Trouble proving Gaussian-like integral

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I came across Gaussian integrals, and was trying to prove them myself. I proved the basics, but am stuck on the following



$$intlimits_-infty^inftyxe^-a(x-b)^2dx = b sqrtfracpia$$



I am having trouble thinking of how to attack this problem. I am currently thinking integration by parts to get rid of the $x$, and then try and coerce the exponent into the general Gaussian integral form but was wondering if there was a better solution.






definite-integrals exponential-function gaussian-integral







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    Substitution with $u=x-b$.
    – Randall
    Sep 11 at 3:04














up vote
-1
down vote

favorite












I came across Gaussian integrals, and was trying to prove them myself. I proved the basics, but am stuck on the following



$$intlimits_-infty^inftyxe^-a(x-b)^2dx = b sqrtfracpia$$



I am having trouble thinking of how to attack this problem. I am currently thinking integration by parts to get rid of the $x$, and then try and coerce the exponent into the general Gaussian integral form but was wondering if there was a better solution.






definite-integrals exponential-function gaussian-integral







share|cite|improve this question

















  • 2




    Substitution with $u=x-b$.
    – Randall
    Sep 11 at 3:04












up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











I came across Gaussian integrals, and was trying to prove them myself. I proved the basics, but am stuck on the following



$$intlimits_-infty^inftyxe^-a(x-b)^2dx = b sqrtfracpia$$



I am having trouble thinking of how to attack this problem. I am currently thinking integration by parts to get rid of the $x$, and then try and coerce the exponent into the general Gaussian integral form but was wondering if there was a better solution.






definite-integrals exponential-function gaussian-integral







share|cite|improve this question













I came across Gaussian integrals, and was trying to prove them myself. I proved the basics, but am stuck on the following



$$intlimits_-infty^inftyxe^-a(x-b)^2dx = b sqrtfracpia$$



I am having trouble thinking of how to attack this problem. I am currently thinking integration by parts to get rid of the $x$, and then try and coerce the exponent into the general Gaussian integral form but was wondering if there was a better solution.






definite-integrals exponential-function gaussian-integral




definite-integrals exponential-function gaussian-integral






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asked Sep 11 at 2:56









Samyak Shah

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  • 2




    Substitution with $u=x-b$.
    – Randall
    Sep 11 at 3:04












  • 2




    Substitution with $u=x-b$.
    – Randall
    Sep 11 at 3:04







2




2




Substitution with $u=x-b$.
– Randall
Sep 11 at 3:04




Substitution with $u=x-b$.
– Randall
Sep 11 at 3:04










1 Answer
1






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up vote
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HINT



If we make the substitution $y = x - b$, we get:
beginalign*
int_-infty^+infty xe^-a(x-b)^2mathrmdx & = int_-infty^+infty(y+b)e^-ay^2mathrmdy \
& = int_-infty^+inftyye^-ay^2mathrmdy + int_-infty^+inftybe^-ay^2mathrmdy\
& = int_-infty^+inftybe^-ay^2mathrmdy
endalign*



once the function $ye^-ay^2$ is odd. Thus we have:
beginalign*
left(int_-infty^+inftybe^-ay^2mathrmdyright)^2 & = left(int_-infty^+inftybe^-ay^2mathrmdyright)timesleft(int_-infty^+inftybe^-az^2mathrmdzright)\
& = int_-infty^+inftyint_-infty^+inftyb^2e^-a(y^2 + z^2)mathrmdymathrmdz
endalign*



Hence, if you make the change of variable $y = rhocos(theta)$ and $z = rhosin(theta)$, you are able to obtain the sought result. Hope this helps.






