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Today in my calculus class, we encountered the function $e^{-x^2}$, and I was told that it was not integrable.

I was very surprised. Is there really no way to find the integral of $e^{-x^2}$? Graphing $e^{-x^2}$, it appears as though it should be.

A Wikipedia page on Gaussian Functions states that

$$\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}$$

This is from -infinity to infinity. If the function can be integrated within these bounds, I'm unsure why it can't be integrated with respect to $(a, b)$.

Is there really no way to find the integral of $e^{-x^2}$, or are the methods to finding it found in branches higher than second semester calculus?

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    There is no antiderivative written in _elementary_ functions (imagine the roots for a polynomial of degree, e.g., five, for which there is no formula).2012-06-07
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    There is no *elementary function* whose derivative is $e^{-x^2}$. By elementary function we mean something obtained using arithmetical operations and composition from the standard functions we all know and love. But this is not a serious problem. A few important *definite integrals* involving $e^{-x^2}$ have pleasant closed form.2012-06-07
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    Try reading [this note](http://math.stanford.edu/~conrad/papers/elemint.pdf) of Brian Conrad's and the article by Rosenlicht referenced therein.2012-06-07
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    I guess this is not really a duplicate of [How to integrate $\int e^{-t^2} dt$ using introductory calculus methods](http://math.stackexchange.com/q/138664/856)?2012-06-07
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    Well, in someway it is no more surprising than stating that $\frac{1}{2}$ cannot be written as an integer. As noted by others, it is integrable, it is just that the collection of 'standard' functions is not rich enough to express the answer.2012-06-07
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    Unfortunately there are three or four different meanings being given to the word "integrable" here: (1) $f(x)$ is Riemann integrable on intervals $[a,b]$ (yes, every continuous function is) (2) $f(x)$ has an antiderivative that is an elementary function (no, it doesn't: the antiderivative $\sqrt{\pi}\ \text{erf}(x)/2$ is not an elementary function) (3) $\int_{-\infty}^\infty |f(x)|\ dx < \infty$ (yes, and this is the usual meaning of "integrable" in analysis) (4) $\int_{-\infty}^\infty f(x)\ dx$ can be expressed in "closed form" (yes, it is $\sqrt{\pi}$).2012-06-07
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    Sure, just use polar coordinates. That's what Gauss did.2012-06-07
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    you can use a trick to say that it = I and find $I^{2}$ then square root it to find a form over any bound for the function2013-03-13
  • 0
    Try [this link](http://www.youtube.com/watch?v=fWOGfzC3IeY) if you guys are still unable to solve it.2013-04-09
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    Hi Vivek, welcome to Math.SE. We prefer not to have answers that just consist of a link with no further explanation. Also, this video is about the definite integral $\int_{-\infty}^\infty e^{-x^2/2}\,dx$, which is not really what the question is about (the asker already knows about this case).2013-04-09
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    the video is legitimate (its title is "Integral of exp(-x^2) | MIT 18.02SC Multivariable Calculus, Fall 2010"), but it would have been better if you explained in the answer what it is.2013-04-09
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    While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes.2013-04-09
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    Integration does produce new kinds of functions. Consider $\ln x=\int_1^x (1/y)\;dy$ for $y>0$, which cannot be expressed in terms of arithmetic combinations, even allowing constant non-integer powers, of rational functions.2016-07-12

2 Answers 2

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That function is integrable. As a matter of fact, any continuous function (on a compact interval) is Riemann integrable (it doesn't even actually have to be continuous, but continuity is enough to guarantee integrability on a compact interval). The antiderivative of $e^{-x^2}$ (up to a constant factor) is called the error function, and can't be written in terms of the simple functions you know from calculus, but that is all.

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    But the evaluation of the integral over the whole real line is relatively easy!2012-06-07
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    Easy for Lord Kelvin.2012-06-07
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    You probably mean that any continuous function is Riemann integrable on a compact interval.2013-03-13
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    How to show that the function is non-elementary? I cannot remember seeing a proof of that.2013-08-10
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    @M.B.: see for example M.P. Wiener's text [here](http://math.stackexchange.com/questions/265780/how-to-determine-with-certainty-that-a-function-has-no-elementary-antiderivative/265884#265884).2013-08-10
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    Why don't you show us what it looks like?2014-05-18
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To build on kee wen's answer and provide more readability, here is an analytic method of obtaining a definite integral for the Gaussian function over the entire real line:

Let $I=\int_{-\infty}^\infty e^{-x^2} dx$.

Then, $$\begin{align} I^2 &= \left(\int_{-\infty}^\infty e^{-x^2} dx\right) \times \left(\int_{-\infty}^{\infty} e^{-y^2}dy\right) \\ &=\int_{-\infty}^\infty\left(\int_{-\infty}^\infty e^{-(x^2+y^2)} dx\right)dy \\ \end{align}$$

Next we change to polar form: $x^2+y^2=r^2$, $dx\,dy=dA=r\,d\theta\,dr$, therefore

$$\begin{align} I^2 &= \iint e^{-(r^2)}r\,d\theta\,dr \\ &=\int_0^{2\pi}\left(\int_0^\infty re^{-r^2}dr\right)d\theta \\ &=2\pi\int_0^\infty re^{-r^2}dr \end{align}$$

Next, let's change variables so that $u=r^2$, $du=2r\,dr$. Therefore, $$\begin{align} 2I^2 &=2\pi\int_{r=0}^\infty 2re^{-r^2}dr \\ &= 2\pi \int_{u=0}^\infty e^{-u} du \\ &= 2\pi \left(-e^{-\infty}+e^0\right) \\ &= 2\pi \left(-0+1\right) \\ &= 2\pi \end{align}$$

Therefore, $I=\sqrt{\pi}$.

Just bear in mind that this is simpler than obtaining a definite integral of the Gaussian over some interval (a,b), and we still cannot obtain an antiderivative for the Gaussian expressible in terms of elementary functions.