Repeat a logarithmic segment with the correct jump values

On the intended base-period region $1\le x\le e$, $V=\pi\int_1^e(\ln x)^2dx$. Two integrations by parts give $\int(\ln x)^2dx=x(\ln x)^2-2x\ln x+2x+C$. Hence $V=\pi(e-2)$. A source intermediate line drops the factor $x$ from $-2x\ln x$, but its following evaluation and final answer use the correct antiderivative.

On $[1+2k,e+2k]$, $f(x)=\ln(x-2k)$; on $(e+2k,3+2k]$, $f(x)=0$. For the requested graph take $k=-2,-1,0$ for three complete periods and $k=1$ for the remaining segment $3\le x\le4$. Each logarithmic top point $(e+2k,1)$ is included, while the zero point immediately below it is open; the log starts $(1+2k,0)$ are included. The last endpoint is $(4,\ln2)$.

$\int_{-3}^4f(x)dx=3\int_1^e\ln x\,dx+\int_1^2\ln x\,dx=3+(2\ln2-1)=2+2\ln2\approx3.39$ (3 s.f.).

The last segment is $f(x)=\ln(x-2)$ on $3\le x\le4$. Thus $3\int_1^e\ln x\,dx+\int_3^4\ln(x-2)dx$ gives the same $2+2\ln2$.