Find the integral of e^(-x^2) with respect to x.

Step-by-step solution

1. 1

   Try the obvious substitution and see it fail. Setting $u=-x^{2}$ gives $du=-2x\,dx$, but there is no $x$ factor in the integrand to absorb, so the substitution cannot be completed. Integration by parts also loops without ever simplifying. This is not a lack of cleverness: the failure is structural.

2. 2

   Know the theoretical statement. By Liouville theorem on integration in finite terms, $e^{-x^{2}}$ has no elementary antiderivative — none can be written with polynomials, roots, exponentials, logarithms and trigonometric functions in finitely many operations. Compare $\int xe^{-x^{2}}dx=-\tfrac12e^{-x^{2}}+C$, which is elementary precisely because the extra $x$ makes the substitution work.

3. 3

   Introduce the error function as a definition. Mathematicians name the missing antiderivative:

   $\operatorname{erf}(x)=\frac{2}{\sqrt{\pi}}\int_{0}^{x}e^{-t^{2}}\,dt$

   The prefactor $\tfrac{2}{\sqrt{\pi}}$ is chosen so that $\operatorname{erf}(\infty)=1$, which makes erf a probability in statistics.

4. 4

   Invert the definition to answer the question. Differentiating the definition gives $\operatorname{erf}'(x)=\tfrac{2}{\sqrt{\pi}}e^{-x^{2}}$, so multiplying by $\tfrac{\sqrt{\pi}}{2}$ undoes the normalisation:

   $\int e^{-x^{2}}\,dx=\frac{\sqrt{\pi}}{2}\operatorname{erf}(x)+C$

5. 5

   Contrast with the definite integral over the whole line. Although no elementary antiderivative exists, the definite integral is famously exact:

   $\int_{-\infty}^{\infty}e^{-x^{2}}dx=\sqrt{\pi}$

   obtained by squaring and switching to polar coordinates. Having a closed-form definite integral is entirely compatible with having no closed-form indefinite one.