I was hoping someone could help me understand the first proof/answer given to this question (I can't make a comment on the post since I don't have a high enough reputation score, so I am posting here.):
how to derive the mean and variance of a Gaussian Random variable?
I follow the proof until the step that follows "and using again integral properties we have." Essentially, I don't fully understand this transition (quoted from the original post):
"$$I_1 = -\int_{0}^{-\infty}x\frac{1}{\sigma\sqrt{2\pi}}\exp\left\{-\frac{x^2}{2\sigma^2}\right\}dx + \int_{0}^{\infty}x\frac{1}{\sigma\sqrt{2\pi}}\exp\left\{-\frac{x^2}{2\sigma^2}\right\}dx$$
and using again integral properties we have
$$I_1 = \int_{0}^{\infty}(-x)\frac{1}{\sigma\sqrt{2\pi}}\exp\left\{-\frac{(-x)^2}{2\sigma^2}\right\}dx + \int_{0}^{\infty}x\frac{1}{\sigma\sqrt{2\pi}}\exp\left\{-\frac{x^2}{2\sigma^2}\right\}dx$$"
It seems the negative sign was added inside the function contained within the integrand (I'll call it $f(x)\equiv x\frac{1}{\sigma\sqrt{2\pi}}\exp\left\{-\frac{x^2}{2\sigma^2}\right\}$), which had the effect of changing the limits from 0 to $-\infty$ to 0 to $\infty$, but it seems the negative sign outside the function disappeared as well. I don't understand how the negative sign disappeared here. Isn't it the case that:
\begin{align*} \intop_{0}^{-\infty}f(x)dx & =\intop_{0}^{\infty}f(-x)dx\\ \end{align*}
and not
\begin{align*} \intop_{0}^{-\infty}f(x)dx & =-\intop_{0}^{\infty}f(-x)dx\\ \end{align*}
So, I am under the impression that the negative sign should have remained in front of the first integral of $I_1$. Am I missing something or forgetting some integration rule? Perhaps something else is going on here that I'm just not seeing. Can someone help me understand how the two $I_1$'s above are equivalent?
Thanks for your help!
This is wrong $$\begin{align*} \intop_{0}^{-\infty}f(x)dx & =\intop_{0}^{\infty}f(-x)dx\\ \end{align*}$$
Let's take this integral $$\intop_{0}^{-\infty}f(x)dx$$
Substitute $x=-t$ thus $dx=-dt$
Also, x goes from 0 to $-\infty$ now t should go form 0 to $\infty$
thus integral becomes
$$\intop_{0}^{\infty}f(-t) \cdot (-dt)$$
thus
$$\intop_{0}^{-\infty}f(x)dx=-\intop_{0}^{\infty}f(-t) \cdot dt$$
Thus we can say that
$$\intop_{0}^{-\infty}f(x)dx=-\intop_{0}^{\infty}f(-x) \cdot dx$$