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Math Help - integral cosine and exponential

  1. #1
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    integral cosine and exponential

    I'm trying to solve the following integral

    int_0^{+infintity} 1/(1+a^2*x^2)*exp(-b^2*x^2)*cos(c*x) dx

    where a,b and c are real.

    thanks a lot

    Marco
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by mathmarco View Post
    I'm trying to solve the following integral

    int_0^{+infintity} 1/(1+a^2*x^2)*exp(-b^2*x^2)*cos(c*x) dx

    where a,b and c are real.

    thanks a lot

    Marco
    \int_0^{\infty}\frac{e^{-b^2x^2}}{1+a^2x^2}\cos(cx)~dx?
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  3. #3
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    yes, it is exactly that.

    I will use \math next time
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by mathmarco View Post
    I'm trying to solve the following integral

    int_0^{+infintity} 1/(1+a^2*x^2)*exp(-b^2*x^2)*cos(c*x) dx

    where a,b and c are real.

    thanks a lot

    Marco
    Well first off I am pretty sure that we must have that b\in\mathbb{N}

    Otherwise the integral would diverge.
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  5. #5
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    I though it was not necessary to have  b\in N
    It is not sufficient that  b\in R


    why do you think it will diverge? Does not the following integral converge?
    \int_0^{+\infty} e^{-b^2 x^2} dx


    I'm maybe missing something...
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by mathmarco View Post
    I though it was not necessary to have  b\in N
    It is not sufficient that  b\in R


    why do you think it will diverge? Does not the following integral converge?
    \int_0^{+\infty} e^{-b^2 x^2} dx


    I'm maybe missing something...
    \int_0^{\infty}e^{-b^2x^2}~dx=\int_0^{\infty}e^{-(bx)^2}

    Now let u=bx\Rightarrow\frac{u}{b}=x

    So dx=\frac{1}{b}

    so we have that when x\to\infty\Rightarrow{u\to\infty}

    and when

    x\to{0}\Rightarrow{u\to{0}}

    So we have

    \frac{1}{b}\int_0^{\infty}e^{-u^2}~du=\frac{\sqrt{\pi}}{2b}


    Yeah you are right, I misread it at first as b not b^2
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  7. #7
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    ok... good. I'm still trying to solve it but I always end up to nothing good!!
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by mathmarco View Post
    ok... good. I'm still trying to solve it but I always end up to nothing good!!
    I can do little more than show it converges, are you sure it is integralbe? Have you tried differentiation under the integral sign after a sub of bx=u?

    \forall{x}\in\mathbb{R}\quad{-1\leq\cos(cx)\leq{1}}

    So

    -\int_0^{\infty}\frac{e^{-b^2x^2}}{1+a^2x^2}~dx\leq\int_0^{\infty}\frac{\cos  (cx)\cdot{e^{-b^2x^2}}}{1+a^2x^2}~dx\leq\int_0^{\infty}\frac{e^{-b^2x^2}}{1+a^2x^2}~dx

    Or in other words

    \bigg|\int_0^{\infty}\frac{\cos(cx)e^{-b^2x^2}}{1+a^2x^2}~dx\bigg|\leq\int_0^{\infty}\fra  c{e^{-b^2x^2}}{1+a^2x^2}~dx

    and obviously since

    0\leq\int_0^{\infty}\frac{e^{-b^2x^2}}{1+a^2x^2}~dx\ll\int_0^{\infty}\frac{dx}{1  +a^2x^2}


    Now

    \int_0^{\infty}\frac{dx}{1+(ax)^2}

    Let ax=\varphi\Rightarrow\frac{\varphi}{a}=x

    So dx=\frac{1}{a}

    And \text{As }x\to\infty\Rightarrow\varphi\to\infty

    and \text{As }x\to{0}\Rightarrow\varphi\to{0}

    So we have

    \frac{1}{a}\int_0^{\infty}\frac{d\varphi}{1+\varph  i^2}=\frac{1}{a}\arctan\left(\varphi\right)\bigg|_  0^{\infty}=\frac{\pi}{2a}


    Therefore \int_0^{\infty}\frac{e^{-b^2x^2}}{1+a^2x^2}~dx

    So we can finally conclude that


    \int_0^{\infty}\frac{e^{-b^2x^2}\cos(cx)}{1+a^2x^2}~dx\quad\text{converges}  ~\forall~(a,b,c)\in\mathbb{R}

    Furthermore we see that

    \bigg|\int_0^{\infty}\frac{e^{-b^2x^2}\cos(cx)}{1+a^2x^2}~dx\bigg|\leq\int_0^{\in  fty}\frac{e^{-b^2x^2}}{1+a^2x^2}~dx\leq\frac{\pi}{2a}

    Hope this has been some help.
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