Differentiation under the integral sign

Statement

Let $f \colon [a, b] \times [t_0, t_1] \to \mathbb{R}$ be continuous, and suppose that its partial derivative $\partial f / \partial t$ exists and is continuous on $[a, b] \times [t_0, t_1]$. Then $F(t) = \int_a^b f(x, t) \, dx$ is continuously differentiable on $[t_0, t_1]$, and $$F'(t) = \int_a^b \frac{\partial f}{\partial t}(x, t) \, dx.$$ If, in addition, $\alpha, \beta \colon [t_0, t_1] \to [a, b]$ are continuously differentiable, then $$\frac{d}{dt} \int_{\alpha(t)}^{\beta(t)} f(x, t) \, dx = \int_{\alpha(t)}^{\beta(t)} \frac{\partial f}{\partial t}(x, t) \, dx + f(\beta(t), t) \, \beta'(t) - f(\alpha(t), t) \, \alpha'(t).$$

Remarks

The second formula follows from the first, the fundamental theorem of calculus and the chain rule, applied to $G(u, v, t) = \int_u^v f(x, t) \, dx$. A measure theory version: let $f(\cdot, t)$ be integrable for each $t$ in an open set, let $\partial f / \partial t$ exist for almost all $x$, and let $|\partial f / \partial t (x, t)| \le g(x)$ for an integrable $g$. Then the same rule holds. Its proof uses the dominated convergence theorem. Example: let $F(t) = \int_0^1 \frac{x^t - 1}{\ln x} \, dx$ for $t \ge 0$. For $t > 0$ the measure version, with $g = 1$, gives $F'(t) = \int_0^1 x^t \, dx = \frac{1}{t + 1}$. Since $F$ is continuous with $F(0) = 0$, $F(t) = \ln(1 + t)$.

Explanations

L1 human

If the integrand depends smoothly on a parameter, you can differentiate the integral by differentiating inside it. When the limits of integration also move, the values of the integrand at the moving limits add extra terms.