About frullani integral 
Possible Duplicate:
Frullani proof integrals 

Let $f:\left[ {0,\infty } \right] \to \mathbb R$ be a a continuous function such that $$
\mathop {\lim }\limits_{x \to0+ } f\left( x \right) = L
$$Prove that $$
\int\limits_0^{\infty}  {\frac{{f\left( {ax} \right) - f\left( {bx} \right)}}
{x}}dx $$ converges and calculate the value.
It is known that $\int_a^\infty (f(x)/x)\,\mathrm{d}x$ converges for all a>0, but nothing of $\lim\limits_{x\to\infty}f(x)$ is told.
Also, what if $a>b$ or $a<b$?

The trick to evaluate the thing is, under suitable hypotheses to assure convergence and also differentiability in $a$ and/or $b$, to differentiate with respect to the parameter $a$. Letting $F(a)$ denote the integral (for fixed $b$), this gives $$F'(a)\;=\;\int_0^\infty xf'(ax)/x\,dx \;=\; \int_0^\infty f'(ax)\,dx
\;=\; {1\over a}\cdot \int_0^\infty f'(x)\,dx
\;=\; -f(0)\cdot {1\over a}
$$
Thus, $F(a)=C-f(0)\cdot \log a\;$, and the integral is of the form $C-f(0)\cdot (\log a-\log b)\;$. Since the integral is visibly $0$ when $a=b$, it is $-f(0)\cdot (\log a - \log b)$.
In fact, use of "Frullani" in the question surprises me, because if one knows such integrals by this name one has a way to look in Whittaker-Watson, etc.
Edit: A careful proof that the integral converges has to look at $\int_\epsilon^T {f(ax)-f(bx)\over x}\;dx$ and show that the limit exists as $\epsilon\rightarrow 0^+$ and $T\rightarrow +\infty$. Justification of differentiation with respect to a parameter is somewhat subtler (although obviously necessary), and the "approved details" depend in a volatile way on your context. 
