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This question may use special integrals.

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r satisfy x=3p+r,p is a non-negative integers,0<r<=3
(edited 6 years ago)
QQ20170722-201011@2x.png r satisfy x=3p+r,0<r<=3
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Zacken
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Original post by ElliotWalton
QQ20170722-201011@2x.png r satisfy x=3p+r,0<r<=3


Where did this come from? What is p?
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Zacken
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Fixed version here: https://www.thestudentroom.co.uk/showthread.php?t=4844016https://www.thestudentroom.co.uk/showthread.php?t=4844016
Original post by ElliotWalton
QQ20170722-201011@2x.png r satisfy x=3p+r,0<r<=3


Not a fan of trivialising integrals by abuse of just one formula, but applying the Abel-Plana Formula with f(x)=exp⁡(−2rtanh⁡−1x)/(1−x2)r=(1+x)−2rf(x) = \exp(-2r\tanh^{-1}x)/(1-x^2)^r = (1+x)^{-2r} and noting that tanh⁡−1(±it)=±itan⁡−1t\tanh^{-1}(\pm it) = \pm i \tan^{-1} t:

Unparseable latex formula:

\begin{array}{lcl}[br]\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 \\[br]\\[br]&=& \displaystyle\sum_{n=0}^{\infty} \dfrac{1}{(n+1)^{2r}} - \displaystyle\int_0^{\infty} (1+x)^{-2r} dx - \dfrac{1}{2} \\[br]\\[br]&=& \boxed{\zeta (2r) - \dfrac{2r+1}{2(2r-1)}}[br]\end{array}



Which is valid for r>1/2r>1/2.
(edited 6 years ago)
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.
QQ20170723-214742@2x.pngCalculate the limits.

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