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.
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6This is an integral wich cannot be expressed in terms of standard functions. – abcdef Apr 19 '15 at 16:29
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8There 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
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See error function, Liouville's theorem, and the Risch algorithm. – Lucian Apr 19 '15 at 17:09
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1Is this really not a duplicate of a previous question? – GEdgar Apr 22 '15 at 17:03
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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
2 Answers
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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.
Jeel Shah
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$\int e^{x^2}~dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{x^{2n}}{n!}dx$
$=\sum\limits_{n=0}^\infty\dfrac{x^{2n+1}}{n!(2n+1)}+C$
Harry Peter
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