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

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    $\begingroup$ +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). $\endgroup$ – Did Sep 4 '11 at 16:45
  • $\begingroup$ @Didier: yes, it's of course $f(0)$ and I have corrected it. Maybe I should add the extension. $\endgroup$ – Davide Giraudo Sep 4 '11 at 16:49
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    $\begingroup$ 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. $\endgroup$ – vesszabo Aug 3 '12 at 20:46
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    $\begingroup$ @Davide Giraudo: awesome (+1) $\endgroup$ – user 1591719 Aug 5 '12 at 14:52
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    $\begingroup$ I'm really stupid. $\endgroup$ – Faraz Masroor Jan 21 '19 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}

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    $\begingroup$ (+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) $\endgroup$ – Alex Apr 17 '15 at 9:04
  • $\begingroup$ Can we use this method to find the integral: for $a <b$ $$\int_0^{\infty} \frac{\cos(ax)- \cos(bx)}{x} dx$$? $\endgroup$ – user008 Jan 9 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.

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  • $\begingroup$ 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 $. $\endgroup$ – user149844 Jun 24 '17 at 4:06
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    $\begingroup$ @user149844 True. If $f$ is continuous then we're good. $\endgroup$ – Pedro Tamaroff Jun 24 '17 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 $g(x)-g(\infty)=\sum_{k=0}^{\infty}\frac{v(k)(-x)^k}{k!}$


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

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

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

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

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

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    $\begingroup$ This is incredible! May I ask is there a source for this? I'd like to know about some related analysis. $\endgroup$ – Lee David Chung Lin Mar 9 '18 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


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

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

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

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

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  • $\begingroup$ Can we use this method to find the integral: for $a < b$ $$ \int_0^{\infty} \frac{\cos(ax)- \cos(bx)}{x} dx$$? $\endgroup$ – user008 Jan 9 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.

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Another generalization is as follows: view formula

I proved it here.

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

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

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