Proof of Frullani's theorem

We will assume $a<b$. Let $x,y>0$. We have: \begin{align*} \int_x^y\dfrac{f(at)-f(bt)}{t}dt&=\int_x^y\dfrac{f(at)}{t}dt- \int_x^y\dfrac{f(bt)}{t}dt\\ &=\int_{ax}^{ay}\dfrac{f(u)}{\frac ua}\frac{du}a- \int_{bx}^{by}\dfrac{f(u)}{\frac ub}\frac{du}b\\ &=\int_{ax}^{ay}\dfrac{f(u)}udu-\int_{bx}^{by}\dfrac{f(u)}udu\\ &=\int_{ax}^{bx}\dfrac{f(u)}udu+\int_{bx}^{ay}\dfrac{f(u)}udu -\int_{bx}^{ay}\dfrac{f(u)}udu-\int_{ay}^{by}\dfrac{f(u)}udu\\ &=\int_{ax}^{bx}\dfrac{f(u)}udu-\int_{ay}^{by}\dfrac{f(u)}udu. \end{align*} Since $\displaystyle\int_0^{+\infty}\dfrac{f(at)-f(bt)}tdt=\lim_{y\to +\infty}\lim_{x\to 0} \int_x^y\dfrac{f(at)-f(bt)}{t}dt$ if these limits exist, we only have to show that the limits $\displaystyle\lim_{x\to 0}\int_{ax}^{bx}\dfrac{f(u)}udu$ and $\displaystyle\lim_{y\to +\infty}\int_{ay}^{by}\dfrac{f(u)}udu$ exists, by computing them.

For the first, we denote $\displaystyle m(x):=\min_{t\in\left[ax,bx\right]}f(t)$ and $\displaystyle M(x):=\max_{t\in\left[ax,bx\right]}f(t)$. We have for $x>0$: $$m(x)\ln\left(\dfrac ba\right)\leq \int_{ax}^{bx}\dfrac{f(u)}udu\leq M(x)\ln\left(\dfrac ba\right) $$ and we get $\displaystyle\lim_{x\to 0}\,m(x)=\lim_{x\to 0}\, M(x)=f(0)$ thanks to the continuity of $f$.

For the second, fix $\varepsilon>0$. We can find $x_0$ such that if $u\geq x_0$ then $|f(u)|\leq \varepsilon$. For $y\geq \frac{x_0}a$, we get $\displaystyle\left|\int_{ay}^{by}\frac{f(u)}udu\right| \leq \varepsilon\ln\left(\dfrac ba\right) $. We notice that we didn't need the differentiability of $f$.

Added later, thanks to Didier's remark: if $f$ has a limit $l$ at $+\infty$, then $g\colon x\mapsto f(x)-l$ is still continuous and has a limit $0$ at $+\infty$. Then $$\int_0^{+\infty}\dfrac{f(at)-f(tb)}tdt = \int_0^{+\infty}\dfrac{g(at)-g(tb)}tdt =g(0)\ln\left(\dfrac ba\right) = \left(f(0)-l\right)\ln\left(\dfrac ba\right).$$

The result is true under weaker assumptions than you state, but under your conditions, there is a cute proof using double integrals. (To be on the safe side, assume that $f$ is monotone, or at least that $f' \in L^1$. This will guarantee that we can change the order of integration.)

Let $D = \{ (x,y) \in \mathbb{R}^2 : x \ge 0, a \le y \le b \}$, and compute the integral $$\iint_D -f'(xy)\,dx\,dy$$ in two different ways.

Firstly \begin{align} \iint_D -f'(xy)\,dx\,dy &= \int_a^b \left( \int_0^\infty -f'(xy)\,dx \right)\,dy \\ &= \int_a^b \left[ \frac{-f(xy)}{y}\right]^\infty_0\,dy \\ &= \int_a^b \frac{f(0)}{y}\,dy = f(0)(\ln b - \ln a). \end{align}

On the other hand, \begin{align} \iint_D -f'(xy)\,dx\,dy &= \int_0^\infty \left( \int_a^b -f'(xy)\,dy \right)\,dx\\ &= \int_0^\infty \left[ \frac{-f(xy)}{x} \right]_a^b\,dx \\ &= \int_0^\infty \frac{f(ax)-f(bx)}{x}\,dx. \end{align}

There is a claim that is slightly more general.

