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Math Help - Number of ways to evaluate an integral

  1. #1
    MHF Contributor chiph588@'s Avatar
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    Number of ways to evaluate an integral

    This is a group challenge:

    It's well known that  \displaystyle \int_0^\infty \frac{\sin x}{x}dx = \frac{\pi}{2} .

    I know of a few ways of showing this. My challenge to all of you is to come up with as many ways as you can to prove this result.

    Let's see if there's any ways I'm not aware of yet!
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  2. #2
    MHF Contributor chisigma's Avatar
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    There are different ways to compute the integral...

    \displaystyle \int_{0}^{\infty} \frac{\sin x}{x}\ dx (1)

    ... even if not all [in my opinion...] are 'fully convincing'... there is however also way to non compute it: if You consider (1) as a Lebesgue integral it diverges... in my opiniom this 'controversial' has to be, sooner or later, clarified...

    Kind regards

    \chi \sigma
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  3. #3
    Behold, the power of SARDINES!
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    I only know of 2 and here they are.

    \displaystyle \int_{0}^{\infty}\frac{\sin(x)}{x}dx=\int_{0}^{\in  fty}\int_{0}^{\infty}e^{-sx}\sin(x)dsdx

    Now switch the order of integration

    \displaystyle \int_{0}^{\infty}\int_{0}^{\infty}e^{-sx}\sin(x)dxds=\int_{0}^{\infty}\frac{1}{s^2+1}ds=  \tan^{-1}(s) \bigg|_{0}^{\infty}=\frac{\pi}{2}

    The other is via the residue theorem

    Let C_1 be the linear arc from [-R,\-\rho] and C_2: \rho e^{it};\pi < t < 2\pi C_3:[\rho,R] and C_4: Re^{it}; 0< t <\pi

    Now notice that \frac{\sin(x)}{x} is an even function so

    \displaystyle \int_{0}^{\infty}\frac{\sin(x)}{x}dx=\frac{1}{2}\i  nt_{-\infty}^{\infty}\frac{\sin(x)}{x}dx

    Now consider

    \displaystyle \oint \frac{e^{iz}}{z}dz=2\pi i(1)

    f(z)=\frac{e^{iz}}{z}=\sum_{n=0}^{\infty}\frac{i^{  n}(z)^{n-1}}{n!}=\frac{1}{z}+g(z) where g(z) is an analytic function.

    Using this and the ML bound we get

    \displaystyle \lim_{\rho \to 0}\int_{C_2}\left( \frac{1}{z}+g(z)\right)dz=\pi i

    Now by Jordan's lemma on C_4

    \displaystyle \lim_{R \to \infty}\int_{C_4}\frac{e^{iz}}{z}dz =0

    This leaves

    \displaystyle \int_{-\infty}^{\infty}\frac{e^{iz}}{z}dz+\pi i=2\pi i \iff \int_{-\infty}^{\infty}\frac{\sin(z)}{z}dz=\pi i

    Since e^{iz}=\cos(z)+i\sin)(z) we take the real and immaginary parts to get

    \int_{-\infty}^{\infty}\frac{\sin(x)}{x}dx=\pi

    This gives

    \int_{0}^{\infty}\frac{\sin(x)}{x}dx=\frac{\pi}{2}
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by TheEmptySet View Post
    I only know of 2 and here they are.

    \displaystyle \int_{0}^{\infty}\frac{\sin(x)}{x}dx=\int_{0}^{\in  fty}\int_{0}^{\infty}e^{-sx}\sin(x)dsdx

    Now switch the order of integration

    \displaystyle \int_{0}^{\infty}\int_{0}^{\infty}e^{-sx}\sin(x)dxds=\int_{0}^{\infty}\frac{1}{s^2+1}ds=  \tan^{-1}(s) \bigg|_{0}^{\infty}=\frac{\pi}{2}
    Haven't seen this one!

    Using this and the ML bound we get

    \displaystyle \lim_{\rho \to 0}\int_{C_2}\left( \frac{1}{z}+g(z)\right)dz=\pi i
    Also, I evaluated  \displaystyle \int_{C_2} differently. What exactly is "the ML bound"?
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  5. #5
    Behold, the power of SARDINES!
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    On the arc \rho=e^{it};\pi < t < 2\pi

    We have

    \displaystyle \int_{\rho}\left(\frac{1}{z}+g(z) \right)dz

    So we get

    \int_{\pi}^{2\pi}idt+\int_{\pi}^{2\pi}g(\rho e^{it})dt

    The 2nd integral is bounded by it's maximum on the arc multiplied by its length so we get

    \pi i +\pi \rho M but as \rho \to 0 we get \pi i
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  6. #6
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by TheEmptySet View Post
    On the arc \rho=e^{it};\pi < t < 2\pi

    We have

    \displaystyle \int_{\rho}\left(\frac{1}{z}+g(z) \right)dz

    So we get

    \int_{\pi}^{2\pi}idt+\int_{\pi}^{2\pi}g(\rho e^{it})dt

    The 2nd integral is bounded by it's maximum on the arc multiplied by its length so we get

    \pi i +\pi \rho M but as \rho \to 0 we get \pi i
    Oh ok, that's what I did actually. I didn't understand what ML meant, but I do now.

