Integrate Gaussian function

I am trying to integrate Gaussian distribution from -m to m to find parametar A. I have done this so far:

$\int_{-m}^{m}\frac{A}{\sqrt(2\pi)\sigma}e^{-(x-m)^{2}/(2\sigma^{2})}dx=1$

after $u=\frac{x-m}{\sigma}$ I got:

$A \frac{1}{\sqrt(2\pi)} \int_{-2m/\sigma}^{0} e^{-0.5 u^{2}}du=1$

In order to solve this integral I want to make Q function but limits of integrals are not matching with limits of Q fnction.

Is anyone know how should solve this integral with Q function to get value of parametar A.

Thank you!

1 Answer

You cannot integrate the Normal Gaussian distribution, because you cannot express it in terms of elementary functions.

Instead, what you can try, is to express your Normal distribution as a Gaussian distibution with mean $0$ and variance $1$: from there you look up the integral table and find an esteem of your parameters.

In your case $$\frac{1}{\sqrt{2\pi}\sigma}e^{-(x-\mu)^{2}/(2\sigma^{2})} \sim N\left(\mu, \sigma^2\right) \sim X$$ but you can normalize it to get $$\frac{X - \mu}{\sigma} \sim Y \sim N(0,1)$$ so the integral you wanted to estimate now comes to be $$\begin{align} \int_{-\mu}^{\mu}\frac{A}{\sqrt{2\pi}\sigma}e^{-(x-\mu)^{2}/(2\sigma^{2})} dx &= A \cdot \int_{-\mu}^{\mu}\frac{1}{\sqrt{2\pi}\sigma}e^{-(x-\mu)^{2}/(2\sigma^{2})} dx\\ &= A \cdot P(-\mu \le X \le \mu)\\ &= A \cdot P\left(\frac{-\mu - \mu}{\sigma} \le \frac{X - \mu}{\sigma} \le \frac{\mu - \mu}{\sigma}\right)\\ &= A \cdot P\left(\frac{-2\mu}{\sigma} \le N(0,1) \le 0\right)\\ &= A \left( \Phi\left( 0 \right) - \Phi\left(\frac{-2\mu}{\sigma}\right) \right)\\ &= A \left( \frac{1}{2} - \Phi\left(\frac{-2\mu}{\sigma}\right) \right)= 1\\ &\Rightarrow A = \frac{1}{\frac{1}{2} - \Phi\left(\frac{-2\mu}{\sigma}\right)} \end{align}$$ where $$\Phi(x) = \text{integral of } N(0,1) = \int_{-\infty}^x \frac{1}{2\pi} e^{-t^2/2} dt = P(X \sim N(0,1) \le x)$$