# 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.

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

• +1. I like your solution. But you should replace lim m = lim M = 0 by lim m = lim M = f(0), to deduce that this part converges to f(0)log(b/a). // An extension is to assume that f has a limit at +oo, say f(+oo). Then your proof shows that the same result holds, with the limit (f(0)+f(+oo))log(b/a).
– Did
Sep 4, 2011 at 16:45
• @Didier: yes, it's of course $f(0)$ and I have corrected it. Maybe I should add the extension. Sep 4, 2011 at 16:49
• It is enough to assume that $f$ is continuous on $(0,\infty)$ and $\lim_{x\to 0+} f(x)=:f(0+)$ is finite (and of course, $f$ has a finite limit in $+\infty$). You can replace $f(0)$ by $f(0+)$ in the answer of @Giraudo. Aug 3, 2012 at 20:46
• @Davide Giraudo: awesome (+1) Aug 5, 2012 at 14:52
• I'm really stupid. Jan 21, 2019 at 2:34

The result is true under weaker assumptions than you state, but under your conditions, there is a cute proof using double integrals. (To be on the safe side, assume that $f$ is monotone, or at least that $f' \in L^1$. This will guarantee that we can change the order of integration.)

Let $D = \{ (x,y) \in \mathbb{R}^2 : x \ge 0, a \le y \le b \}$, and compute the integral $$\iint_D -f'(xy)\,dx\,dy$$ in two different ways.

Firstly \begin{align} \iint_D -f'(xy)\,dx\,dy &= \int_a^b \left( \int_0^\infty -f'(xy)\,dx \right)\,dy \\ &= \int_a^b \left[ \frac{-f(xy)}{y}\right]^\infty_0\,dy \\ &= \int_a^b \frac{f(0)}{y}\,dy = f(0)(\ln b - \ln a). \end{align}

On the other hand, \begin{align} \iint_D -f'(xy)\,dx\,dy &= \int_0^\infty \left( \int_a^b -f'(xy)\,dy \right)\,dx\\ &= \int_0^\infty \left[ \frac{-f(xy)}{x} \right]_a^b\,dx \\ &= \int_0^\infty \frac{f(ax)-f(bx)}{x}\,dx. \end{align}

• (+1), but why do you need $f$ to be monotone? It's enough for $g=f'$ to be continuous, which is what the OP wrote (continuously differentiable)
– Alex
Apr 17, 2015 at 9:04
• Can we use this method to find the integral: for $a <b$ $$\int_0^{\infty} \frac{\cos(ax)- \cos(bx)}{x} dx$$? Jan 9, 2020 at 16:25

There is a claim that is slightly more general.

Let $$f$$ be such that $$\int_a^b f$$ exists for each $$a,b>0$$. Suppose that $$A=\lim_{x\to 0^+}x\int_{x}^1 \frac{f(t)}{t^2}dt\\B=\lim_{x\to+\infty}\frac 1 x\int_1^x f(t)dt$$ exist.

Then $$\int_0^\infty\frac{f(ax)-f(bx)}xdx=(B-A)\log \frac ab$$

PROOF Define $$xg(x)=\displaystyle \int_1^x f(t)dt$$. Since $$g'(x)+\dfrac{g(x)}x=\dfrac{f(x)}x$$ we have $$\int_a^b \frac{f(x)}xdx=g(b)-g(a)+\int_a^b\frac{g(x)}xdx$$

Thus for $$T>0$$

$$\int_{Ta}^{Tb} \frac{f(x)}xdx=g(Tb)-g(Ta)+\int_{Ta}^{Tb}\frac{g(x)}xdx$$

But $$\int_{Ta}^{Tb}\frac{g(x)}xdx-B\int_a^b \frac{dx}x=\int_a^b\frac{g(Tx)-B}xdx$$