    Also for your first method not to be picky but I think you might need to justify switching the order of integration.
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  7. #7
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    Here is a proof , first note that

     \tan(\frac{x}{2}) = \frac{x}{2} \displaystyle{ \prod_{k=1}^{\infty} \frac{ 1- \frac{x^2}{(2k\pi)^2 } }{ 1 - \frac{x^2}{((2k-1)\pi)^2} } }

    By taking log. differentiation , we obtain this equation :

     \frac{1}{\sin(x)} = \frac{1}{x} + 2x \displaystyle{ \sum_{k=1} \frac{(-1)^k }{ x^2 - k^2 \pi^2 } }


    I would like to stop here and back to the integral  \int_0^{\infty} \frac{\sin(x)}{x} ~dx .

    We find that the integrand is even , i write it as

     \frac{1}{2} \int_{-\infty}^{\infty} \frac{\sin(x)}{x} ~dx


     = \frac{1}{2} \displaystyle{ \left( \int_0^{\pi} + \sum_{k=1}^{\infty} [ \int_{k\pi}^{(k+1)\pi} + \int_{-k\pi}^{(-k+1)\pi} ] \right) \frac{\sin(x)}{x} ~dx }



    Consider  \displaystyle{ \sum_{k=1}^{\infty} ( \int_{k\pi}^{(k+1)\pi} + \int_{-k\pi}^{(-k+1)\pi} ) \right) \frac{\sin(x)}{x} ~dx }

    Sub.  x = t + k\pi and  x = t - k\pi on the first and the second integral respectively ,we obtain :

     \displaystyle{ \sum_{k=1}^{\infty}  (-1)^k \left( \int_{0}^{\pi} \frac{\sin(t)}{t+k\pi} ~dt +  \int_{0}^{\pi} \frac{\sin(t)}{t-k\pi} ~dt \right)   }

     = \displaystyle{ \sum_{k=1}^{\infty}  (-1)^k \int_{0}^{\pi} \frac{2t \sin(t) }{ t^2 -k^2\pi^2 }~ dt \right)   }

    Therefore ,

     \int_0^{\infty} \frac{\sin(x)}{x} ~dx = \frac{1}{2} \displaystyle{ \int_0^{\pi} \left( \frac{1}{x} + 2x \sum_{k=1}^{\infty} \frac{ (-1)^k  }{ x^2 -k^2\pi^2 } \right) ~\sin(x)~dx   }

     =  \frac{1}{2} \int_0^{\pi} \frac{\sin(x)}{\sin(x)}~dx = \frac{\pi}{2}
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  8. #8
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    Quote Originally Posted by TheEmptySet View Post
    I only know of 2 and here they are.

    \displaystyle \int_{0}^{\infty}\frac{\sin(x)}{x}dx=\int_{0}^{\in  fty}\int_{0}^{\infty}e^{-sx}\sin(x)dsdx

    Now switch the order of integration
    the problem here is that you don't have absolute convergence.

    in order to justify the interchange, it's better to integrate by parts first by selecting 1-\cos x as an antiderivative for \sin x, from there you introduce the parameter you said and the interchange is justified now by Tonelli's theorem since the function is no negative.
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  9. #9
    Super Member General's Avatar
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    You could use the following laplace transform's property:

    \displaystyle L\left( \dfrac{f(t)}{t} \right) = \int_s^{\infty} F(\beta) \, d\beta
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  10. #10
    Super Member Random Variable's Avatar
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    Quote Originally Posted by Krizalid View Post
    the problem here is that you don't have absolute convergence.

    in order to justify the interchange, it's better to integrate by parts first by selecting 1-\cos x as an antiderivative for \sin x, from there you introduce the parameter you said and the interchange is justified now by Tonelli's theorem since the function is no negative.


