Results 1 to 5 of 5

Math Help - Integrating by Polar Substitution?

  1. #1
    Newbie
    Joined
    Apr 2007
    Posts
    12

    Integrating by Polar Substitution?

    Hey,

    How would you solve the following integral:

    <br />
\int_{-\infty}^{\infty} e^{-x^2/2}dx<br />

    I've been trying to use a substitution, but I kind of forgot how to do it exactly. I've tried both polar and cylindrical coordinates, but I can't crunch it to the right answer. Any help? Sorry its been a while since I've done this type of integral.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Santa Cruz, CA
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by mbaboy View Post
    Hey,

    How would you solve the following integral:

    <br />
\int_{-\infty}^{\infty} e^{-x^2/2}dx<br />

    I've been trying to use a substitution, but I kind of forgot how to do it exactly. I've tried both polar and cylindrical coordinates, but I can't crunch it to the right answer. Any help? Sorry its been a while since I've done this type of integral.
    Note that e^{-\frac{x^2}{2}} is symmetric about the y axis, so \int_{-\infty}^{\infty}e^{-\frac{x^2}{2}}\,dx=2\int_0^{\infty}e^{-\frac{x^2}{2}}\,dx.

    Now, let I=\int_0^{\infty}e^{-\frac{x^2}{2}}\,dx=\int_0^{\infty}e^{-\frac{y^2}{2}}\,dy. Thus, I^2=\int_0^{\infty}e^{-\frac{x^2}{2}}\,dx\int_0^{\infty}e^{-\frac{y^2}{2}}\,dy=\int_0^{\infty}\int_0^{\infty}e  ^{-\frac{1}{2}\left(x^2+y^2\right)}\,dy\,dx

    Now, make the conversion over to polar.

    I leave you to verify that \int_0^{\infty}\int_0^{\infty}e^{-\frac{1}{2}\left(x^2+y^2\right)}\,dy\,dx\implies\i  nt_0^{\frac{\pi}{2}}\int_0^{\infty}e^{-\frac{r^2}{2}}r\,dr\,d\theta

    Now its simple integration from there.

    In the end, you will end up with I^2=\dots\implies I=\sqrt{\dots}. But to get your final answer, multiply your value for I by two.

    Can you take it from here?
    Last edited by Chris L T521; January 14th 2009 at 01:53 PM. Reason: small typo... XD
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2007
    Posts
    12
    Thank you this is a very beautiful solution. I can now solve it algorithmically, but can you possibly provide any explanation to
    I=\int_0^{\infty}e^{-\frac{x^2}{2}}dx=\int_0^{\infty}e^{-\frac{y^2}{2}}dy

    I know it works, but I'm thinking that this implies x=y which then wouldn't make much sense when converting into polar. Could you basically explain out the line where you introduce I?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by mbaboy View Post
    Thank you this is a very beautiful solution. I can now solve it algorithmically, but can you possibly provide any explanation to
    I=\int_0^{\infty}e^{-\frac{x^2}{2}}dx=\int_0^{\infty}e^{-\frac{y^2}{2}}dy

    I know it works, but I'm thinking that this implies x=y which then wouldn't make much sense when converting into polar. Could you basically explain out the line where you introduce I?
    I don't know how to call it in English... Actually, x is a variable you'll use only for the integral. It's just a name, it doesn't matter how it is called. You can call it y, z, a, b, whatever.
    It just has to design the same thing in the integral.

    The aim of writing I with x and with y is because we'll associate the two integrals.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,347
    Thanks
    30
    Quote Originally Posted by Moo View Post
    I don't know how to call it in English.
    I think the english word is dummy variable.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Integrating correct substitution,,
    Posted in the Calculus Forum
    Replies: 10
    Last Post: November 5th 2008, 01:19 PM
  2. Integrating 'sec(x)' Using T-Substitution
    Posted in the Calculus Forum
    Replies: 4
    Last Post: August 3rd 2008, 02:41 PM
  3. Integrating using Trig Substitution
    Posted in the Calculus Forum
    Replies: 7
    Last Post: June 14th 2008, 03:42 PM
  4. Integrating with U-Substitution
    Posted in the Calculus Forum
    Replies: 10
    Last Post: June 14th 2008, 09:51 AM
  5. integrating in polar co-ordinates
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 15th 2007, 04:56 AM

Search Tags


/mathhelpforum @mathhelpforum