Thus $$\lim_{T\to+\infty}\int_{Ta}^{Tb}\frac{g(x)}xdx=B\log\frac ba$$ so

$$\lim_{T\to+\infty}\int_{Ta}^{Tb}\frac{f(x)}xdx=B\log\frac ba$$

It follows, since $$\int_1^T\frac{f(ax)-f(bx)}xdx=\int_{bT}^{aT}\frac{f(x)}xdx+\int_a^b \frac{f(x)}xdx$$ (note $$a,b$$ are swapped) that $$\int_1^\infty \frac{f(ax)-f(bx)}xdx=B\log\frac ab+\int_a^b \frac{f(x)}xdx$$

Let $$\varepsilon >0$$, $$\hat f(x)=f(1/x)$$. Then $$\int\limits_\varepsilon ^1 {\frac{{f\left( x \right)}}{x}dx} = \int\limits_1^{{\varepsilon ^{ - 1}}} {\frac{{\hat f\left( x \right)}}{x}dx}$$ and $$x\int\limits_x^1 {\frac{{f\left( t \right)}}{{{t^2}}}dt} = \frac{1}{{{x^{ - 1}}}}\int\limits_1^{{x^{ - 1}}} {\hat f\left( t \right)dt} = g\left( {{x^{ - 1}}} \right)$$

So $$\hat f(t)$$ is in the hypothesis of the preceding work. It follows that $$\lim_{T\to+\infty}\int\limits_1^T {\frac{{\hat f\left( {x{a^{ - 1}}} \right) - \hat f\left( {x{b^{ - 1}}} \right)}}{x}} dx = A\log \frac ba + \int\limits_{{a^{ - 1}}}^{{b^{ - 1}}} {\frac{{\hat f\left( x \right)}}{x}dx}$$

and by a change of variables $$x\mapsto x^{-1}$$ we get $$\int\limits_0^1 {\frac{{f\left( {ax} \right) - f\left( {bx} \right)}}{x}} dx = A\log \frac ba - \int\limits_a^b {\frac{{f\left( x \right)}}{x}dx}$$ and summing gives the desired $$\int\limits_0^\infty {\frac{{f\left( {ax} \right) - f\left( {bx} \right)}}{x}} dx = \left( {B - A} \right)\log \frac ab$$

This is due to T.M. Apostol.

OBS By L'Hôpital, if the limits at $$x=0^+$$ and $$x=+\infty$$ exist, they equal $$A$$ and $$B$$ respectively.

• Sorry for the (absurdly) late comment, but $g$ may not be differentiable, so your justification of $\int_{a}^{b}f(x)/x\,dx=g(b)-g(a)+\int_{a}^{b}g(x)/x\,dx$, I think, is not quite right. It can be made more accurate, though. We have $\int_{a}^{b} g(x)/x\,dx=\int_{a}^{b} x\,g(x)\,d(-1/x)$, the last integral being Riemann-Stieltjes. Now, by integration by parts, $\int_{a}^{b} g(x)/x\,dx=g(a)-g(b)+\int_{a}^{b}x^{-1} \,d(x\,g(x))$. Now, invoke Theorem 7.26 (Apostol's Analysis) to get $\int_{a}^{b} x^{-1}\,d(x\,g(x))=\int_{a}^{b} f(x)/x\,dx$. Jun 24, 2017 at 4:06
• @user149844 True. If $f$ is continuous then we're good.
– Pedro
Jun 24, 2017 at 4:09

The following theorem is a beautiful generalization of Frullani’s integral theorem.

Let $f(x)-f(\infty)=\sum_{k=0}^{\infty}\frac{u(k)(-x)^k}{k!}$ and $g(x)-g(\infty)=\sum_{k=0}^{\infty}\frac{v(k)(-x)^k}{k!}$

$Theorem1$:

