0
$\begingroup$

I know solution to:

$$\int_{\mathbb{R}^n} e^{x'Ax}\mathrm{d}x=\frac{\pi^{n/2}}{det(A)}$$

Where $$A=I+\sum_{t=1}^Ty_ty_t'$$

But If I have:

$$\int_{\mathbb{R}^n} e^{x'Bx}\mathrm{d}x$$ Where $$B=D^{-1}+\sum_{t=1}^Ty_ty_t'$$And $$D=diag(|x_n|^{1/2})$$ What is the answer?

Suggestion

Can I use the fact:

$$\int_{\mathbb{R}^n} e^{Q(x)}\mathrm{d}x=e^{Q_0}\frac{\pi^{n/2}}{det(A)}$$Where $$Q_0=min_x(Q(x))$$

Since $Q_0=0$, for both cases, thus, the answer should be similar (not sure if this statement is true). And $Q(x)=x'Ax$ or $Q(x)=x'Bx$. $A$ and $B$ both are symmetric positive definite.

1 Answers 1

1

If $A$ has an orthogonal base of eigenvectors and every eigenvalue has a positive real part,

$$ \int_{\mathbb{R}^n}\exp\left(-x' A x\right)\,d\mu = \int_{\mathbb{R}^n}\exp\left(-\sum_{k=1}^{n}\lambda_k x_k^2\right)\,d\mu \stackrel{\text{Fubini}}{=}\prod_{k=1}^{n}\sqrt{\frac{\pi}{\lambda_1}}$$ hence the LHS equals $\color{red}{\large\frac{\pi^{n/2}}{\sqrt{\det A}}}$. This holds, for instance, for every symmetric and positive definite matrix. Can you check the involved matrices are symmetric and positive definite? Can you compute their determinants?

  • 0
    Yes they are symmetric and positive definite I think, since $A$ is positive definite and symmetric (that is known), and $B$ involves just horizontal translation. $D$ is diagonal with absolute values. – 2017-02-01
  • 0
    @Waqas: hence the values of their determinants give the values of the integrals, too. – 2017-02-01
  • 0
    So $\frac{\pi^{n/2}}{\sqrt{\det B}}$ is true! – 2017-02-01
  • 0
    @Waqas: exactly :D – 2017-02-01