Show that $\int_0^{\infty}\frac{f(ax)-f(bx)}{x}dx=[f(0)-L]\ln\frac{b}{a}$ Let $f:[0,\infty)\to\mathbb{R}$ be continuous and $\lim_{x\to\infty}f(x)=L$. Show that
$$\int_0^{\infty}\frac{f(ax)-f(bx)}{x}dx=[f(0)-L]\ln\frac{b}{a}$$
where $0<a<b$.
I don't even know how to start.
 A: It is straightforward to derive the Frullani integral formula on a finite interval 
$$\int_0^1 \frac{f(ax)-f(bx)}{x} \, dx  =f(0) \ln \frac{b}{a} -\int_a^b \frac{f(x)}{x} \, dx,$$
by splitting the integral.
Note that (using the continuity of $f$ and the IVT for integrals)
$$\int_0^1 \frac{f(ax)-f(bx)}{x} \, dx = \lim_{\epsilon \rightarrow 0}\left[\int_{\epsilon}^1\frac{f(ax)}{x}\, dx- \int_{\epsilon}^1\frac{f(bx)}{x}\,dx\right]\\=\lim_{\epsilon \rightarrow 0}\left[\int_{a\epsilon}^a\frac{f(t)}{t}\, dt- \int_{b\epsilon}^b\frac{f(t)}{t}\,dt\right]\\=\lim_{\epsilon \rightarrow 0}\int_{a\epsilon}^{b\epsilon}\frac{f(t)}{t} \, dt- \int_{a}^{b}\frac{f(t)}{t} \, dt\\=\lim_{\epsilon \rightarrow 0}f(\xi)\int_{a\epsilon}^{b\epsilon}\frac{1}{t} \, dt- \int_{a}^{b}\frac{f(t)}{t} \, dt\\=f(0) \ln \frac{b}{a} -\int_a^b \frac{f(x)}{x} \, dx.$$
Then use the substitution $x = Ru$ to get
$$\int_0^R\frac{f(ax)-f(bx)}{x} \, dx=\int_0^1\frac{f(aRu)-f(bRu)}{u} \, du\\=f(0) \ln \frac{bR}{aR}-\int_{aR}^{bR} \frac{f(u)}{u} \, du.$$
Using the IVT for integrals, there is a point $\xi$ between $aR$ and $bR$ such that
$$\int_0^R\frac{f(ax)-f(bx)}{x} \, dx= f(0)\ln \frac{b}{a}-f(\xi)\int_{aR}^{bR} \frac{1}{u} \, du=[f(0) - f(\xi)]\ln \frac{b}{a}.$$
Take the limit as $R \rightarrow \infty$, to obtain
$$\int_0^{\infty}\frac{f(ax)-f(bx)}{x} \, dx=[f(0) - L]\ln \frac{b}{a}.$$
A: Given the definition of improper integrals, it makes sense to look at
$$\int_{\epsilon}^R {f(ax) - f(bx) \over x}\,dx$$
$$= \int_{\epsilon}^R {f(ax) \over x}\,dx - \int_{\epsilon}^R {f(bx) \over x}\,dx$$
Changing variables to $ax$ in the first integral and $bx$ in the second, this becomes
$$ \int_{a\epsilon}^{aR} {f(x) \over x}\,dx - \int_{b\epsilon}^{bR} {f(x) \over x}\,dx$$
$$= \int_{a\epsilon}^{b\epsilon} {f(x) \over x}\,dx - \int_{aR}^{bR} {f(x) \over x}\,dx $$
The first integral can be written as 
$$\int_{a\epsilon}^{b\epsilon} {f(0) \over x}\,dx + \int_{a\epsilon}^{b\epsilon} {f(x) - f(0)\over x}\,dx$$
$$= f(0)\ln(b/a) + \int_{a\epsilon}^{b\epsilon} {f(x) - f(0)\over x}\,dx$$
Since $f(x)$ is continuous at $x = 0$, the second term goes to zero as $\epsilon$ goes to zero and therefore the first integral goes to $f(0)\ln(b/a)$.
Similarly, since $\lim_{x \rightarrow \infty} f(x) = L$, the second integral converges 
to $L\ln(b/a)$ as $R \rightarrow \infty$. Hence we have
$$\int_{0}^{\infty} {f(ax) - f(bx) \over x}\,dx = (f(0) - L)\ln(b/a)$$
A: The key observation here is that the integrand itself looks like the value of a definite integral evaluated at endpoints $a$ and $b$. So, we first suppose $f$ is differentiable, and we undo the integration, which FTC tells us is essentially differentiation:
$$\int_0^{\infty} \frac{f(ax) - f(bx)}{x} dx = \int_0^{\infty} \int_b^a f'(xy) \,dy\,dx.$$

 Then, reverse the order of integration, after which the inner integral becomes $$\int_0^{\infty} f'(xy) \,dy = \left.\frac{1}{y} f(xy)\right\vert_0^{\infty}.$$

Next, we can eliminate the differentiability hypothesis by estimating the integral, see my comment below for a rough sketch of how you might achieve this.
