# Impossible integral

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

Delta2
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.

tiny-tim
Science Advisor
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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Gib Z
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.

Delta2
Homework Helper
Gold Member
Well despite what i originally said wikipedia says there is an analytic expression with infinite terms though.
$$\int e^{-x^2}dx=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{n!(2n+1)}+c$$

Gib Z
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*.

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.

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??