Let f, and g be continuous function on $[0,\infty),$ assume that $f(0)=g(0)$ and $f(\infty)=g(\infty)$. Then if $a,b>0$

$$\lim_{n \to 0+}I_{n}\equiv \lim_{n \to 0+} \int_{0}^{\infty}x^{n-1}\lbrace f(ax)-g(bx) \rbrace dx=\lbrace f(0)-f(\infty)\rbrace \bigg \lbrace \log \bigg(\frac{b}{a} \bigg)+\frac{d}{ds}\bigg(\log\bigg(\frac{v(s)}{u(s)}\bigg) \bigg)_{s=0} \bigg \rbrace$$

if $f(x)=g(x),$ this theorem reduces to the Frullani’s theorem

$$\int_{0}^{\infty} \frac{f(ax)-f(bx)}{x}dx=\lbrace f(0)-f(\infty) \rbrace \log \bigg(\frac{b}{a} \bigg).$$

Let prove $Theorem1$, To do this we need to use Ramanujan's master theorem , Which lies in the fact that

$$\int_0^\infty x^{n-1}\sum_{k=0}^\infty \frac {\phi(k)(-x)^k}{k!}dx= \Gamma(n)\phi(-n).$$

Applying the Master Theorem with $0<n<1,$ we find

$$I_n=\int_{0}^{\infty} x^{n-1}( f(ax)-g(bx))dx=\int_{0}^{\infty} x^{n-1}( \lbrace f(ax)-f(\infty) \rbrace-\lbrace g(bx)-g(\infty) \rbrace) dx$$

$$=\Gamma(n)\lbrace a^{-n}u(-n)-b^{-n}v(-n) \rbrace$$

$$=\Gamma(n+1) \bigg \lbrace \frac{a^{-n}u(-n)-b^{-n}v(-n)}{n} \bigg \rbrace$$

Letting $n$ tend to $0$, using L'Hospital's Rule and fact that $u(0)=v(0)=f(0)-f(\infty).$ we deduce that

$$\lim_{n \to \infty}I_n=\lim_{n \to \infty} \bigg \lbrace \frac{b^nv(n)-a^nu(n)}{n} \bigg \rbrace$$

$$=\lim_{n \to \infty} \lbrace b^nv(n) \log b+ b^nv'(n)-a^nu(n)\log a-a^nu'(n)\rbrace$$

$$= \lbrace f(0)-f(\infty) \rbrace \log \bigg(\frac{b}{a} \bigg)+v'(0)-u'(0)$$

$$=\lbrace f(0)-f(\infty) \rbrace \bigg \lbrace \log \bigg(\frac{b}{a} \bigg)+\frac{d}{ds}\bigg(\log\bigg(\frac{v(s)}{u(s)}\bigg) \bigg)_{s=0} \bigg \rbrace$$

• This is incredible! May I ask is there a source for this? I'd like to know about some related analysis. Mar 9, 2018 at 11:29

You might be interested in an approach to Frullani's theorem I came across online. It is proven for the Lebesgue integral and the Denjoy-Perron integral. We are looking to prove the integral
\begin{equation*} \int^{\infty}_{0}\frac{f(ax)-f(bx)}{x}dx=A\ln(\frac{a}{b}) \end{equation*} where $A$ is a constant. For the Lebesgue integral, the substitution $x=e^t,~\alpha=\ln(\alpha),~\beta=\ln(b)$ is used to get \begin{equation*} \int^{+\infty}_{-\infty}\{ f(e^{t+\alpha})-f(e^{t+\beta})\}dt=A(\alpha-\beta) \end{equation*} which is equivalent to Frullani's theorem. Then verifying the integral \begin{equation*} \int^{+\infty}_{-\infty}\{ g(x+\alpha)-g(x+\beta)\}dx=A(\alpha-\beta) \end{equation*} for a Lebesgue integrable function $g:\mathbb{R}\to\mathbb{R}~\forall \alpha,\beta\in \mathbb{R}$ will suffice. This is proved by setting an integrable function on the real line \begin{equation*} h_{\alpha}(x)=g(x+\alpha)-g(x)~\forall\alpha\in\mathbb{R} \end{equation*} and applying the Fourier transform (as well as a little manipulation).

The Denjoy-Perron integral is used instead of the Lebesgue integral to avoid the problem of a locally integrable function $f:\mathbb{R}\to\mathbb{C}$ admitting a derivative $f'(x)~\forall x\in\mathbb{R}$ without $f'$ being locally integrable. The case for the Denjoy-Perron integral is proved in a similar fashion.

