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I have trouble using a change of variable in order to compute the following integral :

$$\int_{\mathbb{R}^m} e^{<-Ax;x>} d\lambda(x)$$

A is a real symmetric and positive definite matrix, that means it can be diagonalised and the elements of the diagonal are non zeros. I proceed therefore to rewrite the integral as follow:

$$\int_{\mathbb{R}^m} e^{<-B^{-1}DBx;B^{-1}Bx>} d\lambda(x) = \int_{\mathbb{R}^m} e^{<-B^{-1}Dy;B^{-1}y>} |det(B^{-1}|d\lambda(y)$$

But i am now stuck here i would have liked to write :

$$\int_{\mathbb{R}^m} e^{}d\lambda(y)= \pi^{\frac{m}{2}}Det(A)^{-\frac{1}{2}}$$

But i fail to see how this second change of variable is correct.

Thx

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    note that $\langle Dy,y\rangle =\sum_ i\alpha_i (y^i)^2$ where the $\alpha_i$ are the eigenvalues of $A$. Also note that $\det A = \Pi_i \alpha_i$. (This comment does not imply that I checked your calculations) – 2017-02-11
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    I am not stuck in the computation of the last equality i wrote but rather in how to pass from the second to the third line – 2017-02-11

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If $A$ is a real, symmetric and positive definite matrix it has an orthonormal base $B$ of eigenvectors and the change of coordinates bringing the canonical base of $\mathbb{R}^n$ into $B$ is an isometry, hence the determinant of its Jacobian is $\pm 1$. It follows that $A$ is similar to the diagonal matrix having the real and positive eigenvalues $\lambda_1,\ldots,\lambda_n$ along the diagonal and $$ \int_{\mathbb{R}^n}\exp\left(-x^T A x\right)\,d\mu = \int_{\mathbb{R}^n}\exp\left(-\lambda_1 x_1^2-\ldots-\lambda_n x_n^2\right)\,d\mu.$$ Since $\int_{-\infty}^{+\infty}e^{-\lambda x^2}\,dx = \sqrt{\frac{\pi}{\lambda}}$, by Fubini's theorem the last integral equals $$ \frac{\pi^{n/2}}{\sqrt{\lambda_1\cdot\ldots\cdot\lambda_n}}=\frac{\pi^{n/2}}{\sqrt{\det A}} $$ as wanted.

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    thx, what i missed is the fact that the eigenvectors are an orthonormal base which allows to write $x^tB^{-1}DBx = (Bx)^tD(Bx)$ – 2017-02-11