Double Integral of exponentiation

Question

I find this example online, and I can't solve it at all.

All I know is that: . Although I tried to transform the problem to my known rules, It's all in vain.

Any help will be appreciated, thanks.

Answer 1

Hint: $$\int_{-\infty}^{\infty} e^{-\frac{(z-\mu)^2}{\sigma^2}} dz = \sqrt{\sigma^2 \pi} $$

so

$$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-5x^2+8xy-5y^2}$$

$$=\int_{-\infty}^{\infty} e^{-5y^2} \int_{-\infty}^{\infty} e^{-5x^2+8xy} dx \ dy$$

$$=\int_{-\infty}^{\infty} e^{-5y^2} \int_{-\infty}^{\infty} e^{-5(x^2-\frac{8}{5}xy)} dx \ dy$$

$$=\int_{-\infty}^{\infty} e^{-5y^2} \int_{-\infty}^{\infty} e^{-5(x^2-2\frac{8}{2*5}xy)} dx \ dy$$

$$=\int_{-\infty}^{\infty} e^{-5y^2} \int_{-\infty}^{\infty} e^{-5(x^2-2\frac{8}{2*5}xy+(.8y)^2 -(.8y)^2)} dx \ dy$$

$$=\int_{-\infty}^{\infty} e^{-5y^2} \int_{-\infty}^{\infty} e^{-5\left((x-.8y)^2 -(.8y)^2)\right)} dx \ dy$$

$$=\int_{-\infty}^{\infty} e^{-5y^2+5*0.64y^2} \int_{-\infty}^{\infty} e^{-5\left((x-.8y)^2 )\right)} dx \ dy$$

$$=\int_{-\infty}^{\infty} e^{-5y^2+5*0.64y^2} \sqrt{\frac{1}{5}\pi} \ dy$$