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Math Help - Yet another integral challenge question

  1. #1
    Super Member Random Variable's Avatar
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    Yet another integral challenge question

    Challenge Question:

     \int^{\infty}_{-\infty} \sin (x) \arctan \Big(\frac{1}{x}\Big) \ dx



    Moderator editor: Approved Challenge question.
    Last edited by mr fantastic; May 11th 2010 at 03:33 AM.
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  2. #2
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    Quote Originally Posted by Random Variable View Post
    Challenge Question:

     \int^{\infty}_{-\infty} \sin (x) \arctan \Big(\frac{1}{x}\Big) \ dx



    Moderator editor: Approved Challenge question.
    Spoiler:
    well, if i've done everything (which starts with by parts and ends very quickly) correctly at 3:38am, the answer should be \pi(1-e^{-1}).
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  3. #3
    Super Member Random Variable's Avatar
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    NonCommAlg

    So are you saying that this problem is way to easy?


    Anyways, here's what I did (which sounds like exactly what you did):


     \int^{\infty}_{-\infty} \sin(x) \arctan\Big(\frac{1}{x}\Big) \ dx

     = \lim_{a \to 0^{-}} \int^{a}_{-\infty} \sin(x)  \arctan\Big(\frac{1}{x}\Big) \ dx + \lim_{b \to 0^{+}}\int^{\infty}_{b}  \sin(x) \arctan\Big(\frac{1}{x}\Big) \ dx

     = \lim_{a \to 0^{-}} \Bigg(-\cos(x) \arctan\Big(\frac{1}{x}\Big)  \Big|^{a}_{-\infty} - \int^{a}_{-\infty} \frac{\cos x}{1+x^{2}} \ dx  \Bigg) + \lim_{b \to 0^{+}} \Bigg(-\cos(x) \arctan\Big(\frac{1}{x}\Big)  \Big|^{\infty}_{b} - \int^{\infty}_{b} \frac{\cos x}{1+x^{2}} \ dx  \Bigg)

    = \frac{\pi}{2} - \lim_{a \to 0^{-}} \int^{a}_{-\infty} \frac{\cos  x}{1+x^{2}} \ dx + \frac{\pi}{2} - \lim_{b \to 0^{+}} \int^{\infty}_{b}  \frac{\cos x}{1+x^{2}} \ dx

     = \pi - \int^{\infty}_{-\infty} \frac{\cos x}{1+x^{2}} \ dx

    let  f(z) = \frac{e^{iz}}{1+z^{2}}

    then  \int^{\infty}_{-\infty} \sin(x) \arctan\Big(\frac{1}{x}\Big) \ dx = \pi +  2\pi \ Im (Res[f,i])

     = \pi + 2 \pi \ Im \Big(\lim_{z \to i} \frac{e^{iz}}{1+i} \Big) =  \pi + 2  \pi Im \Big(\frac{-i}{2e} \Big) = \pi \Big(1-\frac{1}{e} \Big)
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  4. #4
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    well, first of all the intagrand is an even function and so we just need to find the integral over (0, \infty). then by parts gets us somewhere that we only need to find I=\int_0^{\infty} \frac{\cos x}{1+x^2} \ dx, as you showed yourself. to find the value of I i didn't use contour integration:

    let f(t)=\int_0^{\infty} \frac{\cos (tx)}{1+x^2} \ dx, \ t>0. first note that \frac{1}{1+x^2}=\int_0^{\infty} e^{-xy} \sin y \ dy and \frac{y}{t^2+y^2}=\int_0^{\infty}e^{-xy} \cos (tx) \ dx. so:

    f(t)=\int_0^{\infty} \int_0^{\infty} e^{-xy} \cos (tx) \sin y \ dy \ dx = \int_0^{\infty} \sin y \int_0^{\infty} e^{-xy} \cos (tx) \ dx \ dy

    =\int_0^{\infty} \frac{y \sin y}{t^2+y^2} \ dy.

    on the other hand, f'(t)=-\int_0^{\infty} \frac{x\sin(tx)}{1+x^2} \ dx=-\int_0^{\infty} \frac{x \sin x}{t^2+x^2} \ dx = -f(t). thus f'(t)=-f(t) with f(0)=\int_0^{\infty} \frac{dx}{1+x^2}=\frac{\pi}{2}. solving this simple differential equation gives us f(t)=\frac{\pi}{2}e^{-t}
    and therefore I=f(1)=\frac{\pi}{2e}.
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  5. #5
    Super Member Random Variable's Avatar
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    well, first of all the intagrand is an even function and so we just need to find the integral over




