Evaluate $\int_{-\infty}^{+\infty} \exp(\frac{-1}{2} x^2) dx$

Question

I was trying to find out what the function

$$ f: x, y, x_0, y_0, \sigma_x, \sigma_y \mapsto \frac{1}{2\pi \sigma_x \sigma_y} \exp (\frac{-1}{2} ((\frac{x-x_0}{\sigma_x})^2 + (\frac{y-y_0}{\sigma_y})^2)) $$

is. (No domains were indicated for any of the six variables.)
So I set $x_0, y_0 = 0$, $\sigma_x, \sigma_y =1$ and viewed $f$ as a function in two variables.

$$ f: x, y \mapsto \frac{1}{2\pi} \exp(\frac{-1}{2}(x^2+y^2)) $$

In this constellation, $f$ takes its maximum at $(x,y) = (0,0)$ and decays rapidly as $(x,y) \to \infty$. It is fast-forward to prove that for general $(x_0, y_0) \in \mathbb{R}^2$ and $\sigma_x, \sigma_y > 0$, the implicitly defined function $f: \mathbb{R}^2 \to \mathbb{R}$ has a unique maximum at $(x_0, y_0)$ and decays rapidly as $(x, y) \to \infty$.

I suspect that such $f$ (with given $x_0, y_0, \sigma_x, \sigma_y$) is a density function. In order to prove this, I have to show that

$$ \int_{(x,y) \in \mathbb{R}^2} f(x,y) \, d(x,y) = 1. $$

This task can be reduced to the task of showing

$$ \int_{-\infty}^{+\infty} \exp( \frac{-1}{2} (\frac{x-x_0}{\sigma_x})^2) \, dx = \sqrt{2\pi} $$

I was confident to show this if $x_0 = 0$ and $\sigma_x = 1$. In this case the integral becomes the

$$ \int_{-\infty}^{+\infty} \exp(\frac{-1}{2} x^2) \, dx $$

from the title of this question.
Does someone have a hint for me? How can I prove (in the last, simplified case) that the integral is $\sqrt{2\pi}$? I thought that I would see an easy primitive, but I am missing a factor $x$ on the exponential for this trick.

Answer 1

Well, we are trying to solve:

$$\mathcal{I}_\text{k}\left(\text{n}\right):=\int_\mathbb{R}\exp\left(-\text{n}x^{2\text{k}}\right)\space\text{d}x\tag1$$

Where $\text{k}\in\mathbb{N}$.

Because the integrand is an even function, we can write:

$$\mathcal{I}_\text{k}\left(\text{n}\right)=2\int_0^\infty\exp\left(-\text{n}x^{2\text{k}}\right)\space\text{d}x\tag2$$

Substitute $\text{u}=x\text{n}^\frac{1}{2\text{k}}$, so we get:

$$\mathcal{I}_\text{k}\left(\text{n}\right)=\frac{2}{\text{n}^\frac{1}{2\text{k}}}\int_0^\infty\exp\left(-\text{u}^{2\text{k}}\right)\space\text{d}x\tag3$$

Which is defined by the incomplete gamma function:

$$\mathcal{I}_\text{k}\left(\text{n}\right)=\frac{2}{\text{n}^\frac{1}{2\text{k}}}\cdot\left[-\frac{\Gamma\left(\frac{1}{2\text{k}},\text{u}^{2\text{k}}\right)}{2\text{k}}\right]_0^\infty=\frac{1}{\text{k}\text{n}^\frac{1}{2\text{k}}}\cdot\left[\Gamma\left(\frac{1}{2\text{k}},\text{u}^{2\text{k}}\right)\right]_\infty^0=\frac{\Gamma\left(\frac{1}{2\text{k}}\right)}{\text{k}\text{n}^\frac{1}{2\text{k}}}\tag4$$

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So, when $\text{n}=\frac{1}{2}$ and $\text{k}=1$ we get:

$$\mathcal{I}_1\left(\frac{1}{2}\right)=\int_\mathbb{R}\exp\left(-\frac{1}{2}\cdot x^2\right)\space\text{d}x=\frac{\Gamma\left(\frac{1}{2}\right)}{\frac{1}{2}^\frac{1}{2}}=\sqrt{2\pi}\tag5$$