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    1 Answer
    1






    active

    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    2
    down vote



    accepted










    HINT



    If we make the substitution $y = x - b$, we get:
    beginalign*
    int_-infty^+infty xe^-a(x-b)^2mathrmdx & = int_-infty^+infty(y+b)e^-ay^2mathrmdy \
    & = int_-infty^+inftyye^-ay^2mathrmdy + int_-infty^+inftybe^-ay^2mathrmdy\
    & = int_-infty^+inftybe^-ay^2mathrmdy
    endalign*



    once the function $ye^-ay^2$ is odd. Thus we have:
    beginalign*
    left(int_-infty^+inftybe^-ay^2mathrmdyright)^2 & = left(int_-infty^+inftybe^-ay^2mathrmdyright)timesleft(int_-infty^+inftybe^-az^2mathrmdzright)\
    & = int_-infty^+inftyint_-infty^+inftyb^2e^-a(y^2 + z^2)mathrmdymathrmdz
    endalign*



    Hence, if you make the change of variable $y = rhocos(theta)$ and $z = rhosin(theta)$, you are able to obtain the sought result. Hope this helps.






    share|cite|improve this answer
























      up vote
      2
      down vote



      accepted










      HINT



      If we make the substitution $y = x - b$, we get:
      beginalign*
      int_-infty^+infty xe^-a(x-b)^2mathrmdx & = int_-infty^+infty(y+b)e^-ay^2mathrmdy \
      & = int_-infty^+inftyye^-ay^2mathrmdy + int_-infty^+inftybe^-ay^2mathrmdy\
      & = int_-infty^+inftybe^-ay^2mathrmdy
      endalign*



      once the function $ye^-ay^2$ is odd. Thus we have:
      beginalign*
      left(int_-infty^+inftybe^-ay^2mathrmdyright)^2 & = left(int_-infty^+inftybe^-ay^2mathrmdyright)timesleft(int_-infty^+inftybe^-az^2mathrmdzright)\
      & = int_-infty^+inftyint_-infty^+inftyb^2e^-a(y^2 + z^2)mathrmdymathrmdz
      endalign*



      Hence, if you make the change of variable $y = rhocos(theta)$ and $z = rhosin(theta)$, you are able to obtain the sought result. Hope this helps.






      share|cite|improve this answer






















        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        HINT



        If we make the substitution $y = x - b$, we get:
        beginalign*
        int_-infty^+infty xe^-a(x-b)^2mathrmdx & = int_-infty^+infty(y+b)e^-ay^2mathrmdy \
        & = int_-infty^+inftyye^-ay^2mathrmdy + int_-infty^+inftybe^-ay^2mathrmdy\
        & = int_-infty^+inftybe^-ay^2mathrmdy
        endalign*



        once the function $ye^-ay^2$ is odd. Thus we have:
        beginalign*
        left(int_-infty^+inftybe^-ay^2mathrmdyright)^2 & = left(int_-infty^+inftybe^-ay^2mathrmdyright)timesleft(int_-infty^+inftybe^-az^2mathrmdzright)\
        & = int_-infty^+inftyint_-infty^+inftyb^2e^-a(y^2 + z^2)mathrmdymathrmdz
        endalign*



        Hence, if you make the change of variable $y = rhocos(theta)$ and $z = rhosin(theta)$, you are able to obtain the sought result. Hope this helps.






        share|cite|improve this answer












        HINT



        If we make the substitution $y = x - b$, we get:
        beginalign*
        int_-infty^+infty xe^-a(x-b)^2mathrmdx & = int_-infty^+infty(y+b)e^-ay^2mathrmdy \
        & = int_-infty^+inftyye^-ay^2mathrmdy + int_-infty^+inftybe^-ay^2mathrmdy\
        & = int_-infty^+inftybe^-ay^2mathrmdy
        endalign*



        once the function $ye^-ay^2$ is odd. Thus we have:
        beginalign*
        left(int_-infty^+inftybe^-ay^2mathrmdyright)^2 & = left(int_-infty^+inftybe^-ay^2mathrmdyright)timesleft(int_-infty^+inftybe^-az^2mathrmdzright)\
        & = int_-infty^+inftyint_-infty^+inftyb^2e^-a(y^2 + z^2)mathrmdymathrmdz
        endalign*



        Hence, if you make the change of variable $y = rhocos(theta)$ and $z = rhosin(theta)$, you are able to obtain the sought result. Hope this helps.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 11 at 3:15









        APC89

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