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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?

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    $\begingroup$ go review the rules of differentiation. $\endgroup$
    – ILoveMath
    May 3, 2017 at 4:02
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    $\begingroup$ Have you learnt about the product rule? $\endgroup$
    – Zain Patel
    May 3, 2017 at 4:03
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    $\begingroup$ If you apply the quotient rule and chain rule on the derivative for $\frac{e^{x^2}}{2x},$ you do not get $e^{x^2}$ $\endgroup$ May 3, 2017 at 4:07
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    $\begingroup$ 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 $e^{-x^2}$): en.wikipedia.org/wiki/… $\endgroup$
    – Tob Ernack
    May 3, 2017 at 4:22
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    $\begingroup$ @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). $\endgroup$ May 3, 2017 at 20:57

1 Answer 1

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

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