Is $\int_{M_{n}(\mathbb{R})} e^{-A^{2}}d\mu$ a convergent integral? Is the  following integral a convergent integral? Can we compute it, precisely?
$$\int_{M_{n}(\mathbb{R})} e^{-A^{2}}d\mu $$
Here $\mu$ is the usual measure of $M_{n}(\mathbb{R})\simeq \mathbb{R}^{n^{2}}$?
So $\mu$  can be  counted  as $\mu=\prod_{i,j} da_{ij}$
Note: If this integral would be convergent , either in Lebesgue or  in Riemann sense, then it would be equal to a scalar matrix. Because for every invertible matrix  $P$ we  have:
$P^{-1}(\int_{M_{n}(\mathbb{R})} e^{-A^{2}}d\mu) P= \int_{M_{n}(\mathbb{R})} e^{-(P^{-1}AP)^{2}}d\mu=\int_{M_{n}(\mathbb{R})} e^{-A^{2}}d\mu$ since the  mapping $A\mapsto P^{-1}AP$  is  a  measure preserving and volum preserving linear map.Now we apply the  change of coordinate formula for integral.
 A: With $n=2$, we can look at matrices $A$ such that $A^2$ has eigenvalues with negative real part. When $T(A)^2-4\Delta(A)<0$, the real part of the eigenvalues of $A^2$ is $\frac{T^2}{4}-\frac{4\Delta-T^2}{4}=\frac{2T^2-4\Delta}{4}$. So $A^2$ has eigenvalues with negative real part provided $T^2<2\Delta$ (a stronger condition than $T^2<4\Delta$). One of the simplest matrices with this property is 
$$Q=\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}.$$
Now consider a small perturbation of $Q$:
$$\begin{bmatrix} \delta_1 & -1+\delta_2 \\ 1+\delta_3 & \delta_4 \end{bmatrix}$$
The inequality now reads
$$\delta_1^2 + 2 \delta_1 \delta_4 + \delta_4^2 < 2 \delta_1 \delta_4 - 2(-1+\delta_2)(1+\delta_3).$$
Equivalently:
$$\delta_1^2+\delta_4^2<2+2\delta_3-2\delta_2-2\delta_2 \delta_3.$$
In view of this inequality we can consider the hypercube $H$ with "radius" $1/12$ around $Q$. If $A \in H$ then $T^2-2\Delta<-1$ and so $\frac{2T^2-4\Delta}{4}<-1/2$. Thus the eigenvalues of $A^2$ will have real part less than $-1/2$. Also the volume of $H$ is $(1/6)^4>0$.
Now the integral of $e^{-A^2}$ over the set of all scalar multiples of elements of $H$ cannot converge absolutely.
