4

So a basic form of the Gauss error function is

$$\int e^{x^2}\,dx$$

and apparently this is not solvable analytically. But why? It seems that I can solve it pretty easily as

$$\int e^{x^2}\,dx = \frac{1}{2x}e^{x^2}$$

since

$$\frac{d}{dx} \frac{1}{2x}e^{x^2} = e^{x^2}.$$

Why is this wrong?

DMcMor
  • 9,407
  • 29
    go review the rules of differentiation. – ILoveMath May 03 '17 at 04:02
  • 2
    Have you learnt about the product rule? – Zain Patel May 03 '17 at 04:03
  • 1
    If you apply the quotient rule and chain rule on the derivative for $\frac{e^{x^2}}{2x},$ you do not get $e^{x^2}$ – Vivek Kaushik May 03 '17 at 04:07
  • 1
    As for the possibility of finding an antiderivative using elementary functions, you might be interested in Liouville's theorem (differential algebra), which can be used to prove certain functions do not have elementary antiderivatives (including $e^{-x^2}$): https://en.wikipedia.org/wiki/Liouville%27s_theorem_(differential_algebra) – Tob Ernack May 03 '17 at 04:22
  • Note that this assumes a precise definition of what is an "elementary function". They consist of the rational functions, algebraic functions (like taking n-th roots, or more generally solutions to polynomial equations over $\mathbb{C}$), exponential functions (including complex exponentials which give the trig functions) and logarithms, and finally compositions of functions of the previous types. – Tob Ernack May 03 '17 at 04:34
  • 1
    @TheGreatDuck "The solution to that integral is the error function". Actually, it is the imaginary error function multiplied by a factor of $\frac{\sqrt{\pi}}{2}$ (plus the constant of integration). – projectilemotion May 03 '17 at 20:57

1 Answers1

25

When you take the derivative of $$\frac{1}{2x}e^{x^{2}}$$ you need to use the product and chain rules. You get $$\frac{d}{dx}\frac{1}{2x}e^{x^{2}} = -\frac{1}{2x^{2}}e^{x^{2}}+\frac{1}{2x}2xe^{x^{2}} = e^{x^{2}}\left(1-\frac{1}{2x^{2}}\right) \neq e^{x^{2}}. $$

DMcMor
  • 9,407