Let $ A$ be a symmetric positive definite matrix. $A \in \mathbb{R}^{n \times n} $.
Show that: $$ \int_{\mathbb{R}^n} e^{-\langle Ax,x\rangle } d\lambda_n(x) = \frac{\sqrt{\pi^n}}{\sqrt{\det(A)}} $$
Proof :
First: Can you tell me if my calculation is correct?
Second: I set a $\star_1$, $\star_2$. Can you explain me in detail $ \star_1, \star_2$? Especially $\star_2$. I know that this is change of variables theorem.
First we know: $A = S^t \cdot D_A \cdot S$, $S$ orthogonal and $D_A$ diagonal matrix.
We know: $1 = |\det(S)| = |det(S^t)|$.
We know that: $D_A$ = diagonal$ (\lambda_{11},...,\lambda_{nn}) $ where $\lambda_{ii} > 0$ for $ i \in $ {$1,...,n$}.
Obviously $\det(A) = \lambda_{11} \cdots \lambda_{nn}$. So far so good.
Now: $$-\langle Ax, x \rangle =- \langle S^ty \cdot D_A \cdot S \cdot x, x \rangle \stackrel{(\star_1)}{=} - \langle D_A \cdot S \cdot x, S\cdot x \rangle.$$
Now: $$ \int_{\mathbb{R}^n} e^{-\langle Ax,x\rangle} d\lambda_n(x) = \int_{\mathbb{R}^n} e^{-\langle D_A \cdot S \cdot x, S \cdot x \rangle } d\lambda_n(x)$$
[Now we have use transformation theorem:] [in general $ \int f(y) dy = \int_U f(\phi(x)) \cdot | \det D\phi(x)| dx $]
$$\int_{\mathbb{R}^n} e^{-\langle D_A \cdot S \cdot x, S \cdot x \rangle } d\lambda_n(x) \stackrel{(\star_2)}{=} \int_ {\mathbb{R}^n} e^{-\langle D_A \cdot y, y \rangle } \cdot |\det S| d\lambda_n(y) = \dots $$ Rest is clear .