Unbounded Integral

Question

Given $\lim_{t \rightarrow \infty} f^\prime(t) = L$ and $c \leq L|b|,$ where $a,b$ are some constants, I want to show that $$\lim_{t \rightarrow \infty} \int_{a}^{a + t|b|} f^\prime(s)ds - ct = \infty$$

So I started with a lower bound, $$\lim_{t \rightarrow \infty} \int_{a}^{a + t|b|} (f^\prime(s) - L)ds \leq \lim_{t \rightarrow \infty} \int_{a}^{a + t|b|} f^\prime(s)ds - ct$$

I also know that for all $\epsilon>0$, there exists $N > 0$ such that $|f^\prime(t)-L|< \epsilon$ for all $t>N$. I was hoping to show that $ 0< \epsilon_0 < |f^\prime(s)-L|$, so that I can show that the improper integral is unbounded. However, I am not sure how to proceed

Answer 1

You should assume that $b \not= 0$ and $c < L|b|$. Then

1. choose $\epsilon > 0$ small enough that $(L - \epsilon)|b| > c$, and
2. select $M > a$ with the property that $s \ge M$ implies $f'(s) > L-\epsilon$.

If $a+t|b| > M$ then $$\int_a^{a+t|b|} f'(s) \, ds = \int_a^M f'(s) \, ds + \int_M^{a + t|b|}f'(s) \, ds > f(M) - f(a) + (L-\epsilon)(a+t|b|-M)$$ which leads to $$\int_a^{a+t|b|} f'(s) \, ds - ct > f(M) - f(a) + (L-\epsilon)(a-M) + t \left[(L - \epsilon)|b| - c\right].$$ The right-hand side of this inequality is unbounded above as $t \to \infty$.