    Here's another approach:

     2 \int^{\infty}_{0} \sin (x) \arctan\Big(\frac{1}{x}\Big) \ dx = 2 \int^{\infty}_{0} \int^{1}_{0} \frac{x \sin x}{t^{2}+x^{2}} \ dt \ dx

    switch the order of integration

     = \pi \int^{1}_{0} e^{-t} \ dt (again I used contour integration)

     = \pi \Big(1-\frac{1}{e} \Big)
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  6. #6
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    it'd be nice to find an elementary proof of \int_0^{\infty} \frac{\cos x}{1+x^2} \ dx = \frac{\pi}{2e}. mine is not elementary yet because i used differentiating an integral with respect to a parameter.
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  7. #7
    Super Member Random Variable's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    it'd be nice to find an elementary proof of \int_0^{\infty} \frac{\cos x}{1+x^2} \ dx = \frac{\pi}{2e}. mine is not elementary yet because i used differentiating an integral with respect to a parameter.
    I'm not exactly sure what you mean by "elementary, but you could do the following:

    The Fourier integral representation of e^{-|x|} is  \int^{\infty}_{0}\big( A(\omega) \cos \omega x  + B(\omega) \sin \omega x\big) d \omega

    where  A(\omega) = \int^{\infty}_{-\infty} e^{-|t|} \cos \omega t \ dt = 2 \int^{\infty}_{0} e^{-t} \cos \omega t \ dt = \frac{2}{\pi} \frac{1}{1+\omega^{2}} (integration by parts)

    and  B (\omega) = \frac{1}{\pi} \int^{\infty}_{-\infty} e^{-|t|}  \sin \omega t \ dt = 0


    so  \int^{\infty}_{0} \frac{\cos \omega x}{1+\omega^{2}} \ d \omega = \frac{\pi}{2e^{|x|}}

    let  x =1

    then  \int^{\infty}_{0} \frac{\cos \omega }{1+\omega^{2}} \ d \omega = \frac{\pi}{2e}
    Last edited by Random Variable; May 11th 2010 at 10:41 AM.
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  8. #8
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    Or take a look if you are free ...

     \int_{ - \infty}^{\infty} \frac{\cos{x}}{x^2 + 1}~dx


     = Im \int_{ - \infty}^{\infty} \frac{\cos{x}}{x-i}~dx

     = Im \int_0^{2\pi} \cos{x} \left[ \frac{1}{x-i} + \sum_{k=1}^{\infty} \frac{ 2(x-i)}{ (x-i)^2 - 4k^2\pi^2} \right]~dx

     = \frac{1}{2} Im \left[ \int_0^{2\pi} \cos{x} \cot(\frac{x-i}{2})~dx \right ]

     = \frac{1}{2} Im \left[  \int_0^{\pi} \cos{x} \cot(\frac{x-i}{2})~dx  + \int_{\pi}^{2\pi} \cos{x} \cot(\frac{x-i}{2})~dx  \right]

    Sub. x= t + \pi in the second integral , we have

      Im \left[ \int_0^{\pi} \cos{x} \frac{dx}{ \sin(x-i) } \right]

     = Im \left[   \int_0^{\pi} \cos{x} \frac{ \sin{x} \cosh{1} + i\cos{x}\sinh{1} }{ (\sin{x} \cosh{1} )^2 + (\cos{x}\sinh{1})^2}~dx  \right]

     = \sinh{1}  \int_0^{\pi} \frac{ \cos^2{x}}{ (\cosh{1} )^2 - (\cos{x}) ^2}~dx

     = 2\sinh{1}  \int_0^{\frac{\pi}{2}} \frac{ \cos^2{x}}{ (\cosh{1} )^2 - (\cos{x}) ^2}~dx

    Not long later , we will find that the integral is equal to :

     \pi \cosh{1} - \pi \sinh{1} = \frac{\pi}{e}
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  9. #9
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    Quote Originally Posted by NonCommAlg View Post
    it'd be nice to find an elementary proof of \int_0^{\infty} \frac{\cos x}{1+x^2} \ dx = \frac{\pi}{2e}. mine is not elementary yet because i used differentiating an integral with respect to a parameter.

    I could 'feel' what your definition to an elementary proof is but perhaps much much sweat is needed for finding the elementary proof (Not sure is this true if you work in a room that the air-conditioner is on ).
    Last edited by simplependulum; May 11th 2010 at 09:21 PM.
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