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Integration with hyperbolic secant

  1. Apr 25, 2017 #1
    1. The problem statement, all variables and given/known data

    Solve ##\displaystyle{d\sigma = \frac{d\rho}{\cosh\rho}.}##

    2. Relevant equations

    3. The attempt at a solution

    The answer is ##\displaystyle{\sigma = 2 \tan^{-1}\text{sinh}(\rho/2)}##. See equation (10.2) in page 102 of the lecture notes in http://www.hartmanhep.net/topics2015/gravity-lectures.pdf. There is a typo in the equation.

    Let us first try to check by differentiation. Using ##\displaystyle{d\tan^{-1}(x) = \frac{dx}{1+x^{2}}}##, we have

    ##\displaystyle{d\sigma = \frac{2\sinh'(\rho/2)d\rho}{1+\sinh^{2}(\rho/2)}}##

    ##\displaystyle{d\sigma = \frac{\cosh(\rho/2)d\rho}{\cosh^{2}(\rho/2)}}##

    ##\displaystyle{d\sigma = \frac{d\rho}{\cosh(\rho/2)}}##

    Is this correct?
     
    Last edited: Apr 25, 2017
  2. jcsd
  3. May 1, 2017 #2

    Ssnow

    User Avatar
    Gold Member

    Hi, I think the error is in your derivation:

    ## d\sigma=\frac{2\sinh'(\rho/2)d\rho/2}{1+\sinh^2{(\rho/2)}}=\frac{d\rho/2}{\cosh{(\rho/2)}}##

    bercause you work with differentials you have ##1/2## in addiction ...
    Ssnow
     
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