# Prove$\lim \limits_{x \to \infty} e^{-Px}\int Q'(x)\frac{e^{Px}}{P} \ dx$ exists and find it, for constant $P>0$ and $Q'(x) \to 0$ as $x \to \infty$

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!

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 $$\lim_{x\to\infty}Q'(x) = 0$$.
• 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 Jul 26, 2021 at 1:58
• @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 $f(x)$) Jul 26, 2021 at 2:02