Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
You might want to know there is a tag called intuition. – Git Gud Jan 12 '14 at 23:26
@GitGud Thank you, I didn't know that. – Eric Auld Jan 12 '14 at 23:34
I was just reading about this from a reference from another question. I'm not sure if there is enough intuition involved in that reference, but here it is:… – user114628 Jan 12 '14 at 23:39
@user2943324 Thanks, I will read it – Eric Auld Jan 12 '14 at 23:40

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.

share|cite|improve this answer
Why are we able to bring the $\lim_{s\to 0}$ under the integral? – Eric Auld Jan 12 '14 at 23:39
@EricAuld That's the "nice enough" condition. It does not always work, only under suitable conditions. It's just the intuition why that is a reasonable expectation. If $f$ and $\partial_s f$ are continuous, and the interval of integration is compact, the convergence is uniform. That's great. If the integral extends over all of $\mathbb{R}$, you need other conditions to ensure the convergence is good enough to interchange limit and integration. – Daniel Fischer Jan 12 '14 at 23:41
I'm sorry, when you say "the convergence is uniform", which convergence? The integral is proper if $f$ is continuous, so there is no issue of convergence there, right? – Eric Auld Jan 12 '14 at 23:44
@EricAuld The convergence of $\dfrac{f(x,s+h)-f(x,s)}{h}$ to $\partial_s \, f(x,s)$ for $h\to 0$ is uniform on $[a,b]$ under the continuity assumption, hence integration and limit can be exchanged then. – Daniel Fischer Jan 12 '14 at 23:46
@EricAuld \dfrac{}{} for display mode frac – user114628 Jan 13 '14 at 6:32

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.