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I'm looking to prove$\lim \limits_{x \to \infty} e^{-Px}\int Q'(x)\frac{e^{Px}}{P} \ dx$ exists and determine what it is, for constant $P>0$ and a continuous $Q'(x)$ s.t. $Q'(x) \to 0$ as $x \to \infty$

I can explain what would happen but I'm not sure how to put it rigorously. Consider the successive applications of integration by parts to $e^{-Px}\int Q'(x)\frac{e^{Px}}{P} \ dx$:

$$\Rightarrow Q'(x)\cdot\frac{1}{P^2}-e^{-Px}\int \left[ Q''(x)\int \frac{e^{Px}}{P} \right]$$

$$\Rightarrow Q'(x)\cdot\frac{1}{P^2}-Q''(x)\cdot\frac{1}{P^3}+ e^{-Px}\int \left[ Q'''(x)\int\int \frac{e^{Px}}{P} \right]$$ And so on. Since the $e^{-Px}$ term will keep on cancelling the effect of $e^{Px}$ for successive integration by parts, it seems therefore the convergence of the function will mainly be determined by our supposition that $Q'(x) \to 0$ as $x \to \infty$, so it seems to me that the function converges to $0$, but I don't know how to state this rigorously. How exactly would I begin to show this, since it seems to me we can't estimate how $Q''(x), Q'''(x), ...$ behave other than the fact they approach $0$ as $x \to \infty$? This got me stuck on estimating the value of the integral at the end of each application of integration by parts, so any help is appreciated!

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Note that $$\begin{align} \lim_{x \to \infty}\frac{1}{P} \frac{\int Q'(x)e^{Px}\,dt}{e^{Px}} &\overset1= \lim_{x \to \infty}\frac{1}{P} \frac{\int_a^x Q'(t)e^{Pt}\,dt}{e^{Px}} \\ &\overset2= \frac1P\lim_{x\to\infty} \frac{Q'(x) e^{Px}}{Pe^{Px}} \\ &\overset3= \frac{1}{P^2}\lim_{x\to\infty} Q'(x) \\ &\overset4= 0\end{align}$$ Explanation:

  1. $\int f(x) dx = \int_a^x f(t)dt$ for some $a$.
  2. Since the denominator goes to $\infty$ as $x\to\infty$, we use L'Hôpital's rule.
  3. Cancel out $e^{Px}$
  4. Use the fact that $\lim_{x\to\infty}Q'(x) = 0$.
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    $\begingroup$ wow, super useful identity on #1, thank you so much! does it follow directly from the fundamental theorem of calculus? I would have thought we'd necessarily need to have $F(a) = 0$ for $\int f(x) \ dx = \int_a^x f(t) \ dt$ to work $\endgroup$
    – shintuku
    Jul 26, 2021 at 1:58
  • $\begingroup$ @shintuku Actually, you don't need that identity here. Just knowing that $$\lim_{x\to\infty}\frac{1}{e^{Px}} \int f(x) dx = \lim_{x\to\infty} \frac{F(x) + C}{e^{Px}} = \lim_{x\to\infty}\frac{F(x)}{e^{Px}}$$ and $$\lim_{x\to\infty}\frac{1}{e^{Px}} \int_a^x f(t) dt =\lim_{x\to\infty}\frac{F(x) - F(a)}{e^{Px}} = \lim_{x\to\infty}\frac{F(x)}{e^{Px}} $$ so they have the same limits if exist is enough. (here, $F(x)$ is an antiderivative of $f(x)$) $\endgroup$
    – VIVID
    Jul 26, 2021 at 2:02
  • $\begingroup$ thank you so much for the help! $\endgroup$
    – shintuku
    Jul 26, 2021 at 2:04
  • $\begingroup$ @shintuku You are welcome. $\endgroup$
    – VIVID
    Jul 26, 2021 at 2:06

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