Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
You can't (except for some approximate expressions if $a$ is either larger or small) – Fabian Sep 4 '12 at 16:39
If you want it to be in closed form, you can define "closed form" to include the error function :) – Snowball Sep 4 '12 at 16:44
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 4 '12 at 17:06
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 4 '12 at 17:06
See this and this question. – Argon Sep 4 '12 at 19:13
up vote 8 down vote accepted

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.

share|cite|improve this answer
Is "number" a typo? – Michael Hardy Sep 4 '12 at 17:07
Absolutely! Thanks Mike :-) – Fly by Night Sep 4 '12 at 17:09
Nice food for thought: what we call elementary at some point is of course arbitrary. But then again, I know quite a few simplification rules for logarithm whereas I do not know what to do with a bunch of error functions and of course I was taught logarithms much before error functions so it feels kind of more natural ;-) – Fabian Sep 4 '12 at 18:13
Thanks for this clarification! – jush Sep 5 '12 at 7:58

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

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.

share|cite|improve this answer
Thanks for the hint to the paper! Very nice – jush Sep 5 '12 at 8:09

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

share|cite|improve this answer
The OP has already mentioned their dissatisfaction of the error function $\text{erf}\,x$. – Fly by Night Sep 4 '12 at 18:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.