# When can one use the Leibniz rule for integration?

If I want to determine determine the following expression \begin{align*} \frac{\partial }{\partial t} \int_{a}^{b} f(x, t)\mathrm{d}x \end{align*} is it sufficient that $$f_n\left(x, t+\frac{1}{n}\right)$$ is uniformly converging to $$f(x, t)$$ in order to be allowed to apply the Leibniz rule, i.e. \begin{align*} \frac{\partial }{\partial t} \int_{a}^{b} f(x, t)\mathrm{d}x = \int_{a}^{b} \frac{\partial }{\partial t} f(x, t)\mathrm{d}x \end{align*} given that $$a, b$$ are independent of $$t$$? If not, which conditions are sufficient for being able to use Leibniz' rule (I have not learned about the dominated convergence theorem yet)?

• What do you mean by “$f_n(x, t + 1 / n)$ converging uniformly to $f(x, c)$”? Do you mean $f(x, t + 1 / n)$ converging uniformly to $f(x, t)$? Nov 27, 2021 at 17:27
• @Frank oh yes, there is a little typo which I corrected now, thanks! Nov 27, 2021 at 17:43

TL;DR, if the partial derivative $$\frac{\partial f}{\partial t}$$ is jointly continuous in the variables $$x$$ and $$t$$, then the Leibniz rule works. If you use the Lebesgue integral (which gives you the dominated convergence theorem), this condition can be relaxed.

### Leibniz rule for Riemann integration

When working with Riemann integrals, the standard criterion for switching a limit and an integral sign is the following statement (this is, in fact, a special case of the dominated convergence theorem), which relies on uniform convergence:

Theorem 1. (Interchanging limits and integrals) If $$g_n : [a, b] \to \mathbb{R}$$ is a sequence of Riemann integrable functions that converges uniformly to a Riemann integrable function $$g : [a, b] \to \mathbb{R}$$, then $$\lim_{n \to \infty} \int_a^b g_n(x)\mathrm{d}x = \int_a^b g(x)\mathrm{d}x.$$

Using this result, we can establish a Leibniz rule for Riemann integration. Because notation with multiple variables can get confusing, let us define $$F : \mathbb{R} \to \mathbb{R}$$ be the function $$F(t) = \int_a^b f(x, t)\mathrm{d}x,$$ where $$f : [a, b] \times \mathbb{R}$$ is the function in your question. For a fixed $$t_0 \in \mathbb{R}$$, we would like to find if $$F'(t_0)$$ exists and whether it can be obtained by the Leibniz rule. The key observation is that we can write differentiation as the limit $$\tag{1} F'(t_0) = \lim_{h \to 0} \frac{F(t_0 + h) - F(t_0)}{h} = \lim_{h \to 0} \int_a^b \frac{f(x, t_0 + h) - f(x, t_0)}{h} \mathrm{d}x.$$ To apply Theorem 1, we would like the difference quotient to converge uniformly. (That is to say, for every sequence $$h_n \to 0$$, the difference quotient $$\frac{f(x, t_0 + h_n) - f(x, t_0)}{h_n}$$ should converge uniformly in $$x$$.) However, the difference quotient is a bit unwieldy to work with, but we can use the mean-value theorem to instead write $$\tag{2} F'(t_0) = \lim_{h \to 0} \int_a^b \frac{\partial f}{\partial t}(x, t_0 + h_x) \mathrm{d}x,$$ where $$|h_x| \leq |h|$$ for all $$x \in [a, b]$$. This leads to the following result.

Theorem 2. (Leibniz rule for Riemann integration) Let $$f, F, t_0$$ be defined as above. If $$\frac{\partial f}{\partial t}$$ is continuous on a rectangle $$[a, b] \times [t_0 - \delta, t_0 + \delta]$$, then $$F'(t_0)$$ exists and is given by the formula $$F'(t_0) = \int_a^b \frac{\partial{f}}{\partial t}(x, t_0) \mathrm{d}x.$$ In particular, if $$\frac{\partial f}{\partial t}$$ is continuous on all of $$[a, b] \times \mathbb{R}$$, $$F$$ is differentiable everywhere and can be determined by the Leibniz rule.

By $$(2)$$, it suffices to show that for all sequences $$h_n \to 0$$, the functions $$g_n(x) := \frac{\partial f}{\partial t}(x, t_0 + (h_n)_x)$$ converge uniformly to $$g(x) := \frac{\partial f}{\partial t}(x)$$. This can be done by utilizing the uniform continuity of $$\frac{\partial f}{\partial t}$$. I'll leave the rest of the proof to you.

Also, (if I'm understanding correctly) your criterion of $$f(x, t + 1 / n)$$ converging uniformly to $$f(x, t)$$ doesn't exactly work. For one, it says nothing about the uniform convergence of the difference quotient in $$(1)$$ since it only works with discrete time steps of $$1 / n$$. Even if $$f(x, t + h)$$ were to converge uniformly to $$f(x, t)$$ as $$h \to 0$$, it would not guarantee uniform convergence of the difference quotient (it does not even guarantee the existence of a derivative!).

### Leibniz rule for Lebesgue integration

Finally, here's a criterion for the Leibniz rule if we are using the Lebesgue integral.

Theorem 3. (Leibniz rule for Lebesgue integration) Let $$X$$ be an open subset of $$\mathbb{R}$$. Let $$f : [a, b] \times X \to \mathbb{R}$$ be a Lebesgue integrable function such that the partial derivative $$\frac{\partial f}{\partial t}(x, t)$$ exists everywhere. Further suppose there is a Lebesgue integrable function $$g : [a, b] \to [0, +\infty]$$ such that $$\int_a^b g(x)\mathrm{d}x$$ is finite and $$\left|\frac{\partial f}{\partial t}(x, t)\right| \leq g(x)$$ for all $$t \in X$$ and $$x \in [a, b]$$. Then $$\frac{\mathrm{d}}{\mathrm{d}t}\int_a^b f(x, t)\mathrm{d}x = \int_a^b \frac{\partial f}{\partial t}(x, t) \mathrm{d}x.$$

Observe that Theorem 3 supersedes Theorem 2 because continuous functions are bounded on compact subsets. Setting $$X = [t_0 - \delta, t_0 + \delta]$$ and setting $$g : X \to \mathbb{R}$$ to be a constant bound of $$\frac{\partial f}{\partial t}$$ on $$X$$, Theorem 2 follows.

The proof of Theorem 3 is arguably easier than in the case of Riemann integration, at least if one is equipped with the machinery of measure theory. After obtaining $$(2)$$, the result follows directly from the dominated convergence theorem.