    Something like this:

     \displaystyle = \int^{\infty}_{0} \frac{1- \cos x}{x^{2}} \ dx  = \int^{\infty}_{0} \int^{\infty}_{0} (1-\cos x) t e^{-xt} \ dt \ dx

    so now the integrand is never negative

     \displaystyle =  \int^{\infty}_{0} \int^{\infty}_{0} (te^{-xt} - t \cos x e^{-xt}) \ dx \ dt

     \displaystyle = \int^{\infty}_{0} \big( 1- t \ \mathca{L} \{ \cos x \} (t) \big) \ dt

     = \displaystyle \int^{\infty}_{0} \Big( 1 - \frac{t^{2}}{1+t^{2}} \Big) \ dt = \int^{\infty}_{0} \frac{1}{1+t^{2}} \ dt = \frac{\pi}{2}
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  11. #11
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    Krizalid's Avatar
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    yes, that approach works fine and is justified.
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  12. #12
    Super Member Random Variable's Avatar
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    I asked a professor, and he said something about that because the issue is at a single point (the origin), that what the TheEmptySet did was OK. I think the absolute convergence requirement is too restrictive. I don't know.
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  13. #13
    Math Engineering Student
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    perhaps is restrictive, but it works, but you'd need a good justification when reversing the integration order by not having absolute convergence.
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  14. #14
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    Although \displaystyle \int_{(0, \infty)} \frac{sin(x)}{x}d\lambda diverges, one can show \displaystyle \lim_{a \to \infty} \int_{(0,a)} \frac{sin(x)}{x} = \frac{\pi}{2} in the following way:

    First, note that since 0 \leq |sin(x)e^{-xy}| is measurable, we have that \displaystyle \int_{(0,a) \times (0,\infty)}|sin(x)e^{-xy}| = \int_{0}^{a}|sin(x)| (\int_{0}^{\infty}|e^{-xy}|dy)dx = \int_{0}^{a} \frac{|sin(x)|}{|x|}dx < \infty and so sin(x)e^{-xy} \in L^1((0,a)\times(0,\infty))

    Now note that using Fubini on f(x,y)=sin(x)e^{-xy} gives \displaystyle \int_{0}^{a}\frac{sin(x)}{x}dx and changing the order of integration the other way gives, after using integration by parts twice, \displaystyle \int_{0}^{a}\frac{sin(x)}{x}dx = \frac{\pi}{2} - cos(a)\int_{0}^{\infty}\frac{e^{-ay}}{1+y^2}dy \displaystyle -sin(a)\int_{0}^{\infty} \frac{ye^{-y}}{1+y^2}dy

    Taking the limit as a \to \infty and using DCT finished off the argument.
    Last edited by Defunkt; January 28th 2011 at 01:34 AM.
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  15. #15
    MHF Contributor Drexel28's Avatar
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    Here's a way using minimum machinery, but it requires the simple lemma that if f\in\mathcal{C}[a,b] then \displaystyle \lim_{n\to\infty}\int_a^b f(x)\sin((2n+1)x)\text{ }dx=0 and the fact that \displaystyle \int_0^{\infty}\frac{\sin(x)}{x} exists, although this follows immediately from Dirichlet's test for integrals.


    Ok, so note that since \displaystyle \int_0^{\infty}\frac{\sin(x)}{x}\text{ }dx exists we have that


    \displaystyle \int_0^{\infty}\frac{\sin(x)}{x}\text{ }dx=\lim_{\lambda\to\infty}\int_0^\lambda \frac{\sin(x)}{x}\text{ }dx=\lim_{\substack{n\to\infty\\ n\in\mathbb{N}}}\int_0^{\frac{\pi n}{2}}\frac{\sin(x)}{x}\text{ }dx=\lim_{n\to\infty}\int_0^{\frac{\pi}{2}}\frac{\  sin(nx)}{x}\text{ }dx


    and thus clearly \displaystyle \int_0^\infty \frac{\sin(x)}{x}\text{ }dx=\lim_{n\to\infty}\int_0^{\frac{\pi}{2}}\frac{\  sin\left((2n+1)x\right)}{x}\text{ }dx. We next claim that \displaystyle \int_0^{\frac{\pi}{2}}\frac{\sin\left((2n+1)x\righ  t)}{\sin(x)}\text{ }dx=\frac{\pi}{2} for every n\in\mathbb{N}. But, this follows immediately from the common trigonometric identity associated with the Dirichlet kernel. We then note that


    \displaystyle \begin{aligned}\lim_{n\to\infty}\left(\int_0^{\fra  c{\pi}{2}}\frac{\sin\left((2n+1)x\right)}{x}\text{ }dx-\int_0^{\frac{\pi}{2}}\frac{\sin\left((2n+1)x\righ  t)}{\sin(x)}\text{ }dx\right) &=\lim_{n\to\infty}\int_0^{\frac{\pi}{2}}\frac{\si  n(x)-x}{\sin(x)}\sin\left((2n+1)x\right)\text{ }dx\\ &=0\end{aligned}


    where we've made use of the lemma mentioned at the beginning of the post. Thus, it follows that


    \displaystyle \int_0^{\infty}\frac{\sin(x)}{x}\text{ }dx=\lim_{n\to\infty}\int_0^{\frac{\pi}{2}}\frac{\  sin\left((2n+1)x\right)}{x}\text{ }dx=\lim_{n\to\infty}\int_0^{\frac{\pi}{2}}\frac{\  sin\left((2n+1)x\right)}{\sin(x)}\text{ }dx=\frac{\pi}{2}
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