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?

  • 5
    $\begingroup$ 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). $\endgroup$ – Artem Jun 7 '12 at 5:11
  • 12
    $\begingroup$ 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. $\endgroup$ – André Nicolas Jun 7 '12 at 5:12
  • 9
    $\begingroup$ Try reading this note of Brian Conrad's and the article by Rosenlicht referenced therein. $\endgroup$ – Dylan Moreland Jun 7 '12 at 5:20
  • 7
    $\begingroup$ 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. $\endgroup$ – copper.hat Jun 7 '12 at 6:08
  • 23
    $\begingroup$ 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 $\sqrt{\pi}$). $\endgroup$ – Robert Israel Jun 7 '12 at 6:54

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.

  • 3
    $\begingroup$ But the evaluation of the integral over the whole real line is relatively easy! $\endgroup$ – Lubin Jun 7 '12 at 6:33
  • 20
    $\begingroup$ Easy for Lord Kelvin. $\endgroup$ – copper.hat Jun 7 '12 at 6:41
  • $\begingroup$ You probably mean that any continuous function is Riemann integrable on a compact interval. $\endgroup$ – T. Eskin Mar 13 '13 at 7:25
  • 1
    $\begingroup$ How to show that the function is non-elementary? I cannot remember seeing a proof of that. $\endgroup$ – M.B. Aug 10 '13 at 16:32
  • 1
    $\begingroup$ @M.B.: see for example M.P. Wiener's text here. $\endgroup$ – Raymond Manzoni Aug 10 '13 at 16:44

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.

  • $\begingroup$ If $e^{-x^2}$ is the area under the curve then $I^2$ should have units of $area^{2}$. But \begin{align} I^2 &= \iint e^{-(r^2)}r\,d\theta\,dr \\ \end{align} has units of volume. How is it possible? $\endgroup$ – user599310 Jan 24 '20 at 16:00
  • $\begingroup$ @user599310 The units of $I^2$ are indeed units of $area^2$, but so too are the units of $\iint e^{-(r^2)}r\,d\theta\,dr$ in $area^2$. First note that $dA = r\,d\theta\,dr = dx\,dy$, all of which are in units of $area$. Second note that $e^{-r^2}$ is equivalent to $e^{-x^2}e^{-y^2}$. In words, the base shift that we performed from $dx\,dy$ to $r\,d\theta\,dr$ changes us from measuring the function in terms of two lengths, to measuring it in terms of chunks of areas. Hence why $e^{-r^2}$ is a function that returns areas, where $e^{-x^2}$ returns lengths. $\endgroup$ – Gerald Feb 6 '20 at 19:04
  • $\begingroup$ @CopaceticMan What I don't understand is if we plot the function $f(x,y)=e^{-(x^2+y^2)}$ then shouldn't the integral give us back the volume under the surface? $\endgroup$ – user599310 Feb 7 '20 at 16:20
  • $\begingroup$ @user599310, I am going to attempt some pseudo math to show it: $$ I^2 = \int e^-x^2 dx \times \int e^-x^2 dx = Area \times Area = Area^2$$ We can replace one x, with a dummy variable, move the dummy copy into the first integral to get a double integral. $$ I^2 = \int \int e^{-x^2-y^2} dA $$ In context, the integrand a function that returns two Area, as it is the product of two functions which return a distance, specifically the height above one axis. So $fdA$ is still in terms of $area^2$. You are correct in your confusion, as the leap may not appear obvious, but it's valid. $\endgroup$ – Gerald Jun 16 '20 at 19:42

Not the answer you're looking for? Browse other questions tagged or ask your own question.