Need help computing the double integral $\int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x + y)}{x + y} \mathop{dy} \mathop{dx}.$
I know that $\int_{0}^{\infty} f(u) \mathop{du}$ equals $1$. The entire integral should come out to be $1$.
I am new to multivariable calculus and would appreciate some help.
Making the substitution $u = x + y$ and $v = y$, we get $\mathop{du} = \mathop{dx} + \mathop{dy}$ and $\mathop{dv} = \mathop{dy}$. So,
$$\int_{0}^{\infty} \int_{0}^{\infty} \frac{f(u)}{u} \mathop{dy} \mathop{dx}$$
I don't know what to change the $dy$ and $dx$ to since $du$ has both $dx$ and $dy$ in it. Also, for the Jacobian, I know that it will equal $1$, but I don't know what I'm taking partial derivatives of in the determinant. Can someone please help me with this integral?
You almost nailed it with your substitution $(x,y)\mapsto (u,v):=(x+y,y)$. That said, you made a mistake when you thought that the region of integration is still $\mathbb{R}_{> 0}^2$. The correct region is $$\Omega:=\big\{(u,v)\in\mathbb{R}_{> 0}^2\,\big|\,u> v\big\}\,.$$ Hence, $$I:=\int_0^\infty\,\int_0^\infty\,\frac{f(x+y)}{x+y}\,\text{d}x\,\text{d}y=\iint_\Omega\,\frac{f(u)}{u}\,\text{d}u\,\text{d}v\,.$$ Assuming that $f$ is sufficiently nice so that Fubini's Theorem applies, we get $$I=\int_0^\infty\,\int_0^u\,\frac{f(u)}{u}\,\text{d}v\,\text{d}u=\int_0^\infty\,u\,\left(\frac{f(u)}{u}\right)\,\text{d}u\,.$$ Thus, $$I=\int_0^\infty\,f(u)\,\text{d}u=1\,.$$