Is $\int_{M_{n}(\mathbb{R})} e^{-A^{2}}d\mu$ a convergent integral?(2) We  identify $M_{n}(\mathbb{R})$ with $\mathbb{R}^{n^{2}}$
We put $\int_{M_{n}(\mathbb{R})} e^{-A^{2}}d\mu=\lim_{r\to \infty}  \int_{D_{r}} e^{-A^{2}}$  where the later is  counted as a Riemann integral not Lebesgue integral. Here $D_{r}$  is the disc of radius $r$ with respect to the Euclidean  norm of $\mathbb{R}^{n^{2}}$. Is the above integral a convergent improper integral? What about if we consider $D_{r}$  with respect to the matrix norm?
The following post shows that this integral is not convergent in the Lebesgue sense. It also shows that if it is Riemann convergent, then the value of integral is  an scalar matrix.
Is $\int_{M_{n}(\mathbb{R})} e^{-A^{2}}d\mu$ a convergent integral?
 A: 
In dimension $n = 2$, we have
  $$ \int_{D_r} e^{-A^2} \, \mu(dA) = \frac{\pi^2}{2}\left( e^{-r^2} - 1 + r^2 + \frac{r^4}{2} \right) I_2, \tag{*} $$
  where $I_2$ is the identity matrix in $M_2(\Bbb{R})$.

In particular, the improper integral does not converge.
I have summarized my solution in my blog posting. But here is my idea: Evaluate the integral of the series expansion
$$ \int_{D_r} e^{-A^2} \, \mu(dA) = \sum_{m=0}^{\infty} \frac{(-1)^m}{m!} \int_{D_r} A^{2m} \, \mu(dA). $$
term by term. Writing $A$ as
$$ A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}, $$
notice that each entry of $A^{2m}$ is a homogeneous polynomial of degree $2m$ in variables $a, b, c, d$. With a little bit of observations, we find that

Lemma 1. Let $D$ be any symmetric domain (in the sense that $-D = D$). Then for any $n \geq 0$,
  $$ \int_{D} A^{2m} \, \mu(dA) = I_2 \sum_{i+j+2k=m} \binom{2k+2i}{2i}\binom{2k+2j-1}{2j} \int_{D} a^{2i}d^{2j}(bc)^{2k} \, \mu(dA). \tag{1} $$
  Here we exploited the convention that $\binom{-1}{0} = 1$.

Sketch of proof. Each entry of $A^{2m}$ admits a combinatorial interpretation in terms of transition path on a 2-state space:

(For example, the path 1→1→1→2→2→2→1→2→1 contributes to the $(1,1)$-entry of $A^8$ with the term $aacddbcb = a^2b^2c^2d^2$.) Also, by the symmetry of $D$, we only need to consider terms with even exponents, i.e., of the form 
$$\text{(coef.)}\cdot a^{2i}b^{2j}c^{2k}d^{2l}.$$
These terms correspond to paths where every transition takes place in even numbers. A moment of thought shows that such term corresponds to closed paths. Thus theses term only appear in $(1,1)$-entry and $(2,2)$-entry of $A^{2m}$, and the exponents of $b$ and $c$ are equal. Then $\text{(1)}$ follows by counting the number of such paths. ////
Now, each term in $\text{(1)}$ can be easily computed when $D = D_r$:

Lemma 2. For any $i+j+2k = m$ we have
  $$ \int_{D_r} a^{2i}d^{2j}(bc)^{2k} \, \mu(dA) = \frac{(i-\frac{1}{2})!(j-\frac{1}{2})!(k-\frac{1}{2})!^2}{(m+2)!} r^{2m+4}. \tag{2} $$

Of course we are using the convention that $(l-\frac{1}{2})! = \Gamma(l+\frac{1}{2})!$ in $\text{(2)}$. We skip the proof since the proof is easy.
Somewhat surprisingly, the egregious formula $\text{(1)}$ yields simple values:

Lemma 3. For $m \geq 1$ we have
  $$ \int_{D_r} A^{2m} \, \mu(dA) = \frac{\pi^2}{2(m+1)(m+2)} r^{2m+4}. $$
  (When $m = 0$ we have different value.)

This amounts to prove that
$$ \sum_{i+j+2k=m} \binom{2k+2i}{2i}\binom{2k+2j-1}{2j} (i-\tfrac{1}{2})!(j-\tfrac{1}{2})!(k-\tfrac{1}{2})!^2 = \frac{m! \pi^2}{2}. $$
Proof of this fact can be found in my blog posting. Now assuming that this is true, $\text{(*)}$ easily follows.
