Results 1 to 8 of 8

Thread: integral cosine and exponential

  1. #1
    Newbie
    Joined
    Jul 2008
    Posts
    4

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    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
    $\displaystyle \int_0^{\infty}\frac{e^{-b^2x^2}}{1+a^2x^2}\cos(cx)~dx$?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jul 2008
    Posts
    4
    yes, it is exactly that.

    I will use \math next time
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    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 $\displaystyle b\in\mathbb{N}$

    Otherwise the integral would diverge.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jul 2008
    Posts
    4
    I though it was not necessary to have $\displaystyle b\in N$
    It is not sufficient that $\displaystyle b\in R $


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


    I'm maybe missing something...
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by mathmarco View Post
    I though it was not necessary to have $\displaystyle b\in N$
    It is not sufficient that $\displaystyle b\in R $


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


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

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

    So $\displaystyle dx=\frac{1}{b}$

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

    and when

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

    So we have

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


    Yeah you are right, I misread it at first as $\displaystyle b$ not $\displaystyle b^2$
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Jul 2008
    Posts
    4
    ok... good. I'm still trying to solve it but I always end up to nothing good!!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    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 $\displaystyle bx=u$?

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

    So

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

    Or in other words

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

    and obviously since

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


    Now

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

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

    So $\displaystyle dx=\frac{1}{a}$

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

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

    So we have

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


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

    So we can finally conclude that


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

    Furthermore we see that

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

    Hope this has been some help.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. definite integral of cosine...
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Dec 8th 2011, 07:43 PM
  2. Solving a cosine integral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Jan 6th 2011, 03:41 PM
  3. Integral of cosine squared x?
    Posted in the Calculus Forum
    Replies: 4
    Last Post: May 6th 2010, 08:30 AM
  4. Integral powers of sine and cosine
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Mar 8th 2010, 12:49 PM
  5. cosine and exponential
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Oct 14th 2008, 07:07 AM

Search Tags


/mathhelpforum @mathhelpforum