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These are standard for a (graduate) course of real analysis — see for instance section 2.3 and exercises after it in Folland, "Real analysis". The reason they

These are standard for a (graduate) course of real analysis — see for instance section 2.3 and exercises after it in Folland, "Real analysis". The reason they are not usually covered in calculus is, to justify such differentiation, one needs the notions of Lebesgue integral and measure. The Riemann integral from calculus courses is just not robust enough. Of course, if the function inside the integral is nice enough, nothing bad happens, and the differentiation is valid.

r-zip 6y ago
Under certain technical conditions, differentiation under the integral sign also works for Riemann integration (see Marsden & Hoffman). There's no need to develop Lebesgue theory to demonstrate this technique in a calculus course, but uniform convergence must be understood.
- judofyr 6y ago
  Alternatively you can study physics and then you don't need to worry about these tiny details. I took a course on "mathematical methods in physics" which covered some complex analysis, and my math friends where shocked at how non-rigorous we were going through the theorems. Luckily for physicists these techniques tends to be valid because functions from the real world are well-behaved. For me personally it was so fun with a course where we did advanced mathematics for "practical" problems.
  - sdenton4 6y ago
    It's all fun and games until you're trying to calculate trajectories over a Cantor set.
    - qubex 6y ago
      Or integrate over an interval of surreals.
  - BeetleB 6y ago
    As someone who loved both math and physics, this was why I always found math a bit easier. Everything rests on a solid foundation and you can justify each step. When I got into higher physics, it was so riddled with intuitive arguments as opposed to rigor that I didn't fare so well. I'm sure one can find mathematical justifications for their methods, but it's not part of the curriculum, and almost none of the professors (in a top 10 physics school) knew them either.
    - TheOtherHobbes 6y ago
      Apart from the occasional Einstein and Newton every couple of centuries, physics seems to advance by throwing a semi-random selection of PhD dissertations at the real world and seeing if any of them happen to match experiment.
      - BeetleB 6y ago
        Case in point: Newton's work was not that rigorous. It was not till the 1800's that calculus was put on a firm foundation. Of course, things were all different back then.
anon_tor_12345 6y ago
>one needs the notions of Lebesgue integral and measure. The Riemann integral from calculus courses is just not robust enough. definitely not the case. leibniz's rule https://en.wikipedia.org/wiki/Leibniz_integral_rule#Proof_of_basic_form https://en.wikipedia.org/wiki/Leibniz_integral_rule#Proof_of... only requires fubini's theorem for exchanging order of integration https://en.wikipedia.org/wiki/Fubini%27s_theorem#Riemann_integrals https://en.wikipedia.org/wiki/Fubini%27s_theorem#Riemann_int... which i'm pretty sure everyone learns in multivariable calc. i personally learned it from apostol's calc (not analysis) books.
- Bahamut 6y ago
  This is not true - it's not necessarily obvious that things should hold when you're dealing with infinity as one of the bounds. In fact, there are cases where interchanging the order is not allowed when you're in that situation depending on the integrand.
  - anon_tor_12345 6y ago
    i'm at a loss. >In fact, there are cases where interchanging the order is not allowed when you're in that situation depending on the integrand. yup exactly in the cases where fubini's theorem doesn't hold and therefore those cases for which the integral theorem doesn't hold. Leibniz Theorem: Let f(x, t) be a function such that both f(x, t) and its partial derivative fx(x, t) are continuous in t and x in some region of the (x, t)-plane, including a(x) ≤ t ≤ b(x), x0 ≤ x ≤ x1. Also suppose that the functions a(x) and b(x) are both continuous and both have continuous derivatives for x0 ≤ x ≤ x1... Fubini's Theorem: If f(x,y) is a *continuous function* on a rectangle R=[a,b]×[c,d]...

r-zip 6y ago

Under certain technical conditions, differentiation under the integral sign also works for Riemann integration (see Marsden & Hoffman). There's no need to develop Lebesgue theory to demonstrate this technique in a calculus course, but uniform convergence must be understood.

judofyr 6y ago
Alternatively you can study physics and then you don't need to worry about these tiny details. I took a course on "mathematical methods in physics" which covered some complex analysis, and my math friends where shocked at how non-rigorous we were going through the theorems. Luckily for physicists these techniques tends to be valid because functions from the real world are well-behaved. For me personally it was so fun with a course where we did advanced mathematics for "practical" problems.
- sdenton4 6y ago
  It's all fun and games until you're trying to calculate trajectories over a Cantor set.
  - qubex 6y ago
    Or integrate over an interval of surreals.
- BeetleB 6y ago
  As someone who loved both math and physics, this was why I always found math a bit easier. Everything rests on a solid foundation and you can justify each step. When I got into higher physics, it was so riddled with intuitive arguments as opposed to rigor that I didn't fare so well. I'm sure one can find mathematical justifications for their methods, but it's not part of the curriculum, and almost none of the professors (in a top 10 physics school) knew them either.
  - TheOtherHobbes 6y ago
    Apart from the occasional Einstein and Newton every couple of centuries, physics seems to advance by throwing a semi-random selection of PhD dissertations at the real world and seeing if any of them happen to match experiment.
    - BeetleB 6y ago
      Case in point: Newton's work was not that rigorous. It was not till the 1800's that calculus was put on a firm foundation. Of course, things were all different back then.

anon_tor_12345 6y ago

>one needs the notions of Lebesgue integral and measure. The Riemann integral from calculus courses is just not robust enough. definitely not the case. leibniz's rule https://en.wikipedia.org/wiki/Leibniz_integral_rule#Proof_of_basic_form https://en.wikipedia.org/wiki/Leibniz_integral_rule#Proof_of... only requires fubini's theorem for exchanging order of integration https://en.wikipedia.org/wiki/Fubini%27s_theorem#Riemann_integrals https://en.wikipedia.org/wiki/Fubini%27s_theorem#Riemann_int... which i'm pretty sure everyone learns in multivariable calc. i personally learned it from apostol's calc (not analysis) books.

Bahamut 6y ago
This is not true - it's not necessarily obvious that things should hold when you're dealing with infinity as one of the bounds. In fact, there are cases where interchanging the order is not allowed when you're in that situation depending on the integrand.
- anon_tor_12345 6y ago
  i'm at a loss. >In fact, there are cases where interchanging the order is not allowed when you're in that situation depending on the integrand. yup exactly in the cases where fubini's theorem doesn't hold and therefore those cases for which the integral theorem doesn't hold. Leibniz Theorem: Let f(x, t) be a function such that both f(x, t) and its partial derivative fx(x, t) are continuous in t and x in some region of the (x, t)-plane, including a(x) ≤ t ≤ b(x), x0 ≤ x ≤ x1. Also suppose that the functions a(x) and b(x) are both continuous and both have continuous derivatives for x0 ≤ x ≤ x1... Fubini's Theorem: If f(x,y) is a *continuous function* on a rectangle R=[a,b]×[c,d]...