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Let $[a,b]$ be any closed interval and let $f,g$ be continuous on $[a,b]$ with $g(x)\geq 0$ for all $x\in[a,b]$. Then the Second Mean Value Theorem says that

$$\int_a^bf(t)g(t)\text{d}t = f(c)\int_a^b g(t)\text{d}t,$$

for some $c\in(a,b)$.

Does this theorem work on the interval $[0,\infty]$ ?

EDIT: Assuming the integrals involved converge.

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No. Consider $g(t)=\frac{1}{t}$, $f=g$ on $[1,\infty)$.

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Let $f$ be integrable over the arbitrary interval $[a,\infty)$ and bounded with $m\leq f\leq M$ for all $x\geq a$. Let $g(x)\geq 0$ or $g(x)\leq 0$ for all $x\geq a$. Assuming all integrals converge, then $$ \int_a^\infty f(x)g(x)\text{d}x=\lambda\int_a^\infty g(x)\text{d}x,$$ where $m<\lambda<M$ (adapted from Gradshteyn et al, 2007)

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I believe you mistated the second mean value theorem, what you stated was the genearlized form of the first mean value theorem. The second mean value theorem states: Let g(x) be a monotonic function over (a,b), and let f(x) not change signs more than a finite number of times over (a,b). Then ∫ b a f(t)g(t)dt=f(a)∫ e a g(t)dt + f(b)∫ b e g(t)dt for some e in (a,b)

And under certain conditions this would hold in an infinite interval as well.

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