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?
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change to polar coordinates! See https://stats.stackexchange.com/questions/181057/bivariate-normal-distribution-probability-calculations/295157#295157 for an example – kjetil b halvorsen Sep 04 '17 at 15:13
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$$ \int_0^\infty \int_0^\infty \exp\left(\frac 1 2(-a y^2-2 a y z-a z^2)\right) \, \mathrm dy \, \mathrm dz $$
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}