# solving the integral of $e^{x^2}$

How to solve following integral?

Any hints for the above integral ?

$$\int{e^{x^2}} dx =?$$

I use change of variable $t=x^{2}$. so, $$\frac{1}{2}\int{\frac{e^{t}}{t^{\frac{1}{2}}}dt}$$

But I could not solve it!

thanks.

• This is an integral wich cannot be expressed in terms of standard functions. – abcdef Apr 19 '15 at 16:29
• There is no closed form, maybe you are looking for: $$\int_{-\infty}^{\infty}e^{-x^2}dx=\sqrt{\pi}$$ Which can be derivated by using polar coordinates on: $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}dx\:dy$$ – Uncountable Apr 19 '15 at 16:30
• See error function, Liouville's theorem, and the Risch algorithm. – Lucian Apr 19 '15 at 17:09
• Is this really not a duplicate of a previous question? – GEdgar Apr 22 '15 at 17:03
• See Dawson function and this article, which I found in an answer to this question. Not the same Dawson, by the way. – Jean-Claude Arbaut Apr 22 '15 at 18:07

The integral which you describe has no closed form which is to say that it cannot be expressed in elementary functions.

For example, you can express $\int x^2 \mathrm{d}x$ in elementary functions such as $\frac{x^3}{3} +C$. However, the indefinite integral from $(-\infty,\infty)$ does exist and it is $\sqrt{\pi}$ so explicitly:

$$\int^{\infty}_{-\infty} e^{-x^2} = \sqrt{\pi}$$

Note the difference in your integral and in the integral above, there is a negative sign in the one above. The integral you have does not converge for the specified bounds.

Also look at Risch Algorithm and the ERF Function.

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

$=\int\sum\limits_{n=0}^\infty\dfrac{x^{2n}}{n!}dx$

$=\sum\limits_{n=0}^\infty\dfrac{x^{2n+1}}{n!(2n+1)}+C$

Don't even think about it, it is known to have no closed form.

• "No closed form" is too vague. Better to say, "not an elementary function". – GEdgar Apr 19 '15 at 17:12
• @GEdgar: Agreed. – Alex M. Apr 19 '15 at 17:12
• @GEdgar: Is there anything that can be done against the monkeys that have just discovered the downvote button? – Alex M. Apr 23 '15 at 21:55