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Math Help - Differentiating through the integral sign

  1. #1
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    Differentiating through the integral sign

    Here's the question:

    By integrating through the integral sign, evaluate \int_{0}^\infty (x^2 + a)^{-n} dx where n is a positive integer and a > 1.

    So the standard idea is to let F(a) = \int_{0}^\infty (x^2 + a)^{-n} dx then use (after checking various conditions hold) F'(a) = \int_{0}^\infty (-n)(x^2 + a)^{-n-1} dx by just differentiating through the integral sign, evaluating the integral and solving the resulting differential equation for F(a).

    But here we cannot easily work out the integral, so what can we do?
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  2. #2
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    Quote Originally Posted by Amanda1990 View Post
    Here's the question:

    By integrating through the integral sign, evaluate \int_{0}^\infty (x^2 + a)^{-n} dx where n is a positive integer and a > 1.

    So the standard idea is to let F(a) = \int_{0}^\infty (x^2 + a)^{-n} dx then use (after checking various conditions hold) F'(a) = \int_{0}^\infty (-n)(x^2 + a)^{-n-1} dx by just differentiating through the integral sign, evaluating the integral and solving the resulting differential equation for F(a).

    But here we cannot easily work out the integral, so what can we do?
    You should try an induction argument: if you let F_n(a)=\int_0^\infty \frac{dx}{(x^2+a)^n}, then you proved that F_n'(a)=-nF_{n+1}(a).

    Notice that F_1(a) can be evaluated. Then you get F_2(a) by differentiating F_1(a), and F_3(a) by differentiating F_2(a), and so on...
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  3. #3
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    Nice idea! I can see exactly how the induction would work but I haven't been able to find a pattern and claim what the result should be in the first place...What would the inductive hypothesis be?
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  4. #4
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    Quote Originally Posted by Amanda1990 View Post
    Nice idea! I can see exactly how the induction would work but I haven't been able to find a pattern and claim what the result should be in the first place...What would the inductive hypothesis be?
    Did you get F_1(a)=\frac{\pi}{2}\frac{1}{\sqrt{a}}? Then F_1'(a)=\frac{\pi}{2}\frac{-\frac{1}{2}}{a^{\frac{1}{2}+1}}, so F_2(a)= \frac{1}{1}\frac{\pi}{2}\frac{\frac{1}{2}}{a^{\fra  c{1}{2}+1}}. Then F_3(a)=\frac{1}{2}F_2'(a)=\frac{1}{1\cdot 2}\frac{\pi}{2}\frac{\frac{1}{2}\left(\frac{1}{2}+  1\right)}{a^{\frac{1}{2}+2}}.

    From there, you can infer F_n(a)=\frac{1}{1\cdot 2\cdots (n-1)}\frac{\pi}{2}\frac{\frac{1}{2}\left(\frac{1}{2}  +1\right)\cdots\left(\frac{1}{2}+(n-2)\right)}{a^{\frac{1}{2}+(n-1)}}. This will be straightforward to prove using the induction formula. Then you can simplify the expression, like F_n(a)=\frac{1\cdot 3\cdot 5\cdots(2n-3)}{2^n (n-1)!}\pi=\frac{(2n-3)!}{2^{2n-1}(n-1)!^2}\pi.
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