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Math Help - Integral of Gaussian multiplied by cos(x)

  1. #1
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    Integral of Gaussian multiplied by cos(x)

    Hey I have a problem that was in quantum mechanics, but is basically a math problem.
    I need to solve the following integral:

    How do you achieve this result analytically?
    I do know how to solve for the integral of the plain Gaussian by converting to polar coordinates, but I have no idea how to achieve this result.

    Any Help would be appreciated!
    [IMG]file:///C:/DOCUME%7E1/AllanB/LOCALS%7E1/Temp/moz-screenshot.jpg[/IMG][IMG]file:///C:/DOCUME%7E1/AllanB/LOCALS%7E1/Temp/moz-screenshot-1.jpg[/IMG]
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by all_an88 View Post
    Hey I have a problem that was in quantum mechanics, but is basically a math problem.
    I need to solve the following integral:

    How do you achieve this result analytically?
    I do know how to solve for the integral of the plain Gaussian by converting to polar coordinates, but I have no idea how to achieve this result.

    Any Help would be appreciated!
    [IMG]file:///C:/DOCUME%7E1/AllanB/LOCALS%7E1/Temp/moz-screenshot.jpg[/IMG][IMG]file:///C:/DOCUME%7E1/AllanB/LOCALS%7E1/Temp/moz-screenshot-1.jpg[/IMG]
    I have never done this one...and I am not completely sure how to do this, but I have a few quick notions that you may want to follow

    A) Use Leibniz's rule of differentiation and see if this turns into a simple ODE

    B) This looks like prime countour integral to me

    C) Rewrite it as

    \int_{-\infty}^{\infty}e^{-ax^2}\cos(kx)~dx=\text{Re}\left[\int_{-\infty}^{\infty}e^{-ax^2}e^{ikx}~dx\right]

    and complete the square. I will try this if I get time and get back to you if none of the other members do.

    Best of luck
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  3. #3
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    Quote Originally Posted by Mathstud28 View Post
    I have never done this one...and I am not completely sure how to do this, but I have a few quick notions that you may want to follow

    A) Use Leibniz's rule of differentiation and see if this turns into a simple ODE

    B) This looks like prime countour integral to me

    C) Rewrite it as

    \int_{-\infty}^{\infty}e^{-ax^2}\cos(kx)~dx=\text{Re}\left[\int_{-\infty}^{\infty}e^{-ax^2}e^{ikx}~dx\right]

    and complete the square. I will try this if I get time and get back to you if none of the other members do.

    Best of luck


    C works like a charm!
    Thank you very much!
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  4. #4
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    Quote Originally Posted by Mathstud28 View Post
    I have never done this one...and I am not completely sure how to do this, but I have a few quick notions that you may want to follow

    A) Use Leibniz's rule of differentiation and see if this turns into a simple ODE

    B) This looks like prime countour integral to me

    C) Rewrite it as

    \int_{-\infty}^{\infty}e^{-ax^2}\cos(kx)~dx=\text{Re}\left[\int_{-\infty}^{\infty}e^{-ax^2}e^{ikx}~dx\right]

    and complete the square. I will try this if I get time and get back to you if none of the other members do.

    Best of luck
    I like method (A) if we let

    I(k) = \int_{-\infty}^{\infty}e^{-ax^2}\cos(kx)~dx

    then

    I' (k) = - \int_{-\infty}^{\infty}xe^{-ax^2}\sin(kx)~dx

    this can be integrated by parts giving

    I' (k) = - \left. \frac{e^{-a x^2} \sin kx}{2a} \right|_{-\infty}^{\infty}- \frac{k}{2a} \int_{-\infty}^{\infty}e^{-ax^2}\cos(kx)~dx

    so the first term is zero and the second involve I. Thus,

    I'(k) = - \frac{k}{2a} I(k)

    which integrates giving I(k) = c\,e^{-\frac{k^2}{4a}}

    Now I(0) = c \;\;\text{and}\;\;\; I(0) = \int_{-\infty}^{\infty}e^{-ax^2}~dx so c = \sqrt{\frac{\pi}{a}} giving

    I(k) = \sqrt{\frac{\pi}{a}} \,e^{-\frac{k^2}{4a}}

    giving your result.
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  5. #5
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    Quote Originally Posted by danny arrigo View Post
    I like method (A) if we let

