:
    [itex]\int g(x)f(x)dx[/itex] where [itex]g(x)=\frac{f'(x)}{f(x)}[/itex]

    And for some reason you couldn't reduce the integral to just be f'(x)dx.
  10. Apr 15, 2008 #9
    what the hell are you guys talking about?
  11. Apr 15, 2008 #10
    This really makes absolutely no sense.

    First, this ([tex]f(x)=e^{ln(f(x))}[/tex] ) only makes sense when f is a strictly positive function.

    Second, maybe you try this "technique" with a specific f, say, f(x)=x^2+1, before doing anything.
    Last edited: Apr 15, 2008
  12. Apr 15, 2008 #11
    To my understanding, the natural log of a value is the number which when used as the exponet of e brings it to that specific value. Hence I figured any output, y, of f(x), is the same as e to the power of the natural log of that value. That would be [tex]e^{ln(y)}[/tex], so I changed out y for f(x), since they're equivalent (f(x)=y.) I might be mistaken, I suppose that might should have been read f(x) ==> y (if that makes a difference?)

    As for negative fuction outputs, those did come to mind. I'm not the formost expert on complex numbers, but I was under the impression that we could use the complex field to solve for any negative values the function yields. I recall, for example, reading that [tex]ln(-1)[/tex] is plus or minus [tex]i * pi[/tex]. That would mean if f(x)=2x, and x= -0.5 then [tex]e^{ln(2(-0.5))} = { e^{ln{(-1)} } = {e^{i*pi}[/tex].

    Unless I'm completely mistaken...?
  13. Apr 15, 2008 #12
    OK, so let's see it: calculate the integral of f(x)=2x from x=-1 to x=1 using your technique.
  14. Apr 15, 2008 #13
    I tried, but I failed. I'm no good with using logorythms. That's a major problem with being self-taught, I don't know what I need to focus on learning and what I don't. Would you mind teaching me how to integrate this, or point me to a decent source? I probably need to invest in a real text book, instead of these terrible study guides I have. As it is, I only understand elementary integration, and I haven't had much time to become creative when applying it either.


    Wait... now I'm REALLY confused. I think I did solve it, but the results were impossible so I dismissed it.

    I don't know now.
    Last edited: Apr 15, 2008
  15. Apr 20, 2008 #14

    Gib Z

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    My tip is to give up on this method, get those real text books and relearn from the basics. To integrate what Doodle Bob said is quite easy if you look at its graph and the bounds. If you don't see that after this hint, you back to basics. No offense intended here.
  16. Apr 21, 2008 #15

    Although this method you came up with proved to be ineffective, don't let everybody telling you you're wrong discourage you. By defenition of psychology, intelligence is "the ability to think abstractly"..paraphrased. In other words, it is that kind of thinking that leads to great discoveries in mathematics; just because you don't discover a new method while you are in Calc I doesn't mean it won't ever happen. Keep it up.

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