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Math Help - Integral

  1. #1
    Moo
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    Integral

    Hi

    So this is something that came from a problem we did in probability. I hope it's a correct result, since I modified it a bit.
    So I know the probabilitistic way to prove it. If you can find, then do it (but you'd have to know where you're going).
    Otherwise, I don't know if there is a calculus approach lol, hence this thread.

    Find \int_{\mathbb{R}} \frac{1}{(1+u^2)^k} ~du
    for any positive integer k.

    Answer should be :
    Spoiler:
    \sqrt{\pi} \cdot \frac{\Gamma\left(k-\frac 12\right)}{\Gamma(k)}
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  2. #2
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    One method would be to adapt the reduction formula in this thread (modifying it so as to work for the interval (-\infty,\infty) instead of [0,1]).
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  3. #3
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    another way: letteing t=\sin^2 \theta in the beta function formula we get B(x,y)=2 \int_0^{\frac{\pi}{2}} (\sin \theta)^{2x-1} (\cos \theta)^{2y-1} \ d \theta, \ \ x > 0, \ y> 0.

    now suppose \frac{1}{2} < k \in \mathbb{R} and let u=\tan \theta. then: I_k=\int_{\mathbb{R}} \frac{du}{(1 + u^2)^k}=2 \int_0^{\frac{\pi}{2}} \cos^{2k-2} \theta \ d \theta=B \left(\frac{1}{2}, k -\frac{1}{2} \right). therefore:

    I_k=\frac{\Gamma (\frac{1}{2}) \Gamma (k - \frac{1}{2})}{\Gamma(k)}= \frac {\sqrt{\pi}\Gamma (k - \frac{1}{2})}{\Gamma(k)}.
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  4. #4
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    Quote Originally Posted by Moo View Post
    Hi

    So this is something that came from a problem we did in probability. I hope it's a correct result, since I modified it a bit.
    So I know the probabilitistic way to prove it. If you can find, then do it (but you'd have to know where you're going).
    Otherwise, I don't know if there is a calculus approach lol, hence this thread.

    Find \int_{\mathbb{R}} \frac{1}{(1+u^2)^k} ~du
    for any positive integer k.

    Answer should be :
    Spoiler:
    \sqrt{\pi} \cdot \frac{\Gamma\left(k-\frac 12\right)}{\Gamma(k)}
    Here is another way (I'm in complex this term so here is goes)

    Consider the closed arc in the complex plane. Let R be real and R > 1

    \gamma_1=-R+2R(t), 0 \le t \le 1 and \gamma_2=Re^{it}, 0 \le t \le \pi

    This forms a simple closed curve in the complex plane.

    By the cauchy integral formula

    \oint \frac{1}{(z^2+1)^k}dz=\oint \frac{\frac{1}{(z+i)^k}}{(z-i)^k}dz=\frac{2\pi i}{(k-1)!}\frac{d^k}{dz^k}\left( \frac{1}{(z+i)^k}\right)_{z=i}=\frac{\pi}{2^{2k-2}}\left( \frac{(2k-2)!}{((k-1)!)^2}\right)

    Now if we consider

    \int_{\gamma_1} \frac{1}{(z^2+1)^k}dz +\int_{\gamma_2} \frac{1}{(z^2+1)^k}dz

    \int_{-R}^{R} \frac{1}{(x^2+1)^k}dx +\int_{\gamma_2} \frac{1}{(z^2+1)^k}dz

    Note that on the circular arc that \left| \frac{1}{(z^2+1)^k}\right| \le \frac{1}{(R^2-1)^k}

    Using the ML estimate

    \left| \int_{\gamma_2} \frac{1}{(z^2+1)^k}dz\right| \le \frac{\pi R}{(R^2-1)^k}

    Now if we take the limit as R \to \infty

    We end up with

    \int_{-\infty}^{\infty}\frac{1}{(1+x^2)^k}=\frac{\pi}{2^{  2k-2}}\left( \frac{(2k-2)!}{((k-1)!)^2}\right)
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