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

2 Answers 2

up vote 24 down vote accepted

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! –  Lubin Jun 7 '12 at 6:33
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Easy for Lord Kelvin. –  copper.hat Jun 7 '12 at 6:41
    
You probably mean that any continuous function is Riemann integrable on a compact interval. –  Thomas E. Mar 13 '13 at 7:25
    
How to show that the function is non-elementary? I cannot remember seeing a proof of that. –  M.B. Aug 10 '13 at 16:32
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@M.B.: see for example M.P. Wiener's text here. –  Raymond Manzoni Aug 10 '13 at 16:44

Int e^(-x^2) dx from a to b = Int e^(-y^2) dy from a to b Let I=Int e^(-x^2) dx from -infinity to infinity I^2= Int e^(-x^2) dx from -infinity to infinity x Int e^(-y^2) dy from -infinity to infinity =Dobule Int e^-(x^2+y^2) dxdyfrom -infinity to infinity change to polar form: x^2+y^2=r^2 you get I^2= Int Int e^-(r^2)dA (as dydx = dA, A=area)

So while you observe from the Polar graph, dA = rdθdr as dθ tends to zero (definition of integration) Hence, I^2=Int(2pi,0)Int(infinity,0) re^(-r^2)dθdr

By manipulation: -2I^2=Int(2pi,0)Int(infinity,0) (-2r)e^(-r^2)dθdr where -2r is the derivative of -r^2

I^2=[Int(2pi,0)dθ x Int(infinity,0) (-2r)e^(-r^2)dr]/-2 =(-pi)x[e^(-r^2),infinity,0] do the math. I=Root pi

You're welcome, Kee Wen Feel free to contact me if you don't understand anything here

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