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Interesting integral formula (please help)

  1. Apr 17, 2012 #1
    Please derive the right-hand side from the left. If anyone could shed some light on this I would be grateful.

    [itex]\frac{d^{2}}{dx^{2}}\int_x^{∞}f(x,z)g(z)dz = \int_x^{∞}f_{xx}(x,z)g(z)dz - g(x)(f_{z}(x,x) + 2f_{x}(x,x)) - f(x,x)g'(x) [/itex]

    Please, if you can, provide a detailed explanation of your steps. An attempt I made at getting to the right-hand side of the equation was to consider a parameter α in [x,∞) and using this to
    split the integral into two. Then apply the fundamental theorem of calculus to the one while assuming that the differentiation and integration may be interchanged freely on the other. Although, it seems at the end of this calculation one should set α=x and repeat the previous process to arrive at the final result after once again setting the new parameter equal to x. Setting the parameters that have been in [x,∞) to x, while seemingly valid since x is in this set, seems a bit peculiar. It seems as though this is only a trick to change the form of the integral to the right-hand side, this may be fine after all but I would be interested to hear if anyone has some suggestions about this.
     
  2. jcsd
  3. Apr 17, 2012 #2

    HallsofIvy

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    This is pretty much a standard application of "Leibniz' rule" which is, itself, an extension of the fundamental theorem of Calculus as you suggest. The full statement of Leibniz' rule is
    [tex]\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} f(x,t)dt= f(x, \beta(x))\frac{d\beta}{dx}- f(x, \alpha(x))\frac{d\alpha}{dx}+ \int_{\alpha}^{\beta}\frac{\partial f}{\partial x}dt[/tex]

    Here, [itex]\beta(x)= \infty[/itex] and [itex]\alpha(x)= x[/itex] so that [itex]d\beta/dx=0[/itex], [itex]d\alpha/dx=1[/itex] so the first derivative is
    [tex]-f(x,x)g(x)+ \int_x^\infty f_x g dx[/tex]

    Now, use that same rule to differentiate again.
     
  4. Apr 17, 2012 #3
    Thanks! Yes, I considered Leibniz's rule. What had thrown me off was the fact that the upper limit of integration is ∞. This appears to work just fine, although I am concerned with the notion that [itex]\frac{d∞}{dx}[/itex] should be zero. Might I instead handle this by integrating on [x,α] and then applying Leibniz's rule? With this, since α is not a function of x, [itex]\frac{dα}{dx}=0[/itex]. Then the first term on the right-hand side will vanish after which the limit as α tends to ∞ can be taken giving the desired result. Does this sound reasonable to you?
     
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