The following function is given:
\begin{align} f(n)=\int\limits_{\mathbb{R}^n}\exp(-\vec{x}\dot{}\vec{x})\mathrm{d}x \end{align}
Because it is known that $f(1)=\sqrt{\pi}$, it is also easy to show that $f(n)=\pi^{\frac{n}{2}}$:
\begin{align} f(n)=\int\limits_{\mathbb{R}^n}\exp(-\vec{x}\dot{}\vec{x})\mathrm{d}x=\prod_{i=1}^n \int\limits_{\mathbb{R}}\exp(-x_i^2)\mathrm{d}x_i=\prod_{i=1}^n\sqrt{\pi}=\pi^{\frac{n}{2}} \end{align}
Now my question: It is now easy to evaluate $f(0)=1$ with the new formula. Written with our initial integral, this would be:
\begin{align} f(0)=\int\limits_{\mathbb{R}^0}\exp(-x^2)\mathrm{d}x=1 \end{align}
But is this integral valid? Or does it even make sense? The $x$-"vector" is now 0-dimensional, which means the product with itself is probably 0:
\begin{align} f(0)=\int\limits_{\mathbb{R}^0}\exp(-0)\mathrm{d}x=\int\limits_{\mathbb{R}^0}1\mathrm{d}x=1 \end{align}
Is this valid? And is it even possible to integrate over $\mathbb{R}^0$? Because as I know $\mathbb{R}^0$ only contains one "empty" element.
And now my second question: What is the case for any other $n$. E.g. $f(\sqrt{2})=\pi^\frac{1}{\sqrt{2}}$? This would be the following integral:
\begin{align} f(0)=\int\limits_{\mathbb{R}^\sqrt{2}}\exp(-\vec{x}\dot{}\vec{x})\mathrm{d}x=\pi^\frac{1}{\sqrt{2}} \end{align}
$\vec{x}$ would now be $\sqrt{2}$-dimensional. Is this valid? Or can't this "back-transformation" to the integral be done?
Thank you very much
The definition you have given for $f(n)$ is for positive integers only. There is a real sense in which it can indeed be extended to $n=0$ to give a value of $1$. Namely, that $\Bbb R^0 = \{0\}$, and "zero-dimensional" measure is just counting, which makes integration just a finite sum, in this case over just one element, so the value you get is that element, which is $1$.
But there is no conventional meaning for $\Bbb R^{1/2}$, much less $\Bbb R^{\sqrt 2}$. There is such a thing as fractional dimension. In fact there are several variants, but none of them (that I am aware of) gives meaning to $\Bbb R^p$ when $p$ is not a non-negative integer.
So before you can discuss whether your formula still holds when $n$ is not integer, you first have to specify what $\int_{R^n}e^{-x\cdot x}dx$ is supposed to even mean when $n$ is not an iteger.