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Integration of a exponential squared function

  1. Jul 8, 2010 #1
    Is it possible to integrate this function
    [tex]\int {e^{x^2}} dx [/tex]

    \int \left (y^2) e^{y^2} dy

    The book says there is no direct method to solve this type of integrals but there is no hint how to solve this.Any one please help.

    Thanks :)

    Sorry for the latex code i have put the integration function code but its no coming up may be there is a mistake in the code.
  2. jcsd
  3. Jul 8, 2010 #2


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    Homework Helper

    The functions are possible to integrate in the sense that given limits for the integral you can find the numerical value of it (as long as the limits are such that the integral doesn't diverge, which both of your examples would if a limit was [itex]\pm \infty[/itex]), though you may need to do so using a numerical method.

    The indefinite integrals are still integrable, however, there is no expression for either in terms of 'elementary functions' like e^(something), tan, cos, etc. Functions can be defined in terms of integrals. For example, the error function is defined as

    [tex]\mbox{erf}(x) = \frac{2}{\sqrt{\pi}}\int_0^x dt~e^{-t^2}[/tex]

    (Note the minus sign in the exponential makes it different from your first example, at least for real valued x.)
  4. Jul 8, 2010 #3
    oh i actually have limits i wanted to have a general solution of these type of problems and the actual problem goes like this
    [tex]\int_{1}^{2} {y^2}{e^{1/y^2}}[/tex]
    Last edited: Jul 8, 2010
  5. Jul 8, 2010 #4


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    No, it is NOT.

    The derivative of

    That is nothing like

    [tex]y^2 e^{1/y^2}[/tex]
    cannot be integrated in terms of elementary functions.
  6. Jul 8, 2010 #5


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    Note that

    \int_{-\infty}^\infty \frac 1 {\sqrt{2 \pi \sigma^2}} e^{-\frac{(x-\mu)^2}{2\sigma^2}} \, dx = 1

    try dressing your integrals up as multiples of a specific normal density.
  7. Jul 8, 2010 #6
    so is it really possible to solve
    [tex]\int_{1}^{2} {y^2}{e^{1/y^2}}[/tex]

    Actually it is a part of double integral the double integral was

  8. Jul 8, 2010 #7


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    It is possible to find the value using a computer to whatever number of digits you want. The corresponding definite integral does not have a solution in terms of elementary functions. The change of variables [itex]u = 1/y^2[/itex] will result in the integral of [itex]\sqrt{u}e^u[/itex], which has no closed form solution (in terms of elementary functions).
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