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Thread: Improper integral

  1. #1
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    Improper integral

    Derive the integration formula
    $\displaystyle \int_0^{\infty}\frac{cos(ax)-cos(bx)}{x^2}dx=\frac{\pi}{2}(b-a)$

    By using the residue theorem, I integrated the function $\displaystyle \frac{e^iaz-e^ibz}{z^2}$ from $\displaystyle -\infty $to $\displaystyle \infty$, I can only obtain the result $\displaystyle 2\pi i(b-a)$, which means my answer is $\displaystyle \pi(b-a)$.
    I couldnt find my mistake.Thank you.
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  2. #2
    Super Member girdav's Avatar
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    Re: Improper integral

    The function you have to integrate is even, so the integral on $\displaystyle \mathbb R$ is two times the integral on $\displaystyle \mathbb R_+$.
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    Re: Improper integral

    Thank you for your reply.
    Yes i did it (divided by 2)and still get $\displaystyle \pi(b-a)$ as answer, but not$\displaystyle \frac{\pi}{2}(b-a)$
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  4. #4
    Super Member girdav's Avatar
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    Re: Improper integral

    Did you wrote the exponentials in terms of $\displaystyle \cos$?
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    Re: Improper integral

    Yes, by integrating $\displaystyle \frac{e^{iaz}-e^{ibz}}{z^2}$ and compare the real part of both sides
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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    Re: Improper integral

    Another way: denote $\displaystyle I(\lambda)=\int_0^{+\infty}\frac{\cos \lambda x}{x^2}dx\;(\lambda >0)$ , then $\displaystyle I'(a)=-\int_0^{+\infty}\frac{\sin \lambda x}{x}dx$ . Using the substitution $\displaystyle t=\lambda x$ we get $\displaystyle I'(\lambda)=-\int_0^{+\infty}\frac{\sin t}{t}dt=-\frac{\pi}{2}$ (Dirichlet's integral) . So, $\displaystyle I(\lambda)=-\frac{\pi}{2}\lambda+C$ . For $\displaystyle \lambda =0$ we get $\displaystyle C=\int_0^{+\infty}\frac{dx}{x^2}$ .

    Then,

    $\displaystyle \int_0^{+\infty}\frac{\cos ax-\cos bx}{x^2}dx=-\frac{\pi}{2}a+C-\left(-\frac{\pi}{2}b+C\right)=\boxed{\frac{\pi (b-a)}{2}}$

    @fareastmovement: Show your complete work by the method of residues (of course if you want) and we can check it.

    Edited: Method not valid, see the posts below.
    Last edited by FernandoRevilla; Dec 18th 2011 at 11:46 PM.
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    Re: Improper integral

    Thank you, I got it now. You are really helpful
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  8. #8
    MHF Contributor chisigma's Avatar
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    Re: Improper integral

    Quote Originally Posted by FernandoRevilla View Post
    Another way: denote $\displaystyle I(\lambda)=\int_0^{+\infty}\frac{\cos \lambda x}{x^2}dx\;(\lambda >0)$ , then $\displaystyle I'(a)=-\int_0^{+\infty}\frac{\sin \lambda x}{x}dx$ . Using the substitution $\displaystyle t=\lambda x$ we get $\displaystyle I'(\lambda)=-\int_0^{+\infty}\frac{\sin t}{t}dt=-\frac{\pi}{2}$ (Dirichlet's integral) . So, $\displaystyle I(\lambda)=-\frac{\pi}{2}\lambda+C$ . For $\displaystyle \lambda =0$ we get $\displaystyle C=\int_0^{+\infty}\frac{dx}{x^2}$ .

    Then,

    $\displaystyle \int_0^{+\infty}\frac{\cos ax-\cos bx}{x^2}dx=-\frac{\pi}{2}a+C-\left(-\frac{\pi}{2}b+C\right)=\boxed{\frac{\pi (b-a)}{2}}$

    @fareastmovement: Show your complete work by the method of residues (of course if you want) and we can check it.
    There is only a minor detail: writing $\displaystyle C=\int_{0}^{+ \infty} \frac{dx}{x^{2}}$ is a nonsense because the integral diverges...



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  9. #9
    MHF Contributor chisigma's Avatar
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    Re: Improper integral

    In order to avoid criticity in the solution of the integral I would suggest to use the trigonometric identity...

    $\displaystyle \cos \alpha - \cos \beta = 2\ \sin \frac{\beta + \alpha}{2}\ \sin \frac{\beta-\alpha}{2}$ (1)

    ... so that the integral becomes...

    $\displaystyle I(a,b)=2\ \int_{0}^{\infty} \frac{\sin (\frac{b+a}{2}\ x)\ \sin (\frac{b-a}{2}\ x)}{x^{2}}\ dx$



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  10. #10
    MHF Contributor FernandoRevilla's Avatar
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    Re: Improper integral

    Quote Originally Posted by chisigma View Post
    There is only a minor detail: writing $\displaystyle C=\int_{0}^{+ \infty} \frac{dx}{x^{2}}$ is a nonsense because the integral diverges...
    Right, just a lapse of concentration. Easily avoided: $\displaystyle I(1/2)$ is finite, so $\displaystyle I(1/2)=-\pi/4+C$ and $\displaystyle C=I(1/2)+\pi/4$ is finite.
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  11. #11
    MHF Contributor chisigma's Avatar
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    Re: Improper integral

    Quote Originally Posted by FernandoRevilla View Post
    Right, just a lapse of concentration. Easily avoided: $\displaystyle I(1/2)$ is finite, so $\displaystyle I(1/2)=-\pi/4+C$ and $\displaystyle C=I(1/2)+\pi/4$ is finite.
    The real problem is in defining the function ...

