The probability integral $\int_{0}^{\alpha}$ e$\nolimits^{{-x^2}}_{}$ dx

The normal distribution is a very common model throughout the whole of science for the situation when things occur ``at random''. In particular, probability theory attempts to predict what will happen ``on average'', perhaps for computing risks and premiums on life insurance; in so doing one is often led to consider an integral of the form

I = $$\int_{0}^{\alpha}$$ e$$\nolimits^{{-x^2}}_{}$$ dx.

It turns out that this integral cannot be evaluated using the usual tricks -- substitution, integration by parts etc. But a power series representation and Corollary 7.10 can help. Thus

$$\nolimits^{x}_{} \sum_{{n=0}}^{\infty} {\frac{{x^n}}{{n!}}}$$

$$\nolimits^{{-x^2}}_{} \sum_{{n=0}}^{\infty} {\frac{{(-x^2)^n}}{{n!}}} \sum_{{n=0}}^{\infty} {\frac{{x^{2n}}}{{n!}}}$$

and we can integrate this term-by term, by Corollary 7.10, to get

$$\int$$ e$$\nolimits^{{-x^2}}_{}$$ dx = K + $$\sum_{{n=0}}^{\infty}$$$${\frac{{(-1)^nx^{2n+1}}}{{(2n+1)n!}}}$$.

Performing a definite integral removes the constant of integration, to give

$$\int_{0}^{\alpha}$$ e$$\nolimits^{-}_{}$$x² dx = $$\sum_{{n=0}}^{\infty}$$$${\frac{{(-1)^n\alpha^{2n+1}}}{{(2n+1)n!}}}$$.

The partial sums of the power series on the right can be computed, and converge quite quickly, so we have a practical method of evaluating the integral, even thought we can't ``do'' the integral.