Can I apply a Leibniz integration rule to $\frac{d}{dr} \int_{-\infty}^{\infty} \int_{h(r)}^\infty \int_{g(r)}^\infty {rf(x,y,z)\, dz\, dy\, dx}$?

Question

Suppose I need to compute the derivative $$ \frac{d}{dr} \int_{-\infty}^{\infty} \int_{h(r)}^\infty \int_{g(r)}^\infty {rf(x,y,z)\, dz\, dy\, dx} $$ Can I apply a Leibniz integration rule of some form? How?

Answer 1

Effectively, the below constitutes a derivation of the Leibniz Integral Rule as applicable to the current setting.

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Assuming sufficient conditions to allow for interchanging integration and differentiation, and existence of the integrals, you can.

Perhaps the most insightful way is to introduce some auxiliary functions. Define:

$$\begin{align} G(x,y,b,p) &:= \int_b^\infty pf(x,y,z)\,\mathrm dz \\ H(x,b,c,p) &:= \int_c^\infty G(x,y,b,p)\,\mathrm dy \\ K(b,c,p) &:= \int_{-\infty}^\infty H(x,b,c,p)\,\mathrm dz \end{align}$$

We want to compute $\dfrac{\mathrm d}{\mathrm dr}K(g(r),h(r),r)$, and start by the general chain rule:

$$\dfrac{\mathrm d}{\mathrm dr}K(b,c,p) = \frac{\partial K(b,c,p)}{\partial b}\frac{\mathrm db}{\mathrm dr}+\frac{\partial K(b,c,p)}{\partial c}\frac{\mathrm dc}{\mathrm dr}+\frac{\partial K(b,c,p)}{\partial p}\frac{\mathrm dp}{\mathrm dr}$$

In this expression, the quantities $\dfrac{\mathrm db}{\mathrm dr},\dfrac{\mathrm dc}{\mathrm dr}$ and $\dfrac{\mathrm dp}{\mathrm dr}$ are easily computed from $b = g(r), c=h(r),p=r$.

For the partial derivatives, we compute as follows:

$$\begin{align} \frac{\partial K(b,c,p)}{\partial b} &= \int_{-\infty}^\infty \frac{\partial}{\partial b}H(x,b,c,p)\,\mathrm dz\\ &= \int_{-\infty}^\infty \int_c^\infty \frac{\partial}{\partial b}G(x,y,b,p)\,\mathrm dy\,\mathrm dz \end{align}$$

Now what is $\dfrac{\partial}{\partial b}G(x,y,b,p)$? It is $-pf(x,y,b)$, as we see from the following computation:

$$\begin{align} \lim_{h\to0} \dfrac{G(x,y,b+h,p)-G(x,y,b,p)}{h} &= \lim_{h\to 0} \frac1h \left(\int_{b+h}^\infty pf(x,y,z)\,\mathrm dz-\int_b^\infty pf(x,y,z)\,\mathrm dz\right)\\ &= \lim_{h\to0} \frac1h \left(-\int_b^{b+h}pf(x,y,z)\,\mathrm dz\right)\\ &= -pf(x,y,b) \end{align}$$

where in the last step, the Fundamental Theorem of Calculus was used (since the penultimate line comes down to the expression of the derivative of a $z$-primitive of $pf(x,y,z)$ at $z = b$).

The other two partial differentiations can be executed in a similar fashion.