Integration by substitution in double integral: Is the following correct?

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I have the follwing for a one-dimensional integral: $$ I = \int_{-\infty}^{\infty}f(x-az)g(z)dz$$ where f() and g() are functions. Using $u = G(z)$, $G'(z) = g(z), G(-\infty)=0, G(\infty)=1$
I get: $$I = \int_{0}^{1}f(x-aG^{-1}(u))du$$ which I can evaluate. I know this first part is correct. Now what about the follwowing, will this be correct or am I missing something?
$$ J = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x-az_1-bz_2)g_1(z_1)g_2(z_2)dz_1 dz_2 $$ $$ = \int_{0}^{1}\int_{0}^{1}f(x-aG_1^{-1}(u_1)-bG_2^{-1}(u_2))du_1 du_2$$ The same as above applies: $u_i = G_i(z_i)$, $G_i'(z_i) = g_i(z_i), G_i(-\infty)=0, G_i(\infty)=1$ for $i=1,2$