It is possible to prove that $$\int_{0}^{\infty}\frac{e^{-ix}-e^{-x}}{x}dx=-i\frac{\pi}{2}$$ and in this case the Frullani's theorem does not hold since, if we consider the function $f(x)=e^{-x}$, we should have $$\int_{0}^{\infty}\frac{e^{-ax}-e^{-bx}}{x}dx$$ where $a,b>0$. But if we apply this theorem, we get $$\int_{0}^{\infty}\frac{e^{-ix}-e^{-x}}{x}dx=\log\left(\frac{1}{i}\right)=-i\frac{\pi}{2}$$ which is the right result.

Questions: is it only a coincidence? Is it possible to generalize the theorem to complex numbers? Is it a known result? And if it is, where can I find a proof of it?

Thank you.

  • 2
    $\begingroup$ with your particular case $f(z) = e^{-z}$ I'd do this : once you proved $\int_0^\infty (ax)^{s-1} e^{-ax} d(ax) = \Gamma(s)$ for any $Re(a) > 0$, you get by continuity that it is still true with $a = \pm i$ whenever $Re(s) \in (0,1)$ and $\int_0^\infty x^{s-1} e^{-ix} dx = e^{-i \pi s/2} \Gamma(s)$. now the same argument applies whenever $f(z)$ is analytic and exponentially decreasing on $Re(z) > 0$ as $ |z| \to \infty$, and that $\int_0^\infty x^{s-1} f( \pm ix) dx$ converges. in that case the Frullani's theorem should be still true with $a,b$ complex in the $Re(z) \ge 0$ plane $\endgroup$
    – reuns
    Jun 1 '16 at 20:36
  • $\begingroup$ The complex version is present as an exercise in the classical book by Whittaker and Watson. $\endgroup$
    – mickep
    May 28 '17 at 14:09
  • $\begingroup$ @mickep Thank you, I didn't know it. $\endgroup$ May 29 '17 at 15:49

The following development provides a possible way forward to generalizing Frullani's Theorem for complex parameters.

Let $a$ and $b$ be complex numbers such that $\arg(a)\ne \arg(b)+n\pi$, $ab\ne 0$, and let $\epsilon$ and $R$ be positive numbers.

In the complex plane, let $C$ be the closed contour defined by the line segments (i) from $a\epsilon$ to $aR$, (ii) from $aR$ to $bR$, (iii) from $bR$ to $b\epsilon$, and (iv) from $b\epsilon$ to $a\epsilon$.

Let $f$ be analytic in and on $C$ for all $\epsilon$ and $R$. Using Cauchy's Integral Theorem, we can write

$$\begin{align} 0&=\oint_{C}\frac{f(z)}{z}\,dz\\\\ &=\int_\epsilon^R \frac{f(ax)-f(bx)}{x}\,dx\\\\ &+\int_0^1 \frac{f(aR+(b-a)Rt)}{a+(b-a)t}\,(b-a)\,dt\\\\ &-\int_0^1 \frac{f(a\epsilon+(b-a)\epsilon t)}{a+(b-a) t}\,(b-a)\,dt\tag1 \end{align}$$

Rearranging $(1)$ reveals that

$$\begin{align} \int_\epsilon^R \frac{f(ax)-f(bx)}{x}\,dx&=\int_0^1 \frac{f(a\epsilon+(b-a)\epsilon t)}{a+(b-a) t}\,(b-a)\,dt\\\\ &-\int_0^1 \frac{f(aR+(b-a)Rt)}{a+(b-a)t}\,(b-a)\,dt \tag 2 \end{align}$$

