Proof of Frullani’s theorem

How can I prove the Theorem of Frullani? I did not even know all the hypothesis that $f$ must satisfy, but I think that this are

Let $\,f:\left[ {0,\infty } \right) \to \mathbb R$ be a a continuously differentiable function such that $$
\mathop {\lim }\limits_{x \to \infty } f\left( x \right) = 0,
$$
and let $
a,b \in \left( {0,\infty } \right)$.
Prove that $$
\int\limits_0^{\infty} {\frac{{f\left( {ax} \right) – f\left( {bx} \right)}}
{x}}dx = f\left( 0 \right)\left[ {\ln \frac{b}
{a}} \right]
$$
If you know a more general version please give it to me )= I can´t prove it.

Answer

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
$$\int_0^{+\infty}\dfrac{f(at)-f(tb)}tdt =
\int_0^{+\infty}\dfrac{g(at)-g(tb)}tdt =g(0)\ln\left(\dfrac ba\right) =
\left(f(0)-l\right)\ln\left(\dfrac ba\right).$$

Attribution
Source : Link , Question Author : August , Answer Author : Davide Giraudo

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