# How to approximate the integral of the standard normal distribution.

So I have this eqn.

$$f(x)= \frac {e^ \frac{-x^2}{2}} {\sqrt{2\pi}}$$

I need to find:

$$\int\limits_{-1}^1 f(x)dx$$

So I want to use this series to integrate. I know that:

$$e^x = \sum\limits_{n=0}^\infty \frac {x^n}{n!}$$

So without considering the coefficient of the function, I can build my eqn into this series as such:

$$e^{\frac{-x^2}{2}} =\sum\limits_{n=0}^\infty \frac{(-1)^n x^{2n}}{2^nn!}$$

This is where I get lost. What do I do with the coefficient of e. How do I build this into a series and include the coefficient of e:

$$\frac {1}{\sqrt{2\pi}}\$$

• you know that total probability is one and standard normal distribution ranges from -1 to 1 so your integral final value is 1
– MRK
Sep 22 '13 at 23:38
• you cannot use elementary methods to integrate normal distribution and even if we use special function that will give the answer as an infinite series
– MRK
Sep 22 '13 at 23:39
• ^ what!? the domain of standard normal distribution is not [-1,1]. Sep 23 '13 at 0:07
• Use linearity of summation. Sep 23 '13 at 0:50
• lol domain for standard normal distribution is $$(-\infty, \infty)$$ Sep 23 '13 at 6:29

Ok I figured it out.

$$f(x)= \frac {e^ \frac{-x^2}{2}} {\sqrt{2\pi}}$$

I need to find:

$$\int\limits_{-1}^1 f(x)dx$$

So I want to use this series to integrate. I know that:

$$e^x = \sum\limits_{n=0}^\infty \frac {x^n}{n!}$$

So without considering the coefficient of the function, I can build my eqn into this series as such:

$$e^{\frac{-x^2}{2}}=\sum\limits_{n=0}^\infty \frac{(-1)^nx^{2n}}{2^nn!}$$

So we can just build up and get the origial.

$$\frac {1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}}=\frac {1}{\sqrt{2\pi}} \sum\limits_{n=0}^\infty \frac{(-1)^n x^{2n}}{2^nn!}$$

Then all we need to do is expand out the series to a few values of n, then we can anti-differentiate the series, and then we can solve using the FTC, where the integral from [a,b] = F(b) - F(a).

• Almost! Note that when you replace $x$ with $-x^2/2$ in the Taylor series, you should get $$e^{\frac{-x^2}{2}}=\sum\limits_{n=0}^\infty \frac{(-1)^nx^{2n}}{2^nn!}$$ which for odd $n$ will be negative, but for even $n$ will be positive. So $$\mathrm{e}^{-x^2} \sim 1-\frac{x^2}{2}+\frac{x^4}{8}+\mathcal{O}(x^6),$$ which should converge when $x \in (-1,1)$. Sep 23 '13 at 0:51
• Yeah my algebra was off in the OP, but I ended up fixing it in my answer there. I ended up with exactly that series. And it did converge. And it ends up as 68%, which was kinda funny. And I ended up using the -1^n in my notes. It's a much clearer notation I agree. Sep 23 '13 at 6:13
• Why is 68% funny? How many terms is this? I think the 'true' value is roughly 0.68, which is good, this is the area under the bell curve from -1 std deviation to +1 std deviation. Sep 23 '13 at 14:02
• Sorry, I didn't mean funny as in incorrect. I like math, so I get a kick out of things like this. Sep 23 '13 at 15:14
• Yep, consistency is the beautiful thing about maths :) Sep 24 '13 at 1:22

Hint: You cannot find an explicit answer for your integration. However the integration of normal distribution can be stated in terms of Q Function. $$Q(x)=\int_{x}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-t^2}{2}}dt$$ There are various estimates for this function which may be useful. For instance for $x>0$: $$\frac{x}{(x^2+1)\sqrt{2\pi}}e^{\frac{-x^2}{2}} \leq Q(t)\leq \frac{1}{x\sqrt{2\pi}}e^{\frac{-x^2}{2}}$$

Remark: If you want to continue working with series, it is enough to take the coefficient of the exponential function out of the integration.

• Thanks for the input. There is actually a very easy way to approximate this integral. Note that, since I'm working on an approximation, we don't need to use an infinite series. Rather, we can consider that n goes to 5 or some arbitrary number, and get an approximation of the integral of f(x). And as you said, you can actually take the coefficient right out. I will answer below. Sep 23 '13 at 0:18