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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 $$A=lim_{xto 0^+}xint_{x}^1 frac{f(t)}{t^2}dt\B=lim_{xto+infty}frac 1 xint_1^x f(t)dt$$ exist.

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

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

Thus for $T>0$

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

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

Thus $$lim_{Tto+infty}int_{Ta}^{Tb}frac{g(x)}xdx=Blogfrac ba$$ so

$$lim_{Tto+infty}int_{Ta}^{Tb}frac{f(x)}xdx=Blogfrac ba$$

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

Let $varepsilon >0$, $hat f(x)=f(1/x)$. Then $$intlimits_varepsilon ^1 {frac{{fleft( x right)}}{x}dx} = intlimits_1^{{varepsilon ^{ - 1}}} {frac{{hat fleft( x right)}}{x}dx} $$ and $$xintlimits_x^1 {frac{{fleft( t right)}}{{{t^2}}}dt} = frac{1}{{{x^{ - 1}}}}intlimits_1^{{x^{ - 1}}} {hat fleft( t right)dt} = gleft( {{x^{ - 1}}} right)$$

So $hat f(t)$ is in the hypothesis of the preceding work. It follows that $$lim_{Tto+infty}intlimits_1^T {frac{{hat fleft( {x{a^{ - 1}}} right) - hat fleft( {x{b^{ - 1}}} right)}}{x}} dx = Alog frac ba + intlimits_{{a^{ - 1}}}^{{b^{ - 1}}} {frac{{hat fleft( x right)}}{x}dx} $$

and by a change of variables $xmapsto x^{-1}$ we get $$intlimits_0^1 {frac{{fleft( {ax} right) - fleft( {bx} right)}}{x}} dx = Alog frac ba - intlimits_a^b {frac{{fleft( x right)}}{x}dx} $$ and summing gives the desired $$intlimits_0^infty {frac{{fleft( {ax} right) - fleft( {bx} right)}}{x}} dx = left( {B - A} right)log frac ab$$

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.

The following theorem is a beautiful generalization of Frullani’s integral theorem.

Let $f(x)-f(infty)=sum_{k=0}^{infty}frac{u(k)(-x)^k}{k!}$ and $g(x)-g(infty)=sum_{k=0}^{infty}frac{v(k)(-x)^k}{k!}$

$Theorem1$:

Let f, and g be continuous function on $[0,infty),$ assume that $f(0)=g(0)$ and $f(infty)=g(infty)$. Then if $a,b>0$

$$lim_{n to 0+}I_{n}equiv lim_{n to 0+} int_{0}^{infty}x^{n-1}lbrace f(ax)-g(bx) rbrace dx=lbrace f(0)-f(infty)rbrace bigg lbrace log bigg(frac{b}{a} bigg)+frac{d}{ds}bigg(logbigg(frac{v(s)}{u(s)}bigg) bigg)_{s=0} bigg rbrace$$

if $f(x)=g(x),$ this theorem reduces to the Frullani’s theorem

$$int_{0}^{infty} frac{f(ax)-f(bx)}{x}dx=lbrace f(0)-f(infty) rbrace log bigg(frac{b}{a} bigg).$$

Let prove $Theorem1$, To do this we need to use Ramanujan's master theorem , Which lies in the fact that

$$int_0^infty x^{n-1}sum_{k=0}^infty frac {phi(k)(-x)^k}{k!}dx= Gamma(n)phi(-n).$$

Applying the Master Theorem with $0<n<1,$ we find

$$I_n=int_{0}^{infty} x^{n-1}( f(ax)-g(bx))dx=int_{0}^{infty} x^{n-1}( lbrace f(ax)-f(infty) rbrace-lbrace g(bx)-g(infty) rbrace) dx$$

$$=Gamma(n)lbrace a^{-n}u(-n)-b^{-n}v(-n) rbrace$$

$$=Gamma(n+1) bigg lbrace frac{a^{-n}u(-n)-b^{-n}v(-n)}{n} bigg rbrace
$$

Letting $n$ tend to $0$, using L'Hospital's Rule and fact that $u(0)=v(0)=f(0)-f(infty).$ we deduce that

$$lim_{n to infty}I_n=lim_{n to infty} bigg lbrace frac{b^nv(n)-a^nu(n)}{n} bigg rbrace$$

$$=lim_{n to infty} lbrace b^nv(n) log b+ b^nv'(n)-a^nu(n)log a-a^nu'(n)rbrace$$

$$= lbrace f(0)-f(infty) rbrace log bigg(frac{b}{a} bigg)+v'(0)-u'(0)$$

$$=lbrace f(0)-f(infty) rbrace bigg lbrace log bigg(frac{b}{a} bigg)+frac{d}{ds}bigg(logbigg(frac{v(s)}{u(s)}bigg) bigg)_{s=0} bigg rbrace$$

You might be interested in an approach to Frullani's theorem I came across online. It is proven for the Lebesgue integral and the Denjoy-Perron integral. We are looking to prove the integral

begin{equation*}
int^{infty}_{0}frac{f(ax)-f(bx)}{x}dx=Aln(frac{a}{b})
end{equation*}
where $A$ is a constant. For the Lebesgue integral, the substitution $x=e^t,~alpha=ln(alpha),~beta=ln(b)$ is used to get
begin{equation*}
int^{+infty}_{-infty}{ f(e^{t+alpha})-f(e^{t+beta})}dt=A(alpha-beta)
end{equation*}
which is equivalent to Frullani's theorem. Then verifying the integral
begin{equation*}
int^{+infty}_{-infty}{ g(x+alpha)-g(x+beta)}dx=A(alpha-beta)
end{equation*}
for a Lebesgue integrable function $g:mathbb{R}tomathbb{R}~forall alpha,betain mathbb{R}$ will suffice. This is proved by setting an integrable function on the real line
begin{equation*}
h_{alpha}(x)=g(x+alpha)-g(x)~forallalphainmathbb{R}
end{equation*}
and applying the Fourier transform (as well as a little manipulation).

