$\int\limits_0^{\infty}\left(a^{-x}-b^{-x}\right) d x=$

$\frac{1}{\log a}-\frac{1}{\log b}$ log a – log b log a + log b $\frac{1}{\log a}+\frac{1}{\log b}$

$\frac{1}{\log a}-\frac{1}{\log b}$

$=\int\left(a^{-x}-b^{-x}\right) d x=\left[\frac{a^{-x}}{-\log a}-\frac{b^{-x}}{-\log b}\right]_0^{\infty}=0+\frac{a^0}{\log a}-\frac{b^0}{\log b}=\frac{1}{\log a}-\frac{1}{\log b}$ Hence (1) is the correct answer.