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Let $f\,\,$ be a continuous function on $[a,\infty)$ such that $\int_a^\infty f(t)\,dt$ converges. Define the function $F\,$ on $[a,\infty)$ with

$$F(x) := -\int_x^\infty f(t)\,dt \qquad\text{for all}\quad x\in[a,\infty).$$

Can we somehow deduce from this—using the regular fundamental theorem of calculus—that $F\,\,$ is continuous and differentiable on $(a,\infty)$, and that $F\,\,'(x) = f(x)$ for all $x\in(a,\infty)$? If so, how? Are such “improper” versions of the fundamental theorem to be found in some book out there that I can reference?

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Like this? –  J. M. Nov 20 '11 at 16:57

1 Answer 1

up vote 2 down vote accepted

$$F(x) := -\int_x^\infty f(t)\,dt = -\int_x^c f(t)\;dt + \int_c^\infty f(t)\;dt$$

The derivative of the first term of this sum with respect to $x$ is $f(x)$, and that of the second term is $0$. You have some leeway as to what $c$ is; you could just choose $a$.

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I could be wrong, but isn't the derivative of the first term f(x)? Since you have to reverse the limits to apply the fundamental theorem of calculus? –  lol Apr 16 '13 at 15:50
Sorry: typo. ${{{{}}}}$ –  Michael Hardy Apr 17 '13 at 2:39
no worries, that was my first math.stackexchange post - sewww excited haha! –  lol Apr 21 '13 at 6:22

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