Leibniz Integral Rule (Differentiating under Integral) + Proof

“Differentiating under the Integral” is a useful trick, and here we describe and prove a sufficient condition where we can use the trick. This is the Measure-Theoretic version, which is more general than the usual version stated in calculus books.

Let X be an open subset of \mathbb{R}, and \Omega be a measure space. Suppose f:X\times\Omega\to\mathbb{R} satisfies the following conditions:
1) f(x,\omega) is a Lebesgue-integrable function of \omega for each x\in X.
2) For almost all w\in\Omega, the derivative \frac{\partial f}{\partial x}(x,\omega) exists for all x\in X.
3) There is an integrable function \Theta: \Omega\to\mathbb{R} such that \displaystyle \left|\frac{\partial f}{\partial x}(x,\omega)\right|\leq\Theta(\omega) for all x\in X.

Then for all x\in X, \displaystyle \frac{d}{dx}\int_\Omega f(x,\omega)\,d\omega=\int_\Omega\frac{\partial}{\partial x} f(x,\omega)\,d\omega.

By definition, \displaystyle \frac{\partial f}{\partial x}(x,\omega)=\lim_{h\to 0}\frac{f(x+h,\omega)-f(x,\omega)}{h}.

Let h_n be a sequence tending to 0, and define \displaystyle \phi_n(x,\omega)=\frac{f(x+h_n,\omega)-f(x,\omega)}{h_n}.

It follows that \displaystyle \frac{\partial f}{\partial x}(x,\omega)=\lim_{n\to\infty}\phi_n(x,\omega) is measurable.

Using the Mean Value Theorem, we have \displaystyle |\phi_n(x,\omega)|\leq\sup_{x\in X}|\frac{\partial f}{\partial x}(x,\omega)|\leq\Theta(w) for each x\in X.

Thus for each x\in X, by the Dominated Convergence Theorem, we have \displaystyle \lim_{n\to\infty}\int_\Omega \phi_n(x,\omega)\,d\omega=\int_\Omega\lim_{n\to\infty}\phi_n(x,\omega)\,dw which implies \displaystyle \lim_{h_n\to 0}\frac{\int_\Omega f(x+h_n,\omega)\,d\omega-\int_\Omega f(x,\omega)\,d\omega}{h_n}=\int_\Omega \frac{\partial f}{\partial x}(x,\omega)\,d\omega.

That is, \displaystyle \frac{d}{dx}\int_\Omega f(x,\omega)\,d\omega=\int_\Omega \frac{\partial}{\partial x}f(x,\omega)\,d\omega.

Author: mathtuition88


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