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In preparation for an exam, I must find the primitive of $e^{-x^2}$. But when I look up the equation in an integral list I get a primitive that seems way overcomplicated for the level of math they ask us. Here is the link to the image of the resulting function.

Deepesh Meena
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Matthieu Meunier
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    Could you clarify whether you mean $e^{(-x^2)}$ or $(e^{-x})^2$? Exponentiation is not associative. – mweiss Aug 24 '18 at 14:27
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    This function has no elementary primitive – Math Student Aug 24 '18 at 14:27
  • That's about as simple as one can get. The antiderivative is not an "elementary function". – Angina Seng Aug 24 '18 at 14:28
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    There must be some mistake! The primitive of $e^{-x^2}$ cannot be expressed in terms of elementary functions. Everybody knows this! So I would not expect to see it asked in an exam. – TonyK Aug 24 '18 at 14:28
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    When you say "In preparation for an exam I must...": Please clarify, is this a problem that was posed to you by an examiner, or one you chose for yourself? – mweiss Aug 24 '18 at 14:28
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    I agree with @TonyK. Exactly what is the statement of the problem? – José Carlos Santos Aug 24 '18 at 14:29
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    Are you sure you need to compute the primitive? If you want to compute, for example, $\int_{-\infty}^{+\infty} e^{-x^2},dx$ you do not need a primitive. – Gibbs Aug 24 '18 at 14:29
  • Also, while @The_lost is correct that there is no elementary primitive, the definite integral $\int_{-\infty}^{\infty} e^{-x^2} dx$ can be found. Is it possible that this is what you are supposed to know? – mweiss Aug 24 '18 at 14:29
  • For the definite integral from $0$ to $\infty$ you could see this answer – Ross Millikan Aug 24 '18 at 14:37
  • Perhaps what is wanted is in terms of error function. – karakfa Aug 24 '18 at 14:41

3 Answers3

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If you are talking about an explicit primitive, like to say that, for example, a primitive of $x$ might be $\frac{1}{2}x^2 + C$, then there is naught.

Risch's algorithm will help you in proving this.

For what concerns instead the possibility of writing a "primitive" in other terms (that is, not through elementary function but with something else, like series or products) then here you are:

$$\int e^{-x^2}\ \text{d}x = \sum_{k = 0}^{+\infty} \frac{(-1)^k}{k!} \int x^{2k} \ \text{d}x = \sum_{k = 0}^{+\infty} \frac{(-1)^k}{k!} \frac{x^{2k+1}}{2k+1}$$

The latter sum is known, and it's called the Error Function, a special function (not an elementary function). Whence:

$$\sum_{k = 0}^{+\infty} \frac{(-1)^k}{k!} \frac{x^{2k+1}}{2k+1} = \frac{1}{2} \sqrt{\pi } \text{erf}(x)$$

Hence if you like:

$$\int e^{-x^2}\ \text{d}x = \frac{1}{2} \sqrt{\pi } \text{erf}(x)$$

As I said: this is not a primitive in the usual sense.

You can find documentation on the Error Function online.

P.s. For the integration, I used nothing but Taylor Series for the exponential.

You can derive the result written in your book in the same way.

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Are you allowed to use that list? If yes, all you have to do is plug in $a=1$ and you get the primitive $\frac{\sqrt{\pi}}2\operatorname{erf}(x)$.

You can be of the opinion that this is useless because $\operatorname{erf}(x)$ is usually defined to be $$\operatorname{erf}(x)=\frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2}dt. $$ On the other hand, $\operatorname{erf}$ is a perfectly nice function that is easily accessible via tables or calculators.

Kusma
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Usually they should only ask you to integrate $\int_0^{\infty}e^{-x^2}$, which is $\frac{\sqrt \pi}{2}$, as can be calculated by multivariable calculus. The function $\mathrm{erf}$ arises exactly because of this integral.

Trebor
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    This is not really an answer. I think the OP should make a more precise question before anyone can answer. – Gibbs Aug 24 '18 at 14:35