Gaussian integral and integration by substitution

Question

I know that to calculate the gaussian integral $$\int_{-\infty}^\infty e^{-x^2}dx$$ you consider $\pmb I^2$, change to polar coordinates and it works out to $\sqrt\pi$. We do this because $e^{-x^2}$ doesn't have an indefinite integral. However I don't understand why $e^{-x^2}$ doesn't have an indefinite integral, and why we can't just integrate by substituion. Can anybody explain to me why this is?

Edit: I meant to say whitout an elementary indefinite integral.

Answer 1

The function is integrable. The value of the indefinite integral is $\sqrt{\pi}$. However, there is no elementary function whose derivative $e^{-x^2}$. To see why this is true, first observe that the antiderivative of $e^{-x^2}$ (up to a constant) is the error function. The error function is defined by

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

and is not an elementary function. If we take the derivative with respect to $x$ on both sides we get

$$\frac{d}{dx}\left(\operatorname{erf}(x)\right)=\frac{2}{\sqrt \pi}e^{-x^2}$$

by the first part of the fundamental theorem of calculus. Then, rearranging and taking the antiderivative forms

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

which differs by a constant from the error function. Since the error function cannot be expressed in terms of elementary functions, it is clear that the error function minus a constant also cannot be expressed in terms of elementary functions.

Therefore, you cannot represent the antiderivative of $e^{-x^2}$ by elementary functions. Instead, the antiderivative is represented by the error function.

Answer 2

Well, $\exp(-x^2)$ does have an indefinite integral.
It is evaluated from the fact that the Taylor series of $\exp(-x^2)$ gives $$\exp(-x^2)=\sum_\limits{n=0}^\infty \dfrac{x^{2n}}{n!}$$ and the radius of curvature is $\infty$.

So, we finally get $$\int e^{-x^2} dx = \int \sum_\limits{n=0}^\infty \dfrac{x^{2n}}{n!} dx = \sum_\limits{n=0}^\infty \dfrac{x^{2n+1}}{n!(2n+1)} + C = \sum_\limits{n=0}^\infty 2^{n}\cdot \dfrac{x^{2n+1}}{(2n+1)!} + C$$

This is the desired indefinite integral, but the problem is that this mathematical expression has no closed form and cannot be expressed in terms of any known elementary functions. Hence it is difficult to use this form of the integral for any proper mathematical analysis.