Proof of Frullani’s Theorem

Frullani’s theorem evaluates a class of improper integrals in which two scaled copies of the same function are subtracted. The subtraction matters: each term on its own may produce a divergent integral near zero or infinity, while their difference can be integrable.

A convenient sufficient set of hypotheses gives a short proof and guarantees absolute convergence. The result also makes clear why the endpoint values of the function, rather than its detailed shape, determine the answer.

The theorem

Suppose $a,b>0$, and let $f$ be locally absolutely continuous on $(0,\infty)$. Assume that $f'\in L^1(0,\infty)$ and that the finite limits

$$f(0+)=\lim_{u\to 0+}f(u), \qquad f(\infty)=\lim_{u\to\infty}f(u)$$

exist. Then

$$\int_0^\infty \frac{f(ax)-f(bx)}{x}\,dx =\bigl(f(0+)-f(\infty)\bigr)\ln\!\left(\frac{b}{a}\right).$$

These are sufficient conditions, not the most general possible ones. Many versions of Frullani’s theorem use other hypotheses that ensure the integral exists. The conditions above are useful because they make the interchange of integrals in the proof straightforward.

Proof by varying the scale

First suppose $a<b$. For each $x>0$, apply the fundamental theorem of calculus to the function $t\mapsto f(tx)$:

$$f(ax)-f(bx)=-\int_a^b x f'(tx)\,dt.$$

Divide by $x$ and integrate over $x$. The change of variables $u=tx$ gives

$$\begin{aligned} \int_0^\infty \frac{f(ax)-f(bx)}{x}\,dx &=-\int_a^b\int_0^\infty f'(tx)\,dx\,dt\\ &=-\int_a^b \frac{1}{t}\left(\int_0^\infty f'(u)\,du\right)dt. \end{aligned}$$

The interchange of integrals is justified by absolute integrability. Indeed,

$$\int_a^b\int_0^\infty |f'(tx)|\,dx\,dt =\left(\int_a^b\frac{dt}{t}\right)\left(\int_0^\infty |f'(u)|\,du\right)<\infty.$$

Since $f'$ is integrable and $f$ has the stated endpoint limits, the fundamental theorem of calculus gives

$$\int_0^\infty f'(u)\,du=f(\infty)-f(0+).$$

Therefore,

$$\begin{aligned} \int_0^\infty \frac{f(ax)-f(bx)}{x}\,dx &=-\bigl(f(\infty)-f(0+)\bigr)\int_a^b\frac{dt}{t}\\ &=\bigl(f(0+)-f(\infty)\bigr)\ln\!\left(\frac{b}{a}\right). \end{aligned}$$

This proves the formula for $a<b$. If $a>b$, reversing the limits in the scale integral gives the same formula; if $a=b$, both sides are zero.

Worked example

Take $f(u)=1/(1+u)$. Its endpoint limits are $f(0+)=1$ and $f(\infty)=0$, so Frullani’s theorem predicts

$$\int_0^\infty \frac{1}{x}\left(\frac{1}{1+ax}-\frac{1}{1+bx}\right)dx =\ln\!\left(\frac{b}{a}\right).$$

For $0<a<b$, combine the fractions inside the integral:

$$\frac{1}{x}\left(\frac{1}{1+ax}-\frac{1}{1+bx}\right) =\frac{b-a}{(1+ax)(1+bx)}.$$

An antiderivative of the right-hand side is $\ln(1+bx)-\ln(1+ax)$. Evaluating from $0$ to $R$ and taking $R\to\infty$ gives

$$\lim_{R\to\infty}\left[\ln\!\left(\frac{1+bR}{1+aR}\right)\right] =\ln\!\left(\frac{b}{a}\right),$$

as predicted. The derivative $f'(u)=-1/(1+u)^2$ is integrable, so the proof’s hypotheses hold.

Common mistakes

• Reversing the logarithm: With the numerator $f(ax)-f(bx)$, the factor is $\ln(b/a)$, not $\ln(a/b)$. Swapping the two terms in the numerator changes the sign.
• Dropping the endpoint order: The multiplier is $f(0+)-f(\infty)$. For a decreasing function with a positive limit at zero and zero at infinity, it is positive.
• Splitting the integral: The separate integrals of $f(ax)/x$ and $f(bx)/x$ need not converge. The theorem applies to their combined difference; do not treat the two pieces as separately finite without additional justification.
• Ignoring convergence conditions: Finite endpoint limits alone do not justify every interchange of limits and integrals. Check a suitable version’s hypotheses before applying the formula.