Hey there! Sign in to join this conversationNew here? Join for free

This question may use special integrals. Watch

Announcements
    • Thread Starter
    Offline

    0
    ReputationRep:
    Attachment 675830r satisfy x=3p+r,p is a non-negative integers,0<r<=3
    • Thread Starter
    Offline

    0
    ReputationRep:
    Name:  QQ20170722-201011@2x.png
Views: 25
Size:  18.3 KB r satisfy x=3p+r,0<r<=3
    Offline

    21
    ReputationRep:
    (Original post by ElliotWalton)
    Name:  QQ20170722-201011@2x.png
Views: 25
Size:  18.3 KB r satisfy x=3p+r,0<r<=3
    Where did this come from? What is p?
    Offline

    21
    ReputationRep:
    Fixed version here: https://www.thestudentroom.co.uk/sho....php?t=4844016
    Offline

    17
    ReputationRep:
    (Original post by ElliotWalton)
    Name:  QQ20170722-201011@2x.png
Views: 25
Size:  18.3 KB 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(-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.
    Offline

    17
    ReputationRep:
    (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.
    • Thread Starter
    Offline

    0
    ReputationRep:
    Name:  QQ20170723-214742@2x.png
Views: 16
Size:  39.1 KBCalculate the limits.
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    How are your GCSEs going so far?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.