Double integral involving exponential of quadratic form. Assuming that $a>0$, Maple shows the following
$$
\int_0^\infty \int_0^\infty \exp\left(\frac 1 2(-a y^2-2 a y z-a z^2)\right) \, \mathrm dy \, \mathrm dz =
\frac 1{a},
$$
whereas
$$
\int_0^\infty \int_0^\infty \exp\left(\frac 1 2(-(a+1) y^2-2 a y z-a z^2)\right) \, \mathrm dy \, \mathrm dz =
\frac {\arctan(1/\sqrt{a})}{\sqrt{a}}.
$$
But how can we do these integrals manually?
 A: $$
\int_0^\infty \int_0^\infty \exp\left(\frac 1 2(-a y^2-2 a y z-a z^2)\right) \, \mathrm dy \, \mathrm dz
$$
Let
\begin{align}
u & = y+z \\
v & = y-z
\end{align}
Then we have $\displaystyle \int_0^\infty \left( \int_{-u}^u \cdots\cdots \, \mathrm d v \right) \, \mathrm du$ and
$$
\mathrm dy\,\mathrm d z = \left| \det\begin{bmatrix} \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} \\[6pt] \dfrac{\partial z}{\partial u} & \dfrac{\partial z}{\partial v} \end{bmatrix} \right| \, \mathrm du\,\mathrm dv = \frac{\mathrm du\,\mathrm dv} 2.
$$
$$
y^2 + 2yz + z^2 = (y+z)^2 = u^2.
$$
So
\begin{align}
& \int_0^\infty \int_0^\infty \exp\left(\frac 1 2(-a y^2-2 a y z-a z^2)\right) \, \mathrm dy \, \mathrm dz \\[10pt]
= {} & \int_0^\infty \int_{-u}^u \exp\left( \frac {-a} 2 u^2 \right) \, \frac{\mathrm dv\, \mathrm du} 2 \\[10pt]
= {} & \int_0^\infty 2u \exp\left( \frac{-a} 2 u^2 \right)\,\frac{\mathrm du} 2 \\[10pt]
= {} & \int_0^\infty \exp\left( \frac{-a} 2 w\right) \, \frac{\mathrm dw} 2 \\[10pt]
= {} & \frac 1 a.
\end{align}
