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Math Help - An interesting integral

  1. #1
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    An interesting integral

    solve  \int_{0}^{\frac{\pi}{2}} \cos{[\ln{(\tan{x})}]} ~dx
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  2. #2
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    Quote Originally Posted by simplependulum View Post
    solve  \int_{0}^{\frac{\pi}{2}} \cos{[\ln{(\tan{x})}]} ~dx
    it's easy to find the answer as an infinite series: put \tan x = e^t. then:  I= \int_{0}^{\frac{\pi}{2}} \cos{[\ln{(\tan{x})}]} ~dx =2 \int_0^{\infty} \frac{e^t \cos t}{1+e^{2t}} \ dt=2 \int_0^{\infty} (e^{-t} - e^{-3t} + e^{-5t} - \cdots ) \cos t \ dt.

    we know that for any s > 0: \ \int_0^{\infty}e^{-st} \cos t \ dt = \frac{s}{s^2 + 1}. therefore I=2 \sum_{n=0}^{\infty} (-1)^n\frac{2n+1}{(2n+1)^2 + 1}. now, does this alternating series have a closed form ?
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  3. #3
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    My answer is in terms of  \pi ~ and hyperbolic cosine ( I hope i am correct ) but not in series form .


    Actually this theorem  | {\gamma( \frac{1}{2} + bi ) |}^2 = \frac{ \pi}{ \cosh( \pi b)} inspired me to think of this question .
    Last edited by simplependulum; June 21st 2009 at 02:19 AM.
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  4. #4
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    Check it whether your results are as same as my result

      \int_{0}^{ \frac{ \pi}{2}} \cos[k \ln( \tan{x} ) ] ~dx =







     \frac{ \pi}{ 2 \cosh{ \pi \frac{k}{2} }}
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  5. #5
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    Yes, the substitution t=\ln(\tan x) transforms the integral I=\int_0^{\pi/2}\cos(k\ln(\tan x))\textrm d x to \int_{-\infty}^\infty\frac{\textrm e^t\cos kt}{1+\textrm e^{2t}}\textrm d t, which can be written as \int_{-\infty}^\infty\frac{\cos kt}{2\cosh t}\textrm d t.

    We may assume that k>0 since cosine is even, and since sine is odd we may replace \cos kt with \textrm e^{\textrm i kt} as well. Thus I=\int_{-\infty}^\infty\frac{\textrm e^{\textrm i kt}}{2\cosh t}\textrm d t, a Fourier transform.

    Now consider the integral of the function f(z)=\frac{\textrm e^{\textrm i kz}}{2\cosh z} around the rectangular contour C with vertices -S, R, R+\textrm i\pi and -S+\textrm i\pi, where R,\, S>0, in the complex plane.

    The function f(z) is holomorphic inside and on C except at a simple pole z_0=\textrm i\pi/2, and its residue there is \lim_{z\to z_0}(z-z_0)f(z)=\frac{\textrm e^{\textrm ikz_0}}2\lim_{z\to z_0}\frac{z-z_0}{\cosh z}<br />
=\frac{\textrm e^{-k\pi/2}}2\frac1{\sinh z_0}=\frac{\textrm e^{-k\pi/2}}{2\textrm i}.

    By the residue theorem \oint_C f(z)\textrm dz=2\pi\textrm i\times\frac{\textrm e^{-k\pi/2}}{2\textrm i}=\pi\textrm e^{-k\pi/2}.

    Also \oint_C f(z)\textrm dz=\int_{-S}^R f(t)\textrm dt+\int_0^\pi f(R+\textrm it)\textrm i\textrm dt+\int_R^{-S}f(t+\textrm i\pi)\textrm dt+\int_\pi^0f(-S+\textrm it)\textrm i\textrm dt upon integrating along each side of C.

    Now f(t+\textrm i\pi)=-\textrm e^{-k\pi}f(t), |f(R+\textrm it)|\leq\frac1{2\sinh R} and |f(-S+\textrm it)|\leq\frac1{2\sinh S} in these integrals. Thus

    \pi\textrm e^{-k\pi/2}=(1+\textrm e^{-k\pi})\int_{-S}^R f(t)\textrm dt+I_R+I_S, where |I_R|\leq\int_0^\pi|f(R+\textrm it)|\textrm dt\leq \frac\pi{2\sinh R} and similarly for I_S. Letting R,\, S\to\infty we have I_R,\,I_S\to 0.

    Therefore \pi\textrm e^{-k\pi/2}=(1+\textrm e^{-k\pi})\int_{-\infty}^\infty f(t)\textrm dt, i.e. I=\frac{\pi\textrm e^{-k\pi/2}}{1+\textrm e^{-k\pi}}=\frac\pi{2\cosh(k\pi/2)}.
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  6. #6
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    Thank you very much , halbard

    Besides solving contour integral , i have another method to solve it

    Since  \Gamma(1-z) \Gamma(z) = \frac{ \pi}{\sin{ \pi z}}~~ (1)

    And  \beta( x , y ) = 2 \int_0^{\frac{\pi}{2}} \sin^{2x-1}(t) \cos^{2y-1}(t)~dt~~ (2)

    Then substitute  z = \frac{1}{2} + ki ~~ into (1) and
     x = \frac{1}{2} + ki  ,  y = \frac{1}{2} - ki ~~ into (2)

     \Gamma( \frac{1}{2} + ki  ) \Gamma( \frac{1}{2} - ki ) = \frac{ \pi}{\cosh{ \pi k}}~~ and

     \beta( \frac{1}{2} + ki , \frac{1}{2} - ki ) = 2 \int_0^{\frac{\pi}{2}} \tan^{2ki}(x)~dx


     \beta( \frac{1}{2} + ki , \frac{1}{2} - ki ) = 2 \int_0^{\frac{\pi}{2}} e^{i 2k\ln[\tan(x)]} ~dx

    But  \beta( \frac{1}{2} + ki , \frac{1}{2} - ki ) =  \frac{\Gamma( \frac{1}{2} + ki  ) \Gamma( \frac{1}{2} - ki )}{ \Gamma(1) }

    So after equating these two equations , we obtain

     2 \int_0^{\frac{\pi}{2}} e^{i 2k\ln[\tan(x)]} ~dx  =  \frac{ \pi}{\cosh{ \pi k }}

    Take the real part of the integral ,

     \int_0^{\frac{\pi}{2}} \cos[2k\ln(\tan{x})]~dx = \frac{\pi}{2 \cosh{ \pi k} }

    Replace  k ~~ by  \frac{k}{2} ~~ ,

      \int_0^{\frac{\pi}{2}} \cos[k\ln(\tan{x})]~dx = \frac{\pi}{2 \cosh{\frac{ \pi k}{2}} }
    Last edited by simplependulum; July 4th 2009 at 02:42 AM.
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