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?


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

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 $a.$

  • $\begingroup$ Why is the second one 0? $\endgroup$ – mavavilj Feb 25 '17 at 12:35
  • 3
    $\begingroup$ @mavavilj : No $\text{“} x \text{''}$ appears in the second integral, so that does not change as $x$ changes. These are derivatives with respect to $x,$ and the derivative with respect to $x$, of a quantity that does not change as $x$ changes is $0. \qquad$ $\endgroup$ – Michael Hardy Mar 18 '17 at 23:56
  • $\begingroup$ You could also say that the second term, if it converges, is just a constant, and derivative of a constant is just a zero. $\endgroup$ – TheCoolDrop May 28 '19 at 7:34

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