# Show that $\int_0^{\infty}\frac{f(ax)-f(bx)}{x}dx=[f(0)-L]\ln\frac{b}{a}$ [duplicate]

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.

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

$$\int_0^{\infty}\frac{f(ax)-f(bx)}{x} \, dx=[f(0) - L]\ln \frac{b}{a}.$$

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 $$\int_{0}^{\infty} {f(ax) - f(bx) \over x}\,dx = (f(0) - L)\ln(b/a)$$

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.

• But here $f$ is only assumed to be continuous, not necessarily differentiable Dec 1, 2014 at 5:37
• Ah, I didn't see that; one can probably estimate the integral arbitrarily closely by splitting into an improper integral over some interval $[x(\epsilon), \infty)$ that you can bound by some given $\epsilon$ and a proper integral over $[0, x(\epsilon)]$, which you can estimate arbitrarily closely with smooth functions. Dec 1, 2014 at 5:44