How can I show that $\int_1^x \frac {O(\frac 1 t)} t \ dt$ converge? How can I show that $\int_1^x \frac {O(\frac 1 t)} t \ dt$ converge? I'm so confused of this material because of big-O notation.
Thanks
 A: $\dfrac1t O\Bigl(\dfrac1t\Bigr)=O\biggl(\dfrac1{t^2}\biggr)\;$ and $\;\displaystyle\int_1^x\dfrac{\mathrm d\mkern 1mu t}{t^2}$ is convergent as $x$ tends to infinity.
A: Assumptions: Let's assume that the Big O notation you're using is "Big O as $t \to \infty$". Let's also assume that the class $O(1/t)$ consists of functions that are integrable on $(1,a)$ for all finite $a > 1$. This is, for instance, true for all continuous functions that are Big O of $1/t$ and (typically) any other functions you'd describe in this way in number theory. I'll also assume by the integral being "convergent" you mean that the function $\frac{1}{t}f(t)$ I discuss below (i.e., the $\frac{1}{t}O(t)$ integrand) is integrable on $[1,x]$ for $x \geq 1$ and the value of the integral converges as $x \to \infty$.
Solution: Suppose $f(t) \in O(1/t)$. Then by definition there exist $C, T > 0$ such that
$$
    |f(t)| \leq \frac{C}{t}\qquad\text{for all }t \geq T.
$$
So here what (I think) you're trying to determine is (a) whether or not $\frac{1}{t}f(t)$ is integrable on $[1,x]$ for all $x \geq 1$ and, assuming this, (b) the improper integral $\int_1^\infty \frac{1}{t}f(t)\,dt$ converges.
Regarding (a), note both $1/t$ and (by one of my assumptions) $f(t)$ are integrable on $[1,x]$ for all $x \geq 1$. Hence their product is integrable, as well. For (b), observe that
$$
    \int_1^x \frac{1}{t} f(t)\,dt = \int_1^T \frac{1}{t} f(t)\,dt + \int_T^x \frac{1}{t}f(t)\,dt,
$$
where the first integral on the RHS is (by (a)) some finite constant. To show that the second integral converges as $x \to \infty$, use the definition of $f(t) \in O(1/t)$ and the integral triangle inequality:
$$
    \left|\int_T^x \frac{1}{t}f(t)\,dt\right| \leq \int_T^x \frac{1}{t}|f(t)|\,dt \leq \int_T^x \frac{C}{t^2}\,dt = C\left(\frac{1}{T} - \frac{1}{x}\right).
$$
Now take a limit to obtain (b).