Let $f$ be such that $\int_a^b f$ exists for each $a,b>0$. Suppose that $$A=\lim_{x\to 0^+}x\int_{x}^1 \frac{f(t)}{t^2}dt\\B=\lim_{x\to+\infty}\frac 1 x\int_1^x f(t)dt$$ exist.

Then $$\int_0^\infty\frac{f(ax)-f(bx)}xdx=(B-A)\log \frac ab$$

PROOF Define $xg(x)=\displaystyle \int_1^x f(t)dt$. Since $g'(x)+\dfrac{g(x)}x=\dfrac{f(x)}x$ we have $$\int_a^b \frac{f(x)}xdx=g(b)-g(a)+\int_a^b\frac{g(x)}xdx$$

Thus for $T>0$

$$\int_{Ta}^{Tb} \frac{f(x)}xdx=g(Tb)-g(Ta)+\int_{Ta}^{Tb}\frac{g(x)}xdx$$

But $$\int_{Ta}^{Tb}\frac{g(x)}xdx-B\int_a^b \frac{dx}x=\int_a^b\frac{g(Tx)-B}xdx$$

Thus $$\lim_{T\to+\infty}\int_{Ta}^{Tb}\frac{g(x)}xdx=B\log\frac ba$$ so

$$\lim_{T\to+\infty}\int_{Ta}^{Tb}\frac{f(x)}xdx=B\log\frac ba$$

It follows, since $$\int_1^T\frac{f(ax)-f(bx)}xdx=\int_{bT}^{aT}\frac{f(x)}xdx+\int_a^b \frac{f(x)}xdx$$ (note $a,b$ are swapped) that $$\int_1^\infty \frac{f(ax)-f(bx)}xdx=B\log\frac ab+\int_a^b \frac{f(x)}xdx$$

Let $\varepsilon >0$, $\hat f(x)=f(1/x)$. Then $$\int\limits_\varepsilon ^1 {\frac{{f\left( x \right)}}{x}dx} = \int\limits_1^{{\varepsilon ^{ - 1}}} {\frac{{\hat f\left( x \right)}}{x}dx} $$ and $$x\int\limits_x^1 {\frac{{f\left( t \right)}}{{{t^2}}}dt} = \frac{1}{{{x^{ - 1}}}}\int\limits_1^{{x^{ - 1}}} {\hat f\left( t \right)dt} = g\left( {{x^{ - 1}}} \right)$$

So $\hat f(t)$ is in the hypothesis of the preceding work. It follows that $$\lim_{T\to+\infty}\int\limits_1^T {\frac{{\hat f\left( {x{a^{ - 1}}} \right) - \hat f\left( {x{b^{ - 1}}} \right)}}{x}} dx = A\log \frac ba + \int\limits_{{a^{ - 1}}}^{{b^{ - 1}}} {\frac{{\hat f\left( x \right)}}{x}dx} $$

and by a change of variables $x\mapsto x^{-1}$ we get $$\int\limits_0^1 {\frac{{f\left( {ax} \right) - f\left( {bx} \right)}}{x}} dx = A\log \frac ba - \int\limits_a^b {\frac{{f\left( x \right)}}{x}dx} $$ and summing gives the desired $$\int\limits_0^\infty {\frac{{f\left( {ax} \right) - f\left( {bx} \right)}}{x}} dx = \left( {B - A} \right)\log \frac ab$$

This is due to T.M. Apostol.

OBS By L'Hôpital, if the limits at $x=0^+$ and $x=+\infty$ exist, they equal $A$ and $B$ respectively.