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Thread: 2 integrals

  1. #1
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    2 integrals

    I have some problems proving that:

    $\displaystyle \int_0^{2\pi}$ d$\displaystyle \theta$ / (a+b cos$\displaystyle \theta$) = $\displaystyle 2\pi$ / $\displaystyle \sqrt{a^2-b^2}$, if a>|b| and a,b in $\displaystyle \mathbb{R}$.

    $\displaystyle \int_0^1$$\displaystyle \log$($\displaystyle \sin$$\displaystyle \pi$x)dx=-$\displaystyle \log $$\displaystyle 2$

    Thanks.
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  2. #2
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    Re: 2 integrals

    Quote Originally Posted by Veve View Post
    I have some problems proving that:

    $\displaystyle \int_0^{2\pi}$ d$\displaystyle \theta$ / (a+b cos$\displaystyle \theta$) = $\displaystyle 2\pi$ / $\displaystyle \sqrt{a^2-b^2}$, if a>|b| and a,b in $\displaystyle \mathbb{R}$.

    $\displaystyle \int_0^1$$\displaystyle \log$($\displaystyle \sin$$\displaystyle \pi$x)dx=-$\displaystyle \log $$\displaystyle 2$

    Thanks.
    For the first you need to use the Weierstrauss Substitution.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Re: 2 integrals

    Quote Originally Posted by Veve View Post
    I have some problems proving that:

    $\displaystyle \int_0^{2\pi}$ d$\displaystyle \theta$ / (a+b cos$\displaystyle \theta$) = $\displaystyle 2\pi$ / $\displaystyle \sqrt{a^2-b^2}$, if a>|b| and a,b in $\displaystyle \mathbb{R}$.

    $\displaystyle \int_0^1$$\displaystyle \log$($\displaystyle \sin$$\displaystyle \pi$x)dx=-$\displaystyle \log $$\displaystyle 2$

    Thanks.
    For the first one, you can do the following $\displaystyle \displaystyle I=\int_0^{2\pi}\frac{dx}{a+b\cos(x)}=-2i\int_0^{2\pi} \frac{ie^{ix}}{2ae^{ix}+b+be^{2ix}}\;=-2i\oint_{|z|=1}\frac{1}{2az+b+bz^2}$. Now, if $\displaystyle \displaystyle f(z)=\frac{-2i}{2az+b+bz^2}$ you can check that $\displaystyle f$ has a poles at precisely one place on or in the interior of $\displaystyle |z|=1$, namely one at the interior point $\displaystyle \displaystyle z_0=\frac{\sqrt{a^2-b^2}-a}{b}$. We see then by Cauchy's Integral Formula then that $\displaystyle \displaystyle I=2\pi i \,\text{Res}(f,z_0)=\frac{2\pi}{\sqrt{a^2-b^2}}$.



    If you look down the solution to this problem you will see I solved precisely the second integral.
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