    $\displaystyle I(\lambda)= \int_{0}^{\infty} \frac{\cos (\lambda x)}{x^{2}}\ dx $ (2)

    ... because the integral diverges for any value of $\displaystyle \lambda$. In my previous post a way to overcame such type of criticity is indicated...



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  12. #12
    MHF Contributor FernandoRevilla's Avatar
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    Re: Improper integral

    Quote Originally Posted by chisigma View Post
    ... because the integral diverges for any value of $\displaystyle \lambda$. In my previous post a way to overcame such type of criticity is indicated...
    Not for all values of $\displaystyle \lambda$ but of course the method I used is not valid. Reviewing the situation I find the explanation: in a foolish way my mind thought about $\displaystyle 1$ to $\displaystyle \infty$ instead of $\displaystyle 0$ to $\displaystyle \infty$ and curiously at the end I made the substitution $\displaystyle \lambda=0$ . A complete disaster.

    Method not valid.
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  13. #13
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Improper integral

    Quote Originally Posted by fareastmovement View Post
    Derive the integration formula
    $\displaystyle \int_0^{\infty}\frac{cos(ax)-cos(bx)}{x^2}dx=\frac{\pi}{2}(b-a)$

    By using the residue theorem, I integrated the function $\displaystyle \frac{e^iaz-e^ibz}{z^2}$ from $\displaystyle -\infty $to $\displaystyle \infty$, I can only obtain the result $\displaystyle 2\pi i(b-a)$, which means my answer is $\displaystyle \pi(b-a)$.
    I couldnt find my mistake.Thank you.


    Perhaps, using integration by parts and aplaying Furlani's integrals we can conclude something.

    $\displaystyle \int_0^{\infty} \frac{\cos{ax}-\cos{bx}}{x^2}dx=\frac{\cos{bx}-\cos{ax}}{x}\big{|}_{\mathbb{R^+} $$\displaystyle + \int_0^{\infty} \frac{b\sin{bx}-a\sin{ax}}{x} dx$

    Now, using the theorem:

    $\displaystyle \int_0^{\infty} \frac{f(ax)-f(bx)}{x} dx=\(f(0)-f(\infty))\ln\frac{b}{a}$


    I stuck on evaluating:


    $\displaystyle \int_0^{\infty} \frac{b\sin{bx}-a\sin{ax}}{x} dx$


    Edit:

    Ok... No Furalani needed.

    Using only the well known Dirichlet's integral we get:


    $\displaystyle \int_0^{\infty} \frac{b\sin{bx}-a\sin{ax}}{x} dx=\frac{\pi}{2}(b-a)$

    (While $\displaystyle \frac{\cos{bx}-\cos{ax}}{x}\big{|}_{\mathbb{R^+}$ $\displaystyle {=0}$)
    Last edited by Also sprach Zarathustra; Dec 19th 2011 at 03:30 AM.
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  14. #14
    MHF Contributor FernandoRevilla's Avatar
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    Re: Improper integral

    (a) $\displaystyle I(a,b)=\int_0^{\infty}\frac{\cos ax-\cos bx}{x^2}dx$ is convergent for $\displaystyle a>0,b>0$ (easily proved using chisigma's answer #9 )

    (b) $\displaystyle \frac{\partial I}{\partial a}=\ldots=-\frac{\pi}{2},\;\frac{\partial I}{\partial b}=\ldots=\frac{\pi}{2}$ (differentiation under the integral sign and Dirichlet's Integral).

    (c) $\displaystyle I(a,b)=-\frac{\pi}{2}a+\varphi(b)$ , so $\displaystyle \frac{\partial I}{\partial b}=\varphi'(b)=\frac{\pi}{2}$ hence $\displaystyle \varphi(b)=\frac{\pi}{2}b+C$ . That is, $\displaystyle I(a,b)=\frac{\pi (b-a)}{2}+C\quad (*)$ .

    (d) Taking limits in $\displaystyle (*)$ as $\displaystyle (a,b)\to (0,0)$ we get $\displaystyle C=0$ . So, $\displaystyle I(a,b)=\frac{\pi (b-a)}{2}$
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  15. #15
    Super Member Random Variable's Avatar
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    Re: Improper integral

    $\displaystyle \int^{\infty}_{0} \frac{\cos ax - \cos bx}{x^{2}} \ dx $


    $\displaystyle = \int_{0}^{\infty} \int_{a}^{b} \frac{\sin tx}{x} \ dt \ dx $


    Fubini's theorem is not satsified. So to justify changing the order of integration, let's write the double integral more formally as a limit.


    $\displaystyle = \lim_{n \to \infty} \int_{0}^{n} \int_{a}^{b} \frac{\sin tx}{x} \ dt \ dx = \lim_{n \to \infty} \int_{a}^{b} \int_{0}^{n} \frac{\sin tx}{x} \ dx \ dt$


    $\displaystyle = \lim_{n \to \infty} \int_{a}^{b} \int_{0}^{nt} \frac{\sin u}{u} \ du \ dt = \lim_{n \to \infty} \int_{a}^{b} \text{Si} (nt) \ dt $


    Since the Sine Integral is bounded above by 2, we can use the Dominated Convergence Theorem to justify bringing the limit inside of the integral.


    $\displaystyle = \frac{\pi}{2} \int_{a}^{b} \ dt = \frac{\pi}{2} (b-a) $
    Last edited by Random Variable; Dec 19th 2011 at 01:11 PM.
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