Prove that $\int \limits _{0}^{\infty}\frac{e^{-2x}-e^{-ax}}{x}\text{d}x$ converges for any $a>0$
Solution 1:
Note that $F'(a) = \int \limits _0 ^\infty \Bbb e ^{-ax} \Bbb d x = \dfrac {e ^{-ax}} {-a} \bigg| _0 ^\infty = \dfrac 1 a$, which implies $F(a) = \ln a + C$ where $C$ is a constant that we must find out. In order to do this, evaluate $F$ in $a=2$: on the one hand, this must be $0$; on the other hand, it is $\ln 2 + C$, so $C = - \ln 2$, which implies $F(a) = \ln \dfrac a 2$.