I'm trying to solve the following taylor series $$\sum_{n=0}^\infty \frac{x^n}{n!} \ln(n+1)$$ so I can regularize the following sum $$\sum_{n=1}^\infty \ln(n)$$
Using Borel Regularizaiton I can use the formula $$\sum_{n=1}^\infty \ln(n) = \lim_{t\to1}\mathcal{L}_x\left[\sum_{n=0}^\infty\frac{x^n}{n!}\ln(n+1)\right](t)$$ but I'm unsure on how to solve the Taylor Expansion sum, the closest I've gotten is that the function $xe^x$ looks similar to the expansion.
Using the Frullani integral $\,\ln(1+n)=\int_0^\infty\frac{e^{-t}-e^{-(n+1)t}}tdt$ and changing the order of summation and integration $$S(x)=\sum_{n=0}^\infty \frac{x^n}{n!} \ln(n+1)=\int_0^\infty\frac{e^{-t}}t\left(\sum_{n=0}^\infty \frac{x^n}{n!}\left(1-e^{-nt}\right)\right)dt$$ I'm not sure that the closed form exists. What we can do for sure is to find the asymptotics at $x\gg1$ $$S(x)=\int_0^\infty\frac{e^{-t}}t\big(e^x-e^{xe^{-t}}\big)dt\overset{e^{-t}=s}{=}-e^x\int_0^1\frac{1-e^{x(s-1)}}{\ln s}ds=-e^x\int_0^1\frac{1-e^{-xt}}{\ln (1-t)}dt$$ $$=e^x\int_0^1\big(1-e^{-xt}\big)\left(\frac1t-\frac1{\ln (1-t)}-\frac1t\right)dt=I_1+I_2$$ where $$I_1=e^x\int_0^1\frac{1-e^{-xt}}tdt=e^x\int_0^x\frac{1-e^{-t}}tdt$$ Integrating by parts $$=e^x\big(1-e^{-t})\big)\ln t\,\bigg|_{t=0}^x-e^x\int_0^xe^{-t}\ln tdt$$ $$=e^x\big(1-e^{-x}\big)\ln x-e^x\int_0^\infty e^{-t}\ln tdt+e^x\int_x^\infty e^{-t}\ln tdt$$ $$I_1=e^x\Big(\ln x+\gamma+O(e^{-x}\ln x)\Big)\tag{1}$$ The second integral, in turn $$I_2=-e^x\int_0^1\frac{t+\ln(1-t)}{t\ln(1-t)}\Big(1-e^{-xt}\Big)dt$$ $$=-e^x\int_0^1\frac{t+\ln(1-t)}{t\ln(1-t)}dt+e^x\int_0^1\frac{t+\ln(1-t)}{t\ln(1-t)}e^{-xt}dt=I_{2a}+I_{2b}$$ where $$I_{2a}=-e^x\int_0^1\left(\frac1t+\frac1{\ln(1-t)}\right)dt=-e^x\int_0^1\left(\frac1{1-t}+\frac1{\ln t}\right)dt$$ $$\overset{t=e^{-s}}{=}-e^x\int_0^\infty\left(\frac1{e^s-1}-\frac{e^{-s}}s\right)ds=-e^x\gamma\tag{2a}$$ $$I_{2b}=e^x\int_0^1\frac{t+\ln(1-t)}{t\ln(1-t)}e^{-xt}dt\overset{tx=s}{=}\frac{e^x}x\int_0^x\frac{s+x\ln\big(1-\frac sx\big)}{s\ln\big(1-\frac sx\big)}e^{-s}ds$$ The main contribution to the integral comes from $s\ll x$, so, dropping exponentially small terms, we can simply decompose the logarithm and integrate term by term (putting $\infty$ as the upper bound): $$\sim\frac{e^x}x\int_0^\infty\frac{-\frac{s^2}{2x}-\frac{s^3}{3x^2}-...}{-\frac{s^2}x-\frac{s^3}{2x^2}-...}e^{-s}ds=\frac{e^x}{2x}\int_0^\infty\left(1+\Big(\frac23-\frac12\Big)\frac sx+...\right)e^{-s}ds$$ $$I_{2b}=e^x\left(\frac1{2x}+\frac1{12x^2}+O\Big(\frac1{x^3}\Big)\right)\tag{2b}$$ Taking together (1), (2a) and (2b) $$S(x)=e^x\left(\ln x+\frac1{2x}+\frac1{12x^2}+O\Big(\frac1{x^3}\Big)\right)$$ Numeric checks confirm the answer.
