Here I have a problem to calculate the probability density function (PDF) of the sum of two independent random variables (RVs), $Z_0,Z_1$, and the PDF of $Z_0,Z_1$ are as follow:
$$f_{Z_0}(x)=\frac{1}{a}\ln\frac{a}{x},\quad x\in(0,a]$$ $$f_{Z_1}(x)=\frac{1}{b}\ln\frac{b}{x},\quad x\in(0,b]$$
where $a,b$ are two positive constants.
PDF of the sum of independent RVs is the convolution of PDF of each component RV. I considered characteristic function (Fourier Transform) and direct convolution. However, how to get the closed-form expression of $\displaystyle \int_0^{x} \ln(\tau)\ln{(x-\tau)}d\tau$ ?
$$\int_0^{x} \ln(\tau)\ln{(x-\tau)}d\tau=\\\int_0^{1} \ln(x\sigma)\ln{(x-x\sigma)}d(x\sigma)=$$ $$\int_0^{1}(\ln(x)+\ln(\sigma))(\ln(x)+\ln{(1-\sigma)})xd\sigma=$$ $$x\ln^2(x)+x\ln(x)\int_0^1\ln(\sigma)d\sigma+x\ln(x)\int_0^1\ln(1-\sigma)d\sigma+x\int_0^{1}\ln(\sigma)\ln{(1-\sigma})d\sigma=\\ x\ln^2(x)-2x\ln(x)+(2-\frac{\pi^2}6)x$$