The Denjoy-Perron integral is used instead of the Lebesgue integral to avoid the problem of a locally integrable function $f:mathbb{R}tomathbb{C}$ admitting a derivative $f'(x)~forall xinmathbb{R}$ without $f'$ being locally integrable.
The case for the Denjoy-Perron integral is proved in a similar fashion.

Check out the following paper by J. Reyna

http://www.ams.org/journals/proc/1990-109-01/S0002-9939-1990-1007485-4/S0002-9939-1990-1007485-4.pdf

On the assumption that $f$ is differentiable and $a,b>0$, an application of Fubini's theorem gives the result.

$$int_{(0,infty)} frac{f(ax)-f(bx)}{x} dx$$

$$=int_{(0,infty)} int_{[bx,ax]} f'(y) frac{1}{x} dy dx$$

Let $0<bx leq y leq ax$.

$$=int_{(0,infty)} int_{frac{1}{a} y}^{frac{1}{b} y} f'(y) frac{1}{x} dx dy$$

$$=int_{(0,infty)} f'(y) ln (frac{a}{b}) dy$$

$$=(f(0)-f(infty)) ln frac{b}{a}$$

The following is just a speeded-up version of the answer by Davide Giraudo.

Let $f(x)$ be a real-valued function defined for $xgeq 0$.
Suppose that $f(x)$ is Riemann integrable on every bounded interval
of nonnegative real numbers,
that $f(x)$ is continuous at $x=0$,
and that the limit $f(infty):=lim_{xtoinfty} f(x)$ exists
(as a finite quantity).
If $a>0$ and $b>0$, then the integral
begin{equation*}
int_{,0}^{,infty} frac{f(ax)-f(bx)}{x} dx tag{1}
end{equation*}
exists and is equal to $bigl(f(infty)-f(0)bigr),ln(a/b)$.

The assertion that the integral $(1)$ exists means
that the following limit exists,
begin{equation*}
lim_{lto0,,htoinfty} int_{,l}^{,h} frac{f(ax)-f(bx)}{x} dx
end{equation*}
where $l$ approaches $0$ independently of $h$ approaching $infty$,
and that this limit is the integral $(1)$.

Proof.$,$ Assume that $ageq b,$; this does not lose us any generality.
Let $0<l<h,$.
The change of variables $ax=by$ shows that
begin{equation*}
int_{,l/a}^{,h/a}frac{f(ax)}{x}dx
~=~ int_{,l/b}^{,h/b}frac{f(by)}{y}dy~,
end{equation*}
so we have
begin{align*}
int_{,l/a}^{,h/a}frac{f(ax)-f(bx)}{x}dx
&~=~ int_{,l/b}^{,h/b}frac{f(bx)}{x}dx
~-~ int_{,l/a}^{,h/a}frac{f(bx)}{x}dx \
&~=~ int_{,h/a}^{,h/b}frac{f(bx)}{x}dx
~-~ int_{,l/a}^{,l/b}frac{f(bx)}{x}dx~;
end{align*}
we write the difference in the second line as $I(h) - I(l)$.

Let $varepsilon>0$.
There exists $l_varepsilon>0$ such that
$f(0)-varepsilonleq f(bx)leq f(0)+varepsilon$
for $0leq xleq l_varepsilon/b$.
Since by assumption $l/aleq l/b$ for every $l>0$, we obtain the estimate
begin{equation*}
int_{,l/a}^{,l/b}frac{f(0)-varepsilon}{x}dx
~leq~ I(l)
~leq~ int_{,l/a}^{,l/b}frac{f(0)+varepsilon}{x}dx
qquadqquad text{for $0<lleq l_varepsilon$}~,
end{equation*}
that is,
begin{equation*}
bigl(f(0)-varepsilonbigr),lnfrac{a}{b}
~leq~ I(l)
~leq~ bigl(f(0)+varepsilonbigr),lnfrac{a}{b}
qquadqquad text{for $0<lleq l_varepsilon$}~.
end{equation*}
In other words, $I(l)$ converges to $f(0),ln(a/b)$ as $l$ approaches $0$.
In the same way we see that $I(h)$ converges to $f(infty)$
as $h$ approaches $infty$.$,$ Done.

Another generalization is as follows:
view formula

I proved it here.

Stackexchange won't let me post pictures, so here is the LaTeX version:

$$int_0^infty frac{qf(qx)}{g(qx)} - frac{pf(px)}{g(px)}dx
= int_0^infty frac{qf(qx)g(px)-pf(px)g(qx)} {g(qx)g(px)}dx =E(f)cdotlogfrac{q}{p}$$