About frullani integral

Question

Possible Duplicate:
Frullani proof integrals

Let $f:\left[ {0,\infty } \right] \to \mathbb R$ be a a continuous function such that $$ \mathop {\lim }\limits_{x \to0+ } f\left( x \right) = L $$Prove that $$ \int\limits_0^{\infty} {\frac{{f\left( {ax} \right) - f\left( {bx} \right)}} {x}}dx $$ converges and calculate the value.

It is known that $\int_a^\infty (f(x)/x)\,\mathrm{d}x$ converges for all a>0, but nothing of $\lim\limits_{x\to\infty}f(x)$ is told.

Also, what if $a>b$ or $a<b$?

Answer 1

The trick to evaluate the thing is, under suitable hypotheses to assure convergence and also differentiability in $a$ and/or $b$, to differentiate with respect to the parameter $a$. Letting $F(a)$ denote the integral (for fixed $b$), this gives $$F'(a)\;=\;\int_0^\infty xf'(ax)/x\,dx \;=\; \int_0^\infty f'(ax)\,dx \;=\; {1\over a}\cdot \int_0^\infty f'(x)\,dx \;=\; -f(0)\cdot {1\over a} $$ Thus, $F(a)=C-f(0)\cdot \log a\;$, and the integral is of the form $C-f(0)\cdot (\log a-\log b)\;$. Since the integral is visibly $0$ when $a=b$, it is $-f(0)\cdot (\log a - \log b)$.

In fact, use of "Frullani" in the question surprises me, because if one knows such integrals by this name one has a way to look in Whittaker-Watson, etc.

Edit: A careful proof that the integral converges has to look at $\int_\epsilon^T {f(ax)-f(bx)\over x}\;dx$ and show that the limit exists as $\epsilon\rightarrow 0^+$ and $T\rightarrow +\infty$. Justification of differentiation with respect to a parameter is somewhat subtler (although obviously necessary), and the "approved details" depend in a volatile way on your context.

Answer 2

Let $a,b>0$, without loss of generality we assume $a<b$. Let $0<\varepsilon<R$. First we split up the integral:

$$\int_{\varepsilon}^R \frac{f(ax)-f(bx)}{x} \, dx = \int_{\varepsilon}^R \frac{f(ax)}{x} \, dx- \int_{\varepsilon}^R \frac{f(bx)}{x} \, dx$$ where $$ \int_{\varepsilon}^R \frac{f(ax)}{x} \, dx \stackrel{z:=a \cdot x}{=} \int_{a \cdot \varepsilon}^{a \cdot R} \frac{f(z)}{\frac{z}{a}} \cdot \frac{1}{a} \, dz = \int_{a \cdot \varepsilon}^{a \cdot R} \frac{f(z)}{z} \, dz$$ (similarily for the second integral), thus

$$\int_{\varepsilon}^R \frac{f(ax)-f(bx)}{x} \, dx = \underbrace{\int_{a \varepsilon}^{b \varepsilon} \frac{f(z)}{z} \, dz}_{=:I_1} - \underbrace{\int_{a \cdot R}^{b \cdot R} \frac{f(z)}{z} \, dz}_{=:I_2}$$

1. We have $$I_1 = \int_{a \varepsilon}^{b \varepsilon} \frac{f(z)}{z} \, dz \stackrel{y:= \frac{z}{\varepsilon}}{=} \int_a^b \frac{f(\varepsilon \cdot y)}{y} \, dy$$ Since $a,b>0$ (thus $[a,b] \ni y \mapsto \frac{1}{y} \in L^1([a,b])$) and $f(\varepsilon \cdot y) \to L$ as $\varepsilon \to 0$ for all $y \in [a,b]$ we can apply dominated convergence and obtain $$I_1 \to \int_a^b \frac{L}{y} \, dy = L \cdot (\log b-\log a) \qquad (\varepsilon \to 0)$$
2. We want to prove $I_2 \to 0$ as $R \to \infty$. Let $\delta>0$. Define $$I_R := \int_a^R \frac{f(z)}{z} \, dz$$ Since $\int_a^{\infty} \frac{f(z)}{z} \, dz$ converges by assumption, we know that $I_R$ is a cauchy-sequence, i.e. there exists $S_0$ such that for all $S,T \geq S_0$: $$|I_S-I_T| \leq \delta \tag{1}$$ Now choose $R_0>0$ such that $a \cdot R_0 \geq S_0$. Then we obtain from (1) for all $R \geq R_0$: $$|I_2| = |I_{b \cdot R}-I_{a \cdot R}| \leq \delta$$ i.e. $I_2 \to 0$ as $R \to \infty$.

Adding all up we obtain

$$\int_0^\infty \frac{f(ax)-f(bx)}{x} \, dx = L \cdot (\log b- \log a) = f(0) \cdot (\log b-\log a)$$