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    Attachment 675830r satisfy x=3p+r,p is a non-negative integers,0<r<=3
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    Name:  QQ20170722-201011@2x.png
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Size:  18.3 KB r satisfy x=3p+r,0<r<=3
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    (Original post by ElliotWalton)
    Name:  QQ20170722-201011@2x.png
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    Where did this come from? What is p?
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    Fixed version here: https://www.thestudentroom.co.uk/sho....php?t=4844016
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    (Original post by ElliotWalton)
    Name:  QQ20170722-201011@2x.png
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    Not a fan of trivialising integrals by abuse of just one formula, but applying the Abel-Plana Formula with f(x) = \exp(-2r\tanh^{-1}x)/(1-x^2)^r = (1+x)^{-2r} and noting that \tanh^{-1}(\pm it) = \pm i \tan^{-1} t:

    \begin{array}{lcl}

\displaystyle\int _0^{\infty} \dfrac{2\sin(2r\tan^{-1}t)}{(1+t^2)^r (e^{2\pi t}-1)} dt &=& i\displaystyle\int _0^{\infty} \dfrac{f(it) - f(-it)}{e^{2\pi t} - 1}dt \\

\\

&=& \displaystyle\sum_{n=0}^{\infty} \dfrac{1}{(n+1)^{2r}} - \displaystyle\int_0^{\infty} (1+x)^{-2r} dx - \dfrac{1}{2} \\

\\

&=& \boxed{\zeta (2r) - \dfrac{2r+1}{2(2r-1)}}

\end{array}

    Which is valid for r&gt;1/2.
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    (Original post by Zacken)
    Where did this come from? What is p?
    The condition is indeed puzzling, and appears to be largely unrelated to the problem at hand. My guess is that this is a case of the question being posted without the solution being fully understood by the OP.
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    Name:  QQ20170723-214742@2x.png
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Size:  39.1 KBCalculate the limits.
 
 
 
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