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I know that

$$I \equiv \int \limits_{-\infty}^\infty e^{−t^2} \, dt=\sqrt{\pi},\text{ and }\int \limits_{-\infty}^0 e^{−t^2} \, dt=\frac{\sqrt{\pi}}{2}.$$

However, I don't understand if (or how) I can find a similar solution for $\int \limits_{-\infty}^{a}e^{−t^2}dt, a \neq 0$, given that the error function actually does not yield closed form solutions.

Any help is greatly appreciated!

Michael Hardy
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jush
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    You can't (except for some approximate expressions if $a$ is either larger or small) – Fabian Sep 04 '12 at 16:39
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    If you want it to be in closed form, you can define "closed form" to include the error function :) – Snowball Sep 04 '12 at 16:44
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    If you prefer hypergeometric functions, you could say ${{\rm erf}\left(x\right)}=\frac{2}{\sqrt{\pi}} x; {\mbox{$_1$F$_1$}(1/2;,3/2;,-{x}^{2})} $ – Robert Israel Sep 04 '12 at 17:06
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    You can't do it in "closed form", but the fact that there are zillions of software packages that do it numerically, and efficiently, might give pause to those who might say "you can't". – Michael Hardy Sep 04 '12 at 17:06
  • See this and this question. – Argon Sep 04 '12 at 19:13
  • Thanks for all your comments and answers! I expected this to be the case but your discussion and hints really helped in getting a grip on it. Thanks! – jush Sep 05 '12 at 07:58

3 Answers3

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Since $e^{-t^2}$ is an even function, and since you know $\int_{-\infty}^{\infty}$, if you were able to find $\int_{-\infty}^b$ then you would be able to find $\int_a^b$ for any $a,b \in \mathbb{R}$. The fact that $e^{-t^2}$ does not have an elementary primative should suggest to you that you probably can't find $\int_a^b$ explicitly.

As an A-level student (UK 16-18 pre-university), I always wondered why we had to use a table of normal distribution values to get approximate probabilities. Now I know that the probability density function is basically a stretched and translated version of $e^{-t^2}$.

Food for thought: the natural logarithm, $\ln x$, has an integral definition:

$$ \ln x := \int_1^x \frac{dt}{t} $$

because $t^{-1}$ has no known primitive, yet we call $\ln x$ an elementary function, while $\text{erf} \, x$ is not.

Fly by Night
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$$ \int_{-\infty}^a \exp(-t^2)\, \text dt = \\ \frac{\sqrt{\pi}}{2}+\int_0^a\exp(-t^2)\, \text dt=\\ \frac{\sqrt{\pi}}{2}+\int_0^a\sum_{n=0}^\infty \frac{(-1)^n t^{2n}}{n!} \, \text dt =\\ \frac{\sqrt{\pi}}{2}+\sum_{n=0}^\infty \frac{(-1)^n a^{2n+1}}{n!(2n+1)} $$

You can use this to estimate the integral to sufficient accuracy (as GEdgar mentioned, however, this sum converges well when $a \approx 0$ but much slower when $a$ is large). As was mentioned, no simple "closed form" for this integral exists.

For example, we may approximate $\int_{-\infty}^1 \exp(-t^2)\, \text dt$ with a truncated (6 terms) series:

$$\int_{-\infty}^1 \exp(-t^2)\, \text dt =1.63305\cdots \approx \frac{\sqrt{\pi}}{2}+\sum_{n=0}^6 \frac{(-1)^n}{n!(2n+1)}=1.66306\cdots$$

See also this paper for more strategies.

Argon
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In probability theory, they sometimes define a function $\Phi$ like this enter image description here (reference) Yours is a simple change of variables, right?

GEdgar
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