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Math Help - Three impossible integrals and two challenging ones

  1. #1
    MHF Contributor Mathstud28's Avatar
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    Three impossible integrals and two challenging ones

    Show that

    \int_0^{\frac{\pi}{2}}\tan^{x}(\theta)d\theta=\fra  c{\pi}{2\cos\bigg(\frac{x\pi}{2}\bigg)}


    and

    \int_0^{\infty}\frac{arctan(ax)-arctan(bx)}{x}=\frac{\pi}{2}\ln\bigg(\frac{a}{n}\b  igg)


    \int_0^{\frac{\pi}{2}}\sqrt{\cos(x)}dx=\frac{(2\pi  )^{\frac{3}{2}}}{[\Gamma(\frac{1}{4})]^2}


    and

    \int_0^{\frac{\pi}{2}}\frac{dx}{(a^2\sin^2(x)+b^2\  cos^2(x))^2}=\frac{\pi(a^2+b^2)}{4a^3b^3}

    and finally

    Show that the area enclosed by a curve defined through the equation

    x^{\frac{b}{c}}+y^{\frac{b}{c}}=a^{\frac{b}{c}}

    where a>0

    c is a positive odd integer and b is a positive even integer is

    \frac{\bigg[\Gamma\bigg(\frac{c}{b}\bigg)\bigg]^2}{\Gamma\bigg(\frac{2c}{b}\bigg)}\cdot\bigg(\fra  c{2ca^2}{b}\bigg)


    Come on people lets see that skill
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  2. #2
    MHF Contributor

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    well, they are actually pretty easy. i'll do the first three for now:

    Quote Originally Posted by Mathstud28 View Post
    Show that

    \int_0^{\frac{\pi}{2}}\tan^{x}(\theta)d\theta=\fra  c{\pi}{2\cos\bigg(\frac{x\pi}{2}\bigg)}

    here we need to have |x| < 1. then we have:

    \int_0^{\frac{\pi}{2}} \tan^{x} \theta \ d\theta = \int_0^{\frac{\pi}{2}} \sin^x \theta \cos ^{-x} \theta \ d\theta = \frac{1}{2}B \left(\frac{1+x}{2}, \frac{1-x}{2} \right)

    =\frac{1}{2}\Gamma \left(\frac{1+x}{2} \right) \Gamma \left(1 - \frac{1+x}{2} \right)=\frac{\pi}{2\sin(\frac{1+x}{2} \pi)} by Euler's Reflection formula

    =\frac{\pi}{2 \cos(\frac{\pi x}{2})}. \ \ \square


    \int_0^{\infty}\frac{arctan(ax)-arctan(bx)}{x}=\frac{\pi}{2}\ln\bigg(\frac{a}{n}\b  igg)
    that n is actually b. i'll assume 0 < b < a. we have:

    \int_0^{\infty} \frac{\arctan(ax) - \arctan(bx)}{x} dx= \int_0^{\infty} \int_b^a \frac{1}{1+x^2y^2}dy dx

    =\int_b^a \int_0^{\infty} \frac{1}{1 + x^2y^2} dxdy=\int_b^a \frac{\pi}{2y}dy=\frac{\pi}{2}\ln(\frac{a}{b}). \ \ \ \square


    \int_0^{\frac{\pi}{2}}\sqrt{\cos(x)} dx=\frac{(2\pi)^{\frac{3}{2}}}{[\Gamma(\frac{1}{4})]^2}
     \int_0^{\frac{\pi}{2}} \sqrt{\cos x} \ dx =\frac{1}{2}B \left(\frac{1}{2},\frac{3}{4}\right)=\frac{\Gamma(  \frac{1}{2}) \Gamma(\frac{3}{4})}{2\Gamma(\frac{5}{4})}=\frac{\  sqrt{\pi} \Gamma(\frac{1}{4}) \Gamma(\frac{3}{4})}{2\Gamma(\frac{5}{4})\Gamma(\f  rac{1}{4})}. \ \ \ \ \ (*)

    now, again, by Euler's Reflection formula: \Gamma(\frac{1}{4}) \Gamma(\frac{3}{4})=\frac{\pi}{\sin(\frac{\pi}{4})  }=\pi \sqrt{2}, and

    clearly \Gamma(\frac{5}{4})=\frac{1}{4} \Gamma(\frac{1}{4}). put these two in (*) to get the result. Q.E.D.
    Last edited by NonCommAlg; May 18th 2008 at 10:15 PM.
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by Mathstud28 View Post

    Show that

    \int_0^{\infty}\frac{arctan(ax)-arctan(bx)}{x}=\frac{\pi}{2}\ln\bigg(\frac{a}{n}\b  igg)
    See here.
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