I'm trying to manually calculate the PDF and the CDF of the normal distribution.
Starting from $$N(x,\mu,\sigma) = e^{\frac{-(x-\mu)^2}{2\sigma^2}}$$
I know the next step is to say $$\int^{\infty}_{-\infty} e^{\frac{-(x-\mu)^2}{2\sigma^2}}\partial x = 1$$
In my case, I can fix $\mu=0$ and $\sigma=1$, due to properties of my problem; meaning this simplifies to $$\int^{\infty}_{-\infty} e^{\frac{-x^2}{2}}\partial x = 1$$
However, I don't exactly know how to follow up on this... I've tried one approach (described in the following), but I got a result that doesn't seem correct:
$$ u = \frac{-x^2}{2} $$ $$ \partial u = -x ~ \partial x $$ $$ \partial x = \frac{1}{-x} \partial u $$ $$\int^{\infty}_{-\infty} e^{\frac{-x^2}{2}}\partial x = 1 = \int^{\infty}_{-\infty} e^{u} \frac{1}{-x}\partial u = \frac{1}{-x} \int^{\infty}_{-\infty} e^u \partial u$$ $$= \frac{1}{-x} e^u = \color{red}{\frac{1}{-x} e^\frac{-x^2}{2}}$$
According to Wikipedia and others, I should get $$\frac{1}{2\pi} e^\frac{-x^2}{2}$$
Could anyone give me directions to where I should look for resources that can teach me what I am missing?
You cannot just take out the "$-\frac{1}{x}$" term from the integral, since you defined it as a function of $u$ when substituting $u=-\frac{x^2}{2}$.
Your question has already been answered here.