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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!

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    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!2012-09-05

3 Answers 3

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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.

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    Thanks for this clarification!2012-09-05
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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.

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    Thanks for the hint to the paper! Very nice2012-09-05
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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?

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    The OP has already mentioned their dissatisfaction of the error function $\text{erf}\,x$.2012-09-04