Solve Impossible Integral: \int exp (-x^2) dx

[Kawakaze]

Hi guys,

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

\int exp (-x^2) dx

 

[Delta2]

This is a well known case , integral that exists but doesn't have an analytical expression. You can try it in all mathematic software and see that they can compute it only numerically.

 

[tiny-tim]

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"

(unless the limits are -∞ to ∞, or 0 to ±∞)

 

[Gib Z]

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]

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]

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.

 

[micromass]

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??