Results 1 to 4 of 4

Thread: an integration problem with exponantial

  1. #1
    Newbie
    Joined
    Nov 2010
    Posts
    3

    an integration problem with exponantial

    I have encountered a problem with the following form:

    $\displaystyle \int x e^{ax^2-bx}dx$

    I can only find formulas for $\displaystyle \int x e^{ax^2}dx$ and $\displaystyle \int x e^{bx}dx$ and don't know how to make use of these two formulas.

    Can someone help me on this? Thanks a lot!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by billzhao View Post
    I have encountered a problem with the following form:

    $\displaystyle \int x e^{ax^2-bx}dx$

    I can only find formulas for $\displaystyle \int x e^{ax^2}dx$ and $\displaystyle \int x e^{bx}dx$ and don't know how to make use of these two formulas.

    Can someone help me on this? Thanks a lot!!
    Complete the square in the exponent to get

    $\displaystyle \displaysytle a(x^2-\frac{b}{a}x+\frac{b^2}{4a^2})-\frac{b^2}{4a}=a(x-\frac{b}{a})^2-\frac{b^2}{4a}$

    Now the integral looks like

    $\displaystyle \displaystyle \int xe^{a(x-\frac{b}{2a})^2-\frac{b^2}{4a}}dx=e^{-\frac{b^2}{4a}}\int xe^{a(x-\frac{b}{2a})^2} dx $

    From here just make a u sub for the exponent and then you will use both of your formula's for the integral
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2010
    Posts
    3

    Thanks for the quick reply!! But...

    Is there a closed form solution for the below integration?

    $\displaystyle \int e^{ax^2}dx$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,718
    Thanks
    3003
    Yes, but you have to use the "special function" $\displaystyle erf(x)= \frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2}dt$.

    Let $\displaystyle iu= \sqrt{a}x$. Then $\displaystyle dx= \frac{i}{\sqrt{a}}du$, $\displaystyle e^{ax^2}= e^{-u^2}$ and the integral becomes $\displaystyle \frac{i}{\sqrt{a}}\int e^{-u^2}du= \frac{i\sqrt{\pi}}{2\sqrt{a}}erf(-i\sqrt{a}x)+ C$.

    The fact that you have $\displaystyle x^2$ in the exponential rather than $\displaystyle -x^2$ complicates things and causes those "i" factors.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: Feb 19th 2010, 10:55 AM
  2. Integration Problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Feb 15th 2010, 01:17 PM
  3. integration problem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Nov 8th 2009, 08:52 PM
  4. Negative Exponantial Equation
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: Oct 13th 2009, 06:44 PM
  5. integration problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Aug 30th 2009, 09:08 PM

Search tags for this page

Click on a term to search for related topics.

Search Tags


/mathhelpforum @mathhelpforum