integrate e^(-(x^2))dx

integrate e^(-(x^2))dx

There's no closed-form solution to the integral, though it can be written as an asymptotic series:

First, rewrite the integral into a form where you can do integration by parts:

\int e^{-x^2}\;dx=\int -\(\frac{1}{2x}\) \; 2x e^{-x^2}\;dx

Now we can use the integration by parts formulation

\int u \; dv = uv-\int v \; du ,

where

u=-\(\frac{1}{2x}\)

and

dv = 2x e^{-x^2}\;dx

That gives you

\int e^{-x^2}\;dx= -\(\frac{1}{2x}\)e^{-x^2}-\int \(\frac{1}{2x^2}\)e^{-x^2} \; dx

Now you have a closed-form term, plus a new integral. If you employ integration by parts to solve the new integral, you'll get yet another closed-form term and another new integral (this time with higher powers of x). If you keep going, integrating by parts over and over and over, you build up a series of closed-form terms, and end up with an integral whose value is approaching zero. In the limit, as the number of iterations goes to infinity, you end up with:

\int e^{-x^2}\;dx=\frac{e^{-x^2}}{2}\(x^{-1}-\frac12 x^{-3} + \frac34 x^{-5} - \frac{15}{8} x^{-7}+\frac{105}{16}x^{-9}- \; ... \;\) = \frac{e^{-x^2}}{2}\sum_{n=0}^\infty \frac{\(-1\)^n\(2n-1\)!!}{2^n}x^{2n+1}