Derivate of double integral

Question

How do I perform the following? $$\frac{\mathrm d}{\mathrm dx}\int_{-\infty}^\infty\int_{y+g(x)}^\infty f(y,z)\,\mathrm dz\mathrm dy.$$

Answer 1

Check out Leibniz integral rule. The formula applied naively would give $$\frac d {dx} \int_{-\infty}^{\infty} \int_{y+g(x)}^{\infty} f(y,z) \,dz\,dy= \int_{-\infty}^{\infty} \left( \frac \partial {\partial x}\int_{y+g(x)}^{\infty} f(y,z) \,dz\right)\,dy=$$ $$=\int_{-\infty}^{\infty}\left(-f(y,y+g(x))\cdot \frac \partial {\partial x}\big(y+g(x)\big) +\int_{y+g(x)}^{\infty} \frac \partial {\partial x} f(y,z) \,dz\right)\,dy=$$ $$=\int_{-\infty}^{\infty}\left( -f(y,y+g(x))\cdot g'(x) +\int_{y+g(x)}^{\infty} 0 \,dz\right)\,dy=- g'(x) \int_{-\infty}^{\infty}f(y,y+g(x))\,dy.$$

But of course, for this to be true there are a series of hypotheses that have to be satisfied (see the link above) and this needs to be done each time you apply Leibniz's Rule.

Moreover, every $\infty$ symbol corresponds actually to a limit (you have to change the symbol for a new variable—usually we use $a$,$b$,$c$, etc.— and consider the limit of the integral as that variable tends to $\infty$ or $-\infty$, depending on the case); and there the hypotheses for interchanging limits and integrals or limits and derivatives may also be relevant.

Answer 2

Assuming the functions $f$ and $g$ are well enough behaved that the integrals and derivatives make sense, we have

$$\begin{align}{d\over dx}\int_{-\infty}^\infty\int_{y+g(x)}^\infty f(y,z)dzdy &\approx{1\over\Delta x}\left(\int_{-\infty}^\infty\int_{y+g(x+\Delta x)}^\infty f(y,z)dzdy-\int_{-\infty}^\infty\int_{y+g(x)}^\infty f(y,z)dzdy\right)\\ &=-{1\over\Delta x}\int_{-\infty}^\infty\int_{y+g(x)}^{y+g(x+\Delta x)} f(y,z)dzdy\\ &\approx-{1\over\Delta x}\int_{-\infty}^\infty\int_{y+g(x)}^{y+g(x)+g'(x)\Delta x} f(y,z)dzdy\\ &\approx-g'(x)\int_{-\infty}^\infty f(y,y+g(x))dy \end{align}$$

If you let $\Delta x\to0$, the $\approx$'s become $=$'s.