# Impossible integral

Kawakaze
Hi guys,

I came across this in a textbook, it says its as good as impossible to integrate this expression. Ive met a lot of smart guys on here, maybe someone can do it?

$$\int exp (-x^2) dx$$

## Answers and Replies

Homework Helper
Gold Member
This is a well known case , integral that exists but doesnt have an analytical expression. You can try it in all mathematic software and see that they can compute it only numerically.

Homework Helper
Hi Kawakaze! Sorry, the only way is to look it up in tables of erf(x) (the "error function") … see http://en.wikipedia.org/wiki/Error_function" [Broken] (unless the limits are -∞ to ∞, or 0 to ±∞)

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Homework Helper
Liouville's Theorem (1835, Differential Galois Theory) states that if f(x) and g(x) are rational functions where f(x) is not identically zero and g(x) is non-constant then $$\int f(x) e^{g(x)} dx$$ is an elementary function if and only if there exists some rational function r(x) satisfying the equation $$f(x) = r'(x) +g'(x) r(x)$$. Applied to this specific problem, the theorem states that $\exp(-x^2)$ has an elementary anti-derivative if and only if there is a rational function r(x) such that $$1= r'(x) - 2x r(x)$$. It is possible to show that r(x) doesn't exist with a proof by contradiction. Maybe you or someone else wants to try it.

Homework Helper
Gold Member
Well despite what i originally said wikipedia says there is an analytic expression with infinite terms though.
$$\int e^{-x^2}dx=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{n!(2n+1)}+c$$

Homework Helper
The sum on the right is not considered an elementary function. Your book must have said that the integral is impossible to find * in terms of elementary functions*.

Phezboy
That's a standard Gaussian Integral

I2=∫e-x2∫e-y2

where y is just a dummy variable. Change to polar co-ordinates

r2 = x2+y2
dxdy=rdrdθ

I2=∫∫re-r2drdθ

which is trivial to calculate. You then take the square root of the answer.

Staff Emeritus
Homework Helper
That's a standard Gaussian Integral

I2=∫e-x2∫e-y2

where y is just a dummy variable. Change to polar co-ordinates

r2 = x2+y2
dxdy=rdrdθ

I2=∫∫re-r2drdθ

which is trivial to calculate. You then take the square root of the answer.

Right. So you won't mind us giving us the value of

$$\int_0^1 e^{-x^2}dx$$

if it is so trivial??