Check out the following paper by J. Reyna

http://www.ams.org/journals/proc/1990-109-01/S0002-9939-1990-1007485-4/S0002-9939-1990-1007485-4.pdf

On the assumption that $f$ is differentiable and $a,b>0$, an application of Fubini's theorem gives the result.

$$\int_{(0,\infty)} \frac{f(ax)-f(bx)}{x} dx$$

$$=\int_{(0,\infty)} \int_{[bx,ax]} f'(y) \frac{1}{x} dy dx$$

Let $0<bx \leq y \leq ax$.

$$=\int_{(0,\infty)} \int_{\frac{1}{a} y}^{\frac{1}{b} y} f'(y) \frac{1}{x} dx dy$$

$$=\int_{(0,\infty)} f'(y) \ln (\frac{a}{b}) dy$$

$$=(f(0)-f(\infty)) \ln \frac{b}{a}$$

• Can we use this method to find the integral: for $a < b$ $$\int_0^{\infty} \frac{\cos(ax)- \cos(bx)}{x} dx$$? Jan 9, 2020 at 17:03

The following is just a speeded-up version of the answer by Davide Giraudo.

Let $f(x)$ be a real-valued function defined for $x\geq 0$. Suppose that $f(x)$ is Riemann integrable on every bounded interval of nonnegative real numbers, that $f(x)$ is continuous at $x=0$, and that the limit $f(\infty):=\lim_{x\to\infty} f(x)$ exists (as a finite quantity). If $a>0$ and $b>0$, then the integral \begin{equation*} \int_{\,0}^{\,\infty} \frac{f(ax)-f(bx)}{x} dx \tag{1} \end{equation*} exists and is equal to $\bigl(f(\infty)-f(0)\bigr)\,\ln(a/b)$.

The assertion that the integral $(1)$ exists means that the following limit exists, \begin{equation*} \lim_{l\to0,\,h\to\infty} \int_{\,l}^{\,h} \frac{f(ax)-f(bx)}{x} dx \end{equation*} where $l$ approaches $0$ independently of $h$ approaching $\infty$, and that this limit is the integral $(1)$.

Proof.$\,$ Assume that $a\geq b\,$; this does not lose us any generality. Let $0<l<h\,$. The change of variables $ax=by$ shows that \begin{equation*} \int_{\,l/a}^{\,h/a}\frac{f(ax)}{x}dx ~=~ \int_{\,l/b}^{\,h/b}\frac{f(by)}{y}dy~, \end{equation*} so we have \begin{align*} \int_{\,l/a}^{\,h/a}\frac{f(ax)-f(bx)}{x}dx &~=~ \int_{\,l/b}^{\,h/b}\frac{f(bx)}{x}dx ~-~ \int_{\,l/a}^{\,h/a}\frac{f(bx)}{x}dx \\ &~=~ \int_{\,h/a}^{\,h/b}\frac{f(bx)}{x}dx ~-~ \int_{\,l/a}^{\,l/b}\frac{f(bx)}{x}dx~; \end{align*} we write the difference in the second line as $I(h) - I(l)$.

Let $\varepsilon>0$. There exists $l_\varepsilon>0$ such that $f(0)-\varepsilon\leq f(bx)\leq f(0)+\varepsilon$ for $0\leq x\leq l_\varepsilon/b$. Since by assumption $l/a\leq l/b$ for every $l>0$, we obtain the estimate \begin{equation*} \int_{\,l/a}^{\,l/b}\frac{f(0)-\varepsilon}{x}dx ~\leq~ I(l) ~\leq~ \int_{\,l/a}^{\,l/b}\frac{f(0)+\varepsilon}{x}dx \qquad\qquad \text{for $0<l\leq l_\varepsilon$}~, \end{equation*} that is, \begin{equation*} \bigl(f(0)-\varepsilon\bigr)\,\ln\frac{a}{b} ~\leq~ I(l) ~\leq~ \bigl(f(0)+\varepsilon\bigr)\,\ln\frac{a}{b} \qquad\qquad \text{for $0<l\leq l_\varepsilon$}~. \end{equation*} In other words, $I(l)$ converges to $f(0)\,\ln(a/b)$ as $l$ approaches $0$. In the same way we see that $I(h)$ converges to $f(\infty)$ as $h$ approaches $\infty$.$\,$ Done.

