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I'm working on a homework problem where I need to do something that looks like this (this isn't the exact question - I just want to check my reasoning for my approach). Given:

$f(t)=\int_a^b{g(x,t)dx}$

I need to find $df/dt$.

The approach I've come up with is:

$\frac{df}{dt}=\lim_{h\to0}\frac{\int_a^b{g(x,t+h)dx}- \int_a^b{g(x,t)dx}}{h}=\lim_{h\to0}\int_a^b{\frac{g(x,t+h)-g(x,t)}{h}}dx=\int_a^b{\lim_{h\to0}\frac{g(x,t+h)-g(x,t)}{h}}dx=\int_a^b{\frac{dg}{dt}dx}$

In other words, to calculate $df/dt$, I can calculate $dg/dt$, and then integrate that with respect to $x$ from $a$ to $b$.

Is this correct? It feels right to me, but I'm not sure about bringing the limit inside the integral.

Apologies if the notation is bad. I wasn't sure how to: (a) make bigger integrals; (b) do multiple lines with the equal signs lining up; or (c) get the limit arrow part under the limit.

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"Liebnitz rule", that zkutch refers to, says, in general, that $\frac{d}{dt}\int_{\alpha(t)}^{\beta(t)} f(x,t) dx= f(\beta(t),t)- f(\alpha(t),t)+ \int_{\alpha(t)}^{\beta(t)} \frac{\partial f}{\partial t} dx$.

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In book John M.H. Olmsted - Advanced calculus-Prentice Hall (1961), on pages 321-324 you can find different variants on how to differentiate under integral sign (Lebnitz's rule). One theorem asserts:

If $f(x,y)$ and $f'_x(x,y)$ are continuous on closed rectangle $a \leqslant x \leqslant b, c \leqslant y \leqslant d$, then function $F(x)=\int\limits_{c}^{d}f(x,y)dy$ is differentiable for $a \leqslant x \leqslant b$ and

$$F'(x)=\int\limits_{c}^{d}f'_x(x,y)dy$$

You can find there generalizations for case with continuous functions for integral bounds and improper integral.

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  • $\begingroup$ Lucky Olmsted! I read his book. Good one! + $\endgroup$ – Mikasa Apr 23 at 13:17
  • $\begingroup$ He has also another excellent book "Counterexamples in Analysis". But why "lucky", @Mikasa? Some story to know? $\endgroup$ – zkutch Apr 23 at 14:00
  • $\begingroup$ Cause there are a few books which did what he did with a bit different approach. Like Adams. I read read the book due to teaching Calculus I $\endgroup$ – Mikasa Apr 23 at 14:43

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