Integration of $\exp[f(x,y)]$

Question

Here is the question i want to solve. $$\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} \exp\left[{-2\over3}(y^2-yz+z^2)\right]\,dy\,dz$$ I know that $\exp$ is $e^{f(x)}$ and i can find $\int e^{f(x)}\,dx$ but with two variables makes me confusing.

Answer 1

\begin{align} \text{Let }u & = y+z, \\ \text{and }v & = y-z. \\[15pt] \text{It follows that }y & = \frac{u+v} 2, \\[6pt] \text{and }z & = \frac{u-v} 2. \end{align} Then \begin{align} y^2-yz+z^2 & = \left( \frac{u+v} 2 \right)^2-\frac{u+v} 2 \cdot \frac{u-v} 2 + \left(\frac{u-v} 2\right)^2 \\[6pt] & = \frac{u^2+3v^2} 4. \end{align} And $dy\,dz = \left|\dfrac{\partial(y,z)}{\partial(u,v)}\right|\,du\,dv= \dfrac 1 2 \,du\,dv$. So \begin{align} & \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \exp\left[{-2\over3}(y^2-yz+z^2)\right]\,dy\,dz \\[8pt] = {} & \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \exp\left[\frac{-2}3 \left(\frac{u^2+3v^2} 4\right) \right] \frac{du\,dv} 2\\[8pt] = {} & \int_{-\infty}^{+\infty} \left(\int_{-\infty}^{+\infty} \exp\left[\frac{-2}3 \frac{u^2}4 \right] \underbrace{\exp\left[\frac{-2}3 \frac{3v^2}4 \right] \frac 1 2}\ \ du\right)\,dv. \end{align} The part over the $\underbrace{\text{underbrace}}$ does not change as $u$ goes from $-\infty$ to $+\infty$ and is a factor of the function being integrated (i.e. one multiplies it by something to get that function). Therefore, it can be pulled out of the inside integral, leaving a standard Gaussian integral on the inside. Once you've reduced that to a number, you've got one more standard Gaussian integral.

PS: Here's another way: \begin{align} & \int_{-\infty}^{+\infty} \exp\left[{-2\over3}(y^2-yz+z^2)\right]\,dy \\[8pt] = {} & \int_{-\infty}^{+\infty} \exp\left[{-2\over3}\left(\left(y^2-yz + \frac{z^2}4\right) + \frac{3z^2}4 \right) \right]\,dy \quad (\text{completing the square}) \\[8pt] = {} & \int_{-\infty}^{+\infty} \exp\left[{-2\over3} \left(y^2- \frac z 2 \right)^2 \right] \exp\left[{-2\over3}\frac{3z^2}4 \right]\,dy \\[8pt] = {} & \exp\frac{-z^2} 2 \int_{-\infty}^{+\infty} \exp\left(\frac{-2}3 \left(y - \frac z 2\right)^2\right) \, dy \\[8pt] = {} & \exp\frac{-z^2} 2 \int_{-\infty}^{+\infty} \exp\left(\frac{-2}3 u^2 \right)\,du \\[8pt] = {} & A\exp\frac{-z^2}2 \end{align} where $A$ is the value of the integral with respect to $u$. Then do the outside integral in which the variable is $z$.