I am trying to check how a density function of a bivariate Pareto distribution f(x,y) goes to its accumulated function F(x,y):
$$ f(x,y) = \frac{a(a+1)}{b_1b_2}\biggl [-1 + \frac x {b_1} + \frac y{b_2}\biggr]^{-(a+2)} $$
with $a, b_1$ and $b_2 > 0$, $x>b_1$ and $y>b_2$.
I tried to solve the double integration of f equal to 1 to find its accumulated version but I am stuck on the point:
To see that the integral of the pdf is equal to 1 you have to adjust the lower limits. I factor out the constants. For less typing work I replace $b_1$ and $b_2$ by $b$ and $c$
$$\frac{a\cdot (a+1)}{b\cdot c} \cdot \int_c^{\infty} \int_b^{\infty} \frac{a(a+1)}{b\cdot c} \left(-1 + \frac{x}{b}+\frac{y}{c} \right)^{-(a+2)} \, dx \, dy$$
The inner integral is
$\int_b^{\infty} \left(-1 + \frac{x}{b}+\frac{y}{c} \right)^{-(a+2)} \, dx= \left[-\frac{b}{a+1}\cdot \left(-1 + \frac{x}{b}+\frac{y}{c} \right)^{-(a+1)}\right]_b^{\infty}$
$=-0-\left[-\frac{b}{a+1}\cdot \left(-1 + \frac{b}{b}+\frac{y}{c} \right)^{-(a+1)}\right]=\frac{b}{a+1}\cdot \left(\frac{y}{c} \right)^{-(a+1)}$
Next, this term has to be integrated w.r.t. $y$
$$ \int_c^{\infty} \frac{b}{a+1}\cdot \left(\frac{y}{c} \right)^{-(a+1)}\, dy$$
It isn´t hard to calculate. The result should be $\frac{b\cdot c}{a\cdot (a+1)}$ since we still have the factored out fraction $\frac{a\cdot (a+1)}{b\cdot c}$.