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Integration by Parametric Differentiation

  1. Oct 12, 2009 #1
    1. The problem statement, all variables and given/known data
    Having trouble understanding this

    the example I saw was;

    Solve [tex]\int^{\infty}_{0} x^3 e^{-9x} dx[/tex] using integration by parametric differentiation.


    3. The attempt at a solution

    well, i do know how to do this, so i set out my integral;

    [tex]\int^{\infty}_{0} e^{-\alpha x} dx[/tex] = [tex]\frac{1}{\alpha}[/tex]

    Now is where i get stuck, i'm told that now I can differentiate both sides with respect to alpha, because alpha is only a parameter, which is fine for the right hand side, however on the left, sureley i'm performing an operation on x, so why don't I get a dx/dalpha ? im also at a loss to know how you can do it with the integral sign still there :O

    this is the main problem i'm having, but anyway could someone check the rest of my working; so differentiating both sides;

    [tex]-\int^{\infty}_{0}x e^{-\alpha x} dx = -\frac{1}{\alpha^2}[/tex]

    [tex]\int^{\infty}_{0}x^2 e^{-\alpha x} dx = \frac{2}{\alpha^3}[/tex]

    [tex]-\int^{\infty}_{0}x^3 e^{-\alpha x} dx = -\frac{6}{\alpha^4}[/tex]

    so the integral of

    [tex]\int^{\infty}_{0} x^3 e^{-9x} dx = \frac{6}{9^4}[/tex]

    thanks again.

    ps: i worked really hard on this LaTeX i hope you like it :biggrin:
     
  2. jcsd
  3. Oct 12, 2009 #2

    Dick

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    Nice LaTex! You are doing it exactly correctly. On the left side, you are just using the Leibniz integral rule. There's no reason to think you should get a dx/dalpha, is there?
     
  4. Oct 13, 2009 #3
    Thanks alot for that, and is it because of leinbiz's rule that you can do that?

    also i know it's alot to ask but could you explain his rule as simply as possible? :P

    [tex]
    \frac{\partial}{\partial z} \int^{b(z)}_{a(z)} f(x,z) dx = \int^{b(z)}_{a(z)} \frac{\partial f}{\partial z} dx + f(b(z),z) \frac{\partial b}{\partial z} - f(a(z),z) \frac{\partial a}{\partial z}
    [/tex]
    (i didn't do THAT latex I found it on someone else's post :P)

    i know i'm not taking university mathematics yet and it's unlikely that i'll need it until then however it's really bugging me and i'd like to be able to do it.

    the trouble is, that the notation is extremely inaccessible (for me).

    the part that's bugging me is that surely you want to define the integral you want/need in terms of other things, however on the left side you're also differentiating the integral you need, so it'll be in another form...(which you don't want)

    OR, is this rule just proving that you CAN differentiate with respect to alpha (as in original question), by bringing it inside?

    the notation is extremely scary, and also another rookie question but what does (a(z), z) mean? wouldn't that just mean ("a function of z", "including z")

    sorry again, maybe i'm just way in over my head here.
     
  5. Oct 13, 2009 #4

    Dick

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    Yes, it's saying that you can differentiate inside the integral. Just think of z as alpha. The other two terms on the RHS apply to cases where the limits also depend on alpha. You don't need them since your limits, 0 and infinity, don't depend on alpha. If your integral were from 0 to alpha^2, for example, you get a contribution from the db/dz (or db/dalpha) part where b(alpha)=alpha^2.
     
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