[Introduction to Analytic Number Theory - Tom M. Apostol, chapter 13, question 7]
Express $$\frac 1{2\pi i}\int_{2-\infty i}^{2+\infty i} \frac{x^s}{s^2}\left(-\frac{\zeta^{\prime}(s)}{\zeta(s)}\right)\text{d}s$$ as a finite sum involving $\Lambda (n)$.
In a previous exercise, I proved that $$\frac 1{2\pi i}\int_{2-\infty i}^{2+\infty i} \frac{x^s}{s^2}\text{d}s$$ is $0$ when $0<y<1$ and $\log y$ when $y\ge 1$.
I am pretty sure that we have use this and the fact that $$-\frac{\zeta^{\prime}(s)}{\zeta(s)}=\sum_{n=1}^\infty \frac{\Lambda (n)}{n^s}$$ but I can't see how.
Hint: You can not necessary change the infinite integral with a series, but if the series is uniformly convergent on compact sets, you can interchange it with the integral between $2-iT$ and $2+iT$. Therefore:
$\frac{1}{2\pi i}\int_{2-iT}^{2+iT}\frac{x^s}{s^2}\sum\limits_{n=1}^{\infty}\frac{\Lambda(n)}{n^s}ds=\sum\limits_{n=1}^{\infty}\Lambda(n)\frac{1}{2\pi i}\int_{2-iT}^{2+iT}(\frac{x}{n})^s\frac{1}{s^2}ds$
Now split the series into $\sum\limits_{n<x}$ and $\sum\limits_{n\geq x}$, and use the previous exercise. The first sum is finite, the limit as $T\to\infty$ is obvious here. As for the second sum, show it tends to $0$ as $T\to\infty$. I don't know how you solved the previous exercise (there are a few ways to do it), but I believe you got some upper bound on $|\frac{1}{2\pi i}\int_{2-iT}^{2+iT}\frac{y^2}{s^2}ds|$ when $0<y<1$. (that might depend on both $y$ and $T$) Use this upper bound in order to bound the second sum.