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Math Help - Integral Section

  1. #1
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    Integral Section

    Well, I decided to create a section so that we may collect some integrals that it seems to us, nice, hard, etc.

    Integral #1

    \forall a>0, we have that I(a)=\int_0^{\pi/2}\frac{\ln(1+a\sin^2x)}{\sin^2x}\,dx=\pi\left(\sq  rt{a+1}-1\right)

    Proof:

    I'(a)=\int_0^{\pi/2}\frac1{1+a\sin^2x}\,dx. Let's set u=\tan x and the integral becomes to

    I'(a)=\int_0^\infty\frac1{1+(a+1)u^2}\,du=\frac\pi  {2\sqrt{a+1}}\,\therefore\,I(a)=\pi\sqrt{a+1}+k. Now plug I(0)=0\implies k=-\pi, which finally yields

    \color{red}\boxed{\color{blue}I(a)=\pi\left(\sqrt{  a+1}-1\right)}
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  2. #2
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    \int_0^{\infty} \frac{dx}{x^n+1} = \frac{\pi/n}{\sin (\pi/n)} for n\geq 2 \mbox{ and }n\in \mathbb{Z}.
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  3. #3
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    But you need to show us the proof
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  4. #4
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    Quote Originally Posted by Krizalid View Post
    But you need to show us the proof
    Okay, but it uses Complex Analysis.

    The brilliant move here is not the computation but the contour we chose. Consider a counter \gamma composed of 3 parts: \gamma_1,\gamma_2,\gamma_3 in the complex plane. Let \gamma_1 be the horizontal section on the real axis from 0\mbox{ to }R. Let \gamma_2 be the counterclockwise rotation by \frac{2\pi}{n}. And \gamma_3 be the line from the ending point of \gamma_1 to the origin. So \gamma is a piecewise smooth simple closed curve.

    Thus, consider f(z) = \frac{1}{z^n+1}.

    By the residue theorem we have,
    \int_0^R \frac{dx}{x^n+1} + \int_0^{2\pi/n} \frac{Rie^{i\theta}d\theta}{R^ne^{in\theta}+1} - \int_0^R \frac{e^{2\pi i/n}dt}{R^n e^{2\pi i/n\cdot n}+1} = 2\pi i\sum \mbox{res}.

    Now the only pole within the contour is when z^n+1=0\implies z=e^{\pi i/n} by de Moiver's theorem. The residue is \frac{1}{nz^{n-1}}\big|_{e^{\pi i/n}} = - \frac{1}{n}e^{\pi i/n}.

    Thus,
    \int_0^R \frac{dx}{x^n+1} + \int_0^{2\pi/n} \frac{Rie^{i\theta}d\theta}{R^ne^{in\theta}+1} - \int_0^R \frac{e^{2\pi i/n}dt}{R^n e^{2\pi i/n\cdot n}+1} =- \frac{2\pi i}{n}e^{\pi i/n}.

    Now, \int_0^R \frac{e^{2\pi i/n}dt}{R^n e^{2\pi i/n\cdot n}+1} = e^{2\pi i/n}\int_0^R \frac{dt}{t^n+1}.

    And, \left| \frac{Rie^{i\theta}}{R^ne^{in\theta}+1} \right|  = \frac{|Rie^{\theta}|}{|R^ne^{in\theta}+1|}\leq \frac{R}{|R^ne^{in\theta}|-1} = \frac{R}{R^n-1} \mbox{ for big }R.
    Since n\geq 2 this "error term" goes to zero.

    Thus, as R\to \mbox{ME},
    (1-e^{2\pi i/n})\int_0^{\infty} \frac{dx}{x^n+1} = -\frac{2\pi i}{n}e^{\pi i/n}.

    Thus,
    \int_0^{\infty} \frac{dx}{x^n+1}dx = \frac{2\pi i e^{\pi i/n}}{e^{2\pi i/n}-1} = \frac{\pi /n}{\sin \pi/n}.


    The point of this is that you (Krizalid) should learn Complex Analysis as quickly as possible if you wanna do some nice integrals.
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