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May I affirm $$\frac{d}{dx}\left(\int_{x_0}^{x} f(t)\cdot g(t) dt\right)=f(x)\cdot g(x)$$ if $f$ and $g$ are well definied?

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  • $\begingroup$ You need at least continuity for this. $\endgroup$
    – Randall
    Feb 21 at 19:23

2 Answers 2

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The requirement is that $f(x)\cdot g(x)$ be continuous function over $[a,b]$ where $a < b$ are constants and $x \in [a,b]$. Counter example is $f(x) = 1, g(x) = \dfrac{1}{x}$ over $[a,b] = [-1,1]$.

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  • $\begingroup$ You can make it a bit weaker than continuous, since all you need is a differentiable antiderivative, which means (as a rather trivial example) $f(x) = 1$ and $g(x)$ a derivative of a function which is differentiable but not continuously so will also work. $\endgroup$
    – Mel
    Feb 21 at 19:29
  • $\begingroup$ @WangYeFei Continuity is not a requirement, it is just a sufficient confition. If $h = f g$ is Lebesgue integrable over $[a,b]$ and continuous at $x_0$ then $F(x)=\int_a^x h(t) dt$ is differentiable at $x_0$ and $F'(x_0) = h(x_0)$ $\endgroup$ Feb 21 at 19:32
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Yes, almost everywhere (by Lebesgue's Fundamental Theorem of Calculus, thanks @peek-a-boo) when $f(t) \cdot g(t)$ is integrable, and in general for continuous functions.

Define $h(t) = f(t) \cdot g(t)$, and let $H(t)$ be an antiderivative for $h(t)$ (this is where the integrability assumption comes into play). Then the Fundamental Theorem of Calculus informs you that $$\frac{\mathrm{d}}{\mathrm{d} x} {\bigg( \int_{x_0}^{x} h(t)\ \mathrm{d} t \bigg)} = \frac{\mathrm{d}}{\mathrm{d} x} {\big( H(x) - H(x_0) \big)}.$$ Now $x_0$ is a constant, so $H(x_0)$ is as well, meaning it vanishes when we take the derivative, so our expression simply becomes $\frac{\mathrm{d}}{\mathrm{d} x} H(x)$, which is (by definition of $H$ as an antiderivative of $h$) the same as $h(x)$. This is by definition of $h$ the same as $f(x) \cdot g(x)$, as desired.

Luckily in general integrability is very easy to satisfy, including for instance all continuous functions. See the other answer for an example where this breaks (mildly) at $x = 0$ because integrability is violated.

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  • $\begingroup$ Integrability alone is NOT enough to guarantee this for all points $x$. By Lebesgue’s version of the FTC, we only get it a.e. Or more simply, one should just assume continuity. Also, integrable functions need not have primitives (i.e functions whose derivative agrees with the original function at every point). $\endgroup$
    – peek-a-boo
    Feb 21 at 19:59
  • $\begingroup$ Thanks for the reminder! I tend to automatically patch irregularities on measure-zero sets in my work so I sometimes forget about them. I'll edit to include that $\endgroup$
    – Mel
    Feb 21 at 20:47

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