Prove with $L(x) = \int_1^x\frac{1}{t}dt$ for $x>0$ that $L(xy) = L(x)+L(y)$

Question

Let $L: (0,\infty)\rightarrow \mathbb{R}$ be $L(x) = \int_1^x\frac{1}{t}dt$ for $x>0$, prove that $L(xy) = L(x)+L(y)$.

I do not know where to start here so hint are very welcome

Answer 1

Let $u=yt$. Then

$$\int_1^{x}\frac{1}{t}dt=\int_y^{xy}\frac{y}{u}\left(\frac{1}{y}\right)du=\int_y^{xy}\frac{1}{t}dt=\int_1^{xy}\frac{1}{t}dt-\int_1^{x}\frac{1}{t}dt$$

Note: Alternatively, as $\displaystyle \int_1^x\frac{1}{t}dt=\ln x$, the result follows.

Answer 2

It's just two changes of variable: $$\int_1^{xy}\frac1t\,dt\stackrel{(s= t/x)}=\int_{1/x}^{y}\frac{1}{s}\,ds=\int_1^y\frac1s\,ds-\int_{1}^{1/x}\frac1u\,du\stackrel{(v=\frac1u)}=\\=L(y)-\int_1^x-\frac{v}{v^2}\,dv=L(y)+L(x)$$