A primitive function of $ e^{x^{2}} $

Question

I made some efforts to set a closed form of primitive function of $ e^{x^{2}} $ i find

this function : $ f(x)=\frac{x}{2x^{2}-1}e^{x^{2}} $ where :

$f'(x)=(\frac{x}{2x^{2}-1}e^{x^{2}})'$= $\frac{4x^{4}-4x^{2}-1}{4x^{4}-4x^{2}+1}e^{x^{2}}$

my question is : can I take $ f(x)=\frac{x}{2x^{2}-1}e^{x^{2}} $ as a primitive function

of $ e^{x^{2}} $ if $x$ hold a big values ?

is there a function better then $ f(x)=\frac{x}{2x^{2}-1}e^{x^{2}} $ ?

thank you for any replies or any comments

Answer 1

As already mentioned, there is "closed form" of $\int e^{x^2}$, at least in terms of a collection of "basic functions". Now, when you ask if it is a good approximation for an antiderivative for large $x$, the answer is that it depends! Note: $$ \frac{4x^4 - 4x^2 - 1}{4x^4 - 4x^2 + 1} = 1 - \frac{2}{4x^4 - 4x^2 + 1} = 1 + O(x^{-4}). $$ Now, as a ratio: $$ \frac{\big(1+O(x^{-4})\big)\, e^{x^2}}{e^{x^2}} = 1+O(x^{-4}) \to 1 ~ \text{ as $x \to \infty$}. $$ However, as a difference $$ \big(1+O(x^{-4})\big)\,e^{x^2} - e^{x^2} = O(x^{-4})\, e^{x^2} \to - \infty ~\text{ as $x \to \infty$} $$ where we note that term $O(x^{-4})$ is negative. So, as a ratio, reasonably good; as a difference, it seems reasonably bad.

Answer 2

$f$ is not a primitive of $e^{x^2}$ because $$f'(x) = \frac{e^{x^2}(4x^4-4x^2-1)}{(2x^2-1)^2}$$

which is not $e^{x^2}$.

In fact, $e^{x^2}$ does not have an antiderivative in terms of elementary functions. For some notes on this topic, see this talk by Martin Leslie.

Answer 3

thank you for any replies or any comments

It can be proven, using either Liouville's theorem or the Risch algorithm, that the anti–

derivative of $e^{x^n}$, with $n\in\mathbb N$, cannot be expressed in terms of elementary functions for

$n\ge2$. Letting $n\in\mathbb Q$ , the case $n=\dfrac1m$ , with $m\in\mathbb N^*$, can also be added to the previous

list of exceptions. However, it can be expressed in terms of special functions. Thus, for

$n=2$, it can be expressed in terms of the error function, $\displaystyle\int e^{x^2}dx=\frac{\sqrt\pi}2\cdot\text{erfi}(x)$, and

for other values of n in terms of the $($incomplete$)$ $\Gamma$ function, $\displaystyle\int e^{x^n}dx=\frac{(-1)^{^{-\tfrac1n}}}n~\cdot$

$\cdot~\Gamma\bigg(\frac1n,-x^n\bigg)$. Also, as far as definite integrals are concerned, we have the beautiful

identity $\displaystyle\int_0^\infty e^{-x^n}dx=\Big(\tfrac1n\Big)!=\Gamma\bigg(1+\frac1n\bigg)$.