Intuition for differentiating beneath the integral

Question

I apologize in advance for a vague question.

There is a theorem:

If both $f(x,s)$ and $\partial _sf(x,s)$ are continuous in $x$ and $s$, then $$\partial_s\int_a^bf(x,s)\,dx=\int_a^b \partial_sf(x,s)\,dx$$

If in addition $\int_{-\infty}^\infty \partial_s f(x,s)\,dx$ converges uniformly in a neighborhood of $s_0$, then $$\partial_s \int_{-\infty}^\infty f(x,s)\,dx=\int_{-\infty}^\infty \partial_s f(x,s)\,dx.$$

The proof I know relies on integrating in $s$ and then switching the order of integration by uniform convergence. But beyond the mechanics of the proof, I am trying to develop an intuition for this fact. It does not seem intuitive to me that $$\partial_s \int f(x,s)\,dx = \int \partial_s f(x,s)\,dx.$$

I think the reason why it seems surprising to me is that you're integrating with respect to a different variable than the integration.

I am familiar with some real analysis and measure theory, so feel free to pitch an answer on that level.

Answer 1

The integral is linear, so, writing $F(s) = \int f(x,s)\,dx$, we have

$$\frac{F(s+h)-F(s)}{h} = \int \frac{f(x,s+h)-f(x,s)}{h}\,dx,$$

and letting $h\to 0$, the integrand converges pointwise to $\partial_s f(x,s)$.

So intuitively, it is to be expected that the derivative of $F$ is obtained by integrating the partial derivative of $f$ if that is nice enough.

Answer 2

Think of the integral $\int_a^bf(x,s)\,dx$ as an infinite sum $\sum_a^b f(x_i, s) \Delta x_i$, where the $x_i$ are the points of a suitable partition of $[a,b]$. You are probably aware of the fact that sums can be differentiated term-by-term; it follows that differentiating the sum with respect to $s$ can be brought into the sum. The good news is that this informal procedure can be formalized over the hyperreals in terms of hyperfinite sums; note that the integral itself it not exactly a hyperfinite sum but rather the standard part thereof, i.e. the result of rounding it off to the nearest real number.