Evaluate $\sum_{j=1}^{+\infty}\left(\log (j)-\log (j+x)+\frac x{2j}+\frac{3x}2\right)$

Question

I've been stuck with calculating the following series:

$$\sum_{j=1}^{+\infty}\left(\log (j)-\log (j+x)+\frac x{2j}+\frac{3x}2\right),$$

for real $x\geq 0$.

My attempt: Using the Frullani integral $\ln x=\int_0^{+\infty}\frac{e^{-t}-e^{-xt}}tdt$ and $\frac1k=\int_0^{+\infty}e^{-kt}dt$, we have \begin{align*} S(x) =& \sum_{j=1}^{+\infty}\left(\int_0^\infty\left(\frac{e^{-t}-e^{-kt}}t-\frac{e^{-t}-e^{-(k+x)t}}t+\frac{xe^{-kt}}2\right)dt+\frac{3x}2\right) \\=& \sum_{j=1}^{+\infty}\left(\int_0^\infty \frac{e^{-kt}}t\left(-1+e^-kt+\frac{xt}2\right)+\frac{3x}2\right) \end{align*} but I encountered a problem to calculate it. Can you help me?

Answer 1

Weierstrass product for $\Gamma$ implies $$ \sum_{j=1}^\infty\left(\log(j)-\log(j+z)+\frac zj\right)=\gamma z+\log\Gamma(1+z). $$ The series in the question (as currently written) diverges for $x\neq0$.

Answer 2

Consider the partial sum $$S_p=\sum_{j=1}^{p}\left(\log (j)-\log (j+x)+\frac x{2j}+\frac{3x}2\right)$$ and use

$$\sum_{j=1}^{p}\log(j+x)=\log \left((x+1)_p\right)=\log \left(\frac{\Gamma (p+x+1)}{\Gamma (x+1)}\right)$$ You are now ready to use Stirling approximation as well as the asymptotic of harmonic number since $$S_p=\frac{x H_p}{2}+\frac{3 p x}{2}+\log \left(\frac{\Gamma (p+1) \Gamma (x+1)}{\Gamma (p+x+1)}\right)$$