Why is this procedure wrong? (Frullani's Integral) It is known that for $a,b>0$,
   $$\int_0^\infty \frac{e^{-ax} - e^{-bx}}{x}\ dx=\ln\left(\frac{b}{a}\right)$$
However, I was asked to show why the following procedure is incorrect:
$$\int_0^\infty \frac{e^{-ax} - e^{-bx}}{x}\ dx=\int_0^\infty \frac{e^{-ax}}{x}\ dx-\int_0^\infty \frac{e^{-bx}}{x}\ dx \stackrel{\text{c.o.v.}}{=}\int_0^\infty \frac{e^{u}}{u}\ dx-\int_0^\infty \frac{e^{u}}{u}\ dx=0 $$
I can't seem to find a mistake with the procedure, maybe I am missing something? I presume that you cannot seperate the integral, but the function is defined everywhere. 

The procedure is incorrect because $$ \int_0^\infty \frac{e^{-ax}}{x}dx$$ does not converge (the integral of $1/x$ diverges at the origin). The equation $$\int (f(x)+g(x)) dx = \int f(x) dx +\int g(x) dx$$ is only valid if both integrals converge.