$$F(a) = \int_0^\infty \frac{f(ax)}x dx,$$ $$F'(a) = \int_0^\infty f'(ax) dx = \frac{f(\infty) - f(0)}{a}.$$

$$\Rightarrow F(a) = (f(\infty) - f(0)) \ln a + C_1.$$

Similarly

$$F(b) = (f(\infty) - f(0)) \ln b + C_2$$

Here $$C_1 = C_2$$ since basically function are the same. Subtracting gives

$$F(a) - F(b) = (f(\infty) - f(0)) \ln (a/b).$$

I think we can prove it by splitting the integral into two, treat each part as functions of a and b respectively, differentiate wrt a and b respectively (after this step finding the integral is trivial). This will be followed by the integration of the obtained functions wrt to a and b. The constants of integration will be the same as the function essentially remains the same, only the variables are changed. So on subtraction, the constants get cancelled, and hence the result! At least for an undergraduate, this should be fine.

• Seems I have to learn it. Mar 8, 2021 at 8:10
• I have edited your post, you can click "edit" and see what I did. Mar 8, 2021 at 18:34
• Thank you Arctic Char Mar 10, 2021 at 1:11

Another generalization is as follows: view formula

I proved it here.

Stackexchange won't let me post pictures, so here is the LaTeX version:

$$\int_0^\infty \frac{qf(qx)}{g(qx)} - \frac{pf(px)}{g(px)}dx = \int_0^\infty \frac{qf(qx)g(px)-pf(px)g(qx)} {g(qx)g(px)}dx =E(f)\cdot\log\frac{q}{p}$$

This is not an answer but motivated by the question Can we use this method to find the integral: for $$a $$\int_0^{\infty} \frac{\cos(ax)- \cos(bx)}{x} dx$$ by user008.

I try to use Laplace method to prove it. Since we have \begin{align} \mathcal{L_s}\left[\frac{f(t)}{t}\right]_{s=0}=\int_0^{\infty}F(s)ds \end{align} then \begin{align} \int_0^{\infty} \frac{\cos(ax)- \cos(bx)}{x} dx&=\int_0^{\infty}\left[\frac{s}{s^2+a^2}-\frac{s}{s^2+b^2}\right]ds\\ &=\frac12\int_0^{\infty}\frac{d(s^2+a^2)}{s^2+a^2}-\frac12\int_0^{\infty}\frac{d(s^2+b^2)}{s^2+b^2}\\ &=\left.\frac12\ln{\frac{s^2+a^2}{s^2+b^2}}\right|_0^\infty\\ &=\ln{\frac{b}{a}} \end{align}

However, when I continue to show the Frullani's theorem, I was stucked... \begin{align} \int\limits_0^\infty{\frac{{f\left({ax} \right) - f\left( {bx} \right)}}{x}} dx&=\int\limits_0^\infty\mathcal{L_s}[f(ax)]-\mathcal{L_s}[f(bx)]ds\\ &=\int\limits_0^\infty\int\limits_0^\infty\left(f(ax)e^{-sx}-f(bx)e^{-sx}\right)dxds\\ &=\int\limits_0^\infty\int\limits_0^\infty\left(f(u)e^{-su/a}-f(u)e^{-su/b}\right)dsdu\\ &=\int\limits_0^\infty f(u)\left(\frac{a^2-b^2}{u}\right)du\\ &=(a^2-b^2)\int\limits_0^\infty\frac{f(u)}{u}du\\ &\overset{?}=(a^2-b^2)\lim_{\epsilon\to0^+}\int_0^\infty\frac{f(\epsilon u)}{u}du~\ldots... \end{align}