How to show that the integral of bivariate normal density function is 1?

Question

How to show the following?

$$\large \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{1}{2 \pi \sqrt{1-\rho^2}} e^{-\frac{x^2+y^2-2 \rho x y}{2(1-\rho^2)}} dx\ dy=1$$

Answer 1

Try to use polar coordinates. Put $x=r\cos\theta$, $y=r\sin\theta$ and Jakobian will be $r$.

Answer 2

Complete the square of the exponent $x^2+y^2-2\rho xy = \left(x-\rho y\right)^2+y^2(1-\rho^2)=:u^2+y^2(1-\rho^2)$

$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{1}{2 \pi \sqrt{1-\rho^2}} e^{-\frac{x^2+y^2-2 \rho x y}{2(1-\rho^2)}} dx dy=\\ =\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{1}{2 \pi \sqrt{1-\rho^2}} e^{-\frac{u^2}{2(1-\rho^2)}} du \; e^{-\frac{y^2(1-\rho^2)}{2(1-\rho^2)}} dy = \\ =\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}} e^{\frac{-y^2}{2}} dy=1$$

PS. Using the well known formula that $$\int_{-\infty}^{\infty}e^{-\frac{u^2}{a}} du=\sqrt{a\pi}$$