    I(k) = \int_{-\infty}^{\infty}e^{-ax^2}\cos(kx)~dx

    then

    I' (k) = - \int_{-\infty}^{\infty}xe^{-ax^2}\sin(kx)~dx

    this can be integrated by parts giving

    I' (k) = - \left. \frac{e^{-a x^2} \sin kx}{2a} \right|_{-\infty}^{\infty}- \frac{k}{2a} \int_{-\infty}^{\infty}e^{-ax^2}\cos(kx)~dx

    so the first term is zero and the second involve I. Thus,

    I'(k) = - \frac{k}{2a} I(k)

    which integrates giving I(k) = c\,e^{-\frac{k^2}{4a}}

    Now I(0) = c \;\;\text{and}\;\;\; I(0) = \int_{-\infty}^{\infty}e^{-ax^2}~dx so c = \sqrt{\frac{\pi}{a}} giving

    I(k) = \sqrt{\frac{\pi}{a}} \,e^{-\frac{k^2}{4a}}

    giving your result.
    Wow that's a really nice way to do it to, i had no idea what Leibniz's rule was....
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by all_an88 View Post
    Wow that's a really nice way to do it to, i had no idea what Leibniz's rule was....
    Yes...its a nice little rule that says that if \int_a^b f(x,\theta)~dx satisfies certain conditions \frac{d}{d\theta}\int_a^b f(x,\theta)~dx=\int_a^b \frac{\partial}{\partial \theta}f(x,\theta)~dx

    To learn more look here

    Leibniz integral rule - Wikipedia, the free encyclopedia

    Prof. Arrigo: Thank you, I was fairly sure that method would work but was not sure how simple the computation would be...thanks again
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  7. #7
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    Quote Originally Posted by all_an88 View Post
    Hey I have a problem that was in quantum mechanics, but is basically a math problem.
    I need to solve the following integral:

    How do you achieve this result analytically?
    I do know how to solve for the integral of the plain Gaussian by converting to polar coordinates, but I have no idea how to achieve this result.
    Here is a solution through complex analysis.

    First by letting t = \frac{x}{\sqrt{a}} the integral can be brought to the form \int \limits_{-\infty}^{\infty} e^{-x^2} \cos bx dx,b>0.
    Also, it is sufficient to solve, \int \limits_0^{\infty} e^{-x^2} \cos bx ~ dx because e^{-x^2}\cos bx is an even function.

    Let f(z) = e^{-z^2} and \Gamma be the rectangular contour with verticles at 0,R,R+i\tfrac{b}{2}, i\tfrac{b}{2}.
    Since the function f is entire it means \oint_{\Gamma} f(z) dz = 0.
    Writing the integral by definition we get,
    \int_0^R e^{-x^2} dx - \int_0^R e^{-(x+ib/2)^2} dx+ \int_0^{b/2} e^{-(R+it)^2} i dt - \int_0^{b/2} e^{t^2} i dt = 0 ... [1]

    Notice that,
    \int_0^R e^{-(x+ib/2)^2} dx = \int_0^R e^{-x^2 - bix + b^2/4} dx = e^{b^2/4} \int_0^R e^{-x^2} \left( \cos bx - i \sin bx\right) dx

    Also,
    \left| \int_0^{b/2} e^{-(R+it)^2} i dt \right| = e^{-R^2} \left| \int_0^{b/2} e^{-2iRt} e^{t^2} dt\right| \leq e^{-R^2} \int_0^{b/2} \left| e^{-2iRt} e^{t^2} \right| dt \leq \frac{b}{2}e^{-R^2} e^{b^2/4}\to 0 \text{ as }R\to \text{ Me }

    Now if we take the limit as R\to \infty in [1] and take the real part we get:
    \int_0^{\infty} e^{-x^2} dx  - e^{b^2/4} \int_0^{\infty} e^{-t^2} \cos bx ~ dx = 0

    Thus, \int_0^{\infty} e^{-t^2}\cos bx ~ dx = e^{-b^2/4}\int_0^{\infty} e^{-x^2} dx = \tfrac{1}{2}\sqrt{\pi}e^{-b^2/4}
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