Calculate $\int_A e^{x^2+y^2-z^2-w^2} \, dx\,dy\,dz\,dw$ Calculate $\int_A e^{(x^2+y^2-z^2-w^2)}\,dx\,dy\,dz\,dw$ where $A=\{x, y, z, w \in \mathbb{R} \mid x^2+y^2+z^2+w^2\leq1\}$
attempt:
$$\int_A e^{x^2+y^2-z^2-w^2} \,dx\,dy\,dz\,dw
  = \int_0^1 \int_{\mathbb{S}^3_r} e^{x^2+y^2-z^2-w^2} \;\mathrm{d}S \;\mathrm{d}r
  \\= \int_0^1 \int_{\mathbb{S}^3_r} e^{2x^2+2y^2-r} \;\mathrm{d}S \;\mathrm{d}r
  = \int_0^1 \int_0^{2\pi} \int_0^{\pi} e^{2r^2 \sin^2\phi - r} r^2 \sin\phi \;\mathrm{d}\phi \;\mathrm{d}\theta \;\mathrm{d}r
$$
I got stuck here. Any help please?
 A: Note that $$A=\{x^2+y^2\leq1, z^2+w^2\leq1-(x^2+y^2)\}$$
So the integral becomes
$$I=\int_{\{x^2+y^2\leq1\}}e^{x^2+y^2}\int_{\{z^2+w^2\leq1-(x^2+y^2)\}}e^{-(z^2+w^2)}$$
Where the inner integral is by Fubini again:
$$(*) = \int_{-\pi}^{\pi}\int_0^cre^{-r^2}drd\theta$$
where $c=\sqrt{1-(x^2+y^2)}$.
Now, this equals $\pi(1-e^{-c^2})$, so we got
$$I =  \pi\int_{\{x^2+y^2\leq1\}}e^{x^2+y^2} \cdot(1-e^{-(1-(x^2+y^2))})$$
Where from here, it is easier to continue by polar coordinates and similar integral as in $(*)$
A: This was my attempt:
$$I=\iiiint\limits_A e^{x^2+y^2-(z^2+w^2)}\,\mathrm dx\,\mathrm dy\,\mathrm dz\,\mathrm dw\qquad A=\left\{x,y,z,w\in\mathbb{R}|x^2+y^2+z^2+w^2\le1\right\}\tag{1}$$

First let:
$$\boxed{\begin{align}
x=&r_1\cos\theta_1\\
y=&r_1\sin\theta_1\\
z=&r_2\cos\theta_2\\
w=&r_2\sin\theta_2
\end{align}}$$
so:
$$A\mapsto B=\left\{(r_1,r_2)\in\mathbb{R}_+^2\,\wedge\,(\theta_1,\theta_2)\in[-\pi,\pi]^2\,|\,r_1^2+r_2^2\le1\right\}$$
this means that:
$$\boxed{\mathrm dx\,\mathrm dy\,\mathrm dz\,\mathrm dw=r_1r_2\,\mathrm dr_1\,\mathrm d\theta_1\,\mathrm dr_2\,\mathrm d\theta_2}$$
and so we can rewrite as:
$$\begin{align}
I=&\int\limits_{-\pi}^{\pi}\mathrm d\theta_1\int\limits_{-\pi}^{\pi}\mathrm d\theta_2\iint\limits_Cr_1r_2e^{r_1^2-r_2^2}\,\mathrm dr_1\,\mathrm dr_2\qquad C=\left\{(r_1,r_2)\in\mathbb{R}_+^2\,|\,r_1^2+r_2^2\le 1\right\}\\
=&4\pi^2\iint\limits_Cr_1r_2e^{r_1^2-r_2^2}\,\mathrm dr_1\,\mathrm dr_2
\end{align}\tag{2}$$

Now for another substitution:
$$\boxed{\begin{align}
r_1=&\rho\cos\varphi\\
r_2=&\rho\sin\varphi
\end{align}}$$
so:
$$C\mapsto D=\left\{0\le\rho\le1\,\wedge\,0\le\varphi\le\pi/2\right\}$$
this means that:
$$\boxed{\mathrm dr_1\,\mathrm dr_2=\rho\,\mathrm d\rho\,\mathrm d\varphi}$$
and so we can rewrite it as:
$$I=2\pi^2\int\limits_0^{\pi/2}\int\limits_0^1\rho^3e^{\rho^2\cos2\varphi}\sin2\varphi\,\mathrm d\rho\,\mathrm d\varphi\tag{3}$$

Now for another substitution:
$$\boxed{\begin{align}
u=&\rho^2\cos2\varphi\\
\frac{\partial u}{\partial\rho}=&2\rho\cos2\varphi\\\
\mathrm d\rho=&\frac{\mathrm du}{2\rho\cos2\varphi}
\end{align}}$$
allowing us to rewrite as:
$$\begin{align}
I=&\pi^2\int\limits_0^{\pi/2}\int\limits_0^{\cos2\varphi}\frac{\sin2\varphi}{\cos^22\varphi}ue^u\,\mathrm du\,\mathrm d\varphi\\
=&\pi^2\int\limits_0^{\pi/2}\frac{\sin2\varphi}{\cos^22\varphi}\left(e^{\cos2\varphi}\left[\cos2\varphi-1\right]+1\right)\,\mathrm d\varphi
\end{align}\tag{4}$$

Now finally making the substitution:
$$\boxed{\begin{align}
v=&\cos2\varphi\\
\Rightarrow \mathrm d\varphi=&-\frac{\mathrm dv}{2\sin2\varphi}
\end{align}}$$
Allows us to finally rewrite as:
$$I=\frac{\pi^2}{2}\int\limits_{-1}^{1}\frac{e^v(v-1)+1}{v^2}\,\mathrm dv\tag{5}$$
Which appears to be solvable using the gamma function