If $\lim_{R\to \infty}\int_0^1 \frac{f(aR+(b-a)Rt)}{a+(b-a)t}\,(b-a)\,dt=0$, then we find that

$$\begin{align} \int_0^\infty \frac{f(ax)-f(bx)}{x}\,dx&=f(0)(b-a)\int_0^1\frac{1}{a+(b-a)t}\,dt\\\\ &=f(0)\log(|b/a|)\\\\ &+if(0)\left(\arctan\left(\frac{|b|^2-\text{Re}(\bar a b)}{\text{Im}(\bar a b)}\right)-\arctan\left(\frac{\text{Re}(\bar a b)-|a|^2}{\text{Im}(\bar a b)}\right)\right) \tag 3 \end{align}$$

Since $(a-b)\int_0^1 \frac{1}{a+(b-a)t}\,dt$, $ab\ne 0$ is continuous in $a$ and $b$, then $(3)$ is valid for $\arg(a)=\arg(b)+n\pi$ also.

Note that the tangent of the term in large parentheses on the right-hand side of $(3)$ is . $$\begin{align} \frac{\text{Im}(\bar a b)}{\text{Re}(\bar a b)}&=\tan\left(\arctan\left(\frac{ |b|^2-\text{Re}(\bar a b)}{\text{Im}(\bar a b)}\right)-\arctan\left(\frac{\text{Re}(\bar a b)-|a|^2}{\text{Im}(\bar a b)}\right)\right)\\\\ &=\tan\left(\arctan\left(\frac{\text{Im}(b)}{\text{Re}(b)}\right)-\arctan\left(\frac{\text{Im}(a)}{\text{Re}(a)}\right)\right) \end{align}$$

  • 1
    $\begingroup$ I get this for the $$\int_{(a \epsilon , b \epsilon)} \frac{f(z)}{z} dz = \int_0^1 \frac{f(a \epsilon + (b\epsilon-a\epsilon) t))}{a \epsilon + (b\epsilon-a\epsilon) t} d(a \epsilon + (b\epsilon-a\epsilon) t) = \int_0^1 \frac{f(\epsilon(a + (b-a) t)}{a + (b-a) t} (b-a)dt $$ and as $\epsilon \to 0$, assuming $f(z)$ is continuous at $z=0$, it $$\to f(0)\int_0^1 \frac{b-a}{a + (b-a) t} dt = f(0)\int_0^1 \frac{1}{\frac{a}{b-a} + t} dt = \log(\frac{a}{b-a} + t)|_0^1$$ $$ = \log(\frac{a}{b-a} + 1)- \log(\frac{a}{b-a}) = \log(1+\frac{b-a}{a}) = \log(\frac{b}{a})$$ $\endgroup$
    – reuns
    Jun 2 '16 at 2:58
  • 1
    $\begingroup$ and in the same way, if $f(z) \to C$ as $|z| \to \infty$ in the corresponding part of the complex plane, the $\int_{(aR,bR)} \frac{f(z)}{z} dz \to C \log(\frac{b}{a})$, so Frullani's theorem would be still true whenever $f(z)$ is analytic on $\Omega = \{ z \ \mid \ arg(z) \in [\theta_1,\theta_2], |z| \ge 0\}$ and that $\lim_{|z| \to \infty} f(z) = C$ on $\Omega$ and that $a,b \in \Omega \setminus \{0\}$ $\endgroup$
    – reuns
    Jun 2 '16 at 3:04
  • 1
    $\begingroup$ @user1952009 The result I obtained is $\log(b/a)$ if $a$ and $b$ are restricted. But if the line segment from $b$ to $a$ crosses the branch cut for the principal branch of the complex logarithm then one needs to add or subtract $i \pi$. And the same result does hold if $\lim_{R\to \infty}\left(\cdot \right)=C$. $\endgroup$
    – Mark Viola
    Jun 2 '16 at 3:12
  • 1
    $\begingroup$ all we need is that $\log(z)$ to be continuous on $\frac{a}{b}-1+t, t \in [0,1]$ i.e. on $\frac{a}{b}-t, t \in [0,1]$, and this is the case whenever $\frac{a}{b} \not \in \mathbb{R}$ with the usual branch of the logarithm ?? that works if $Re(a/b) > 1$ I think $\endgroup$
    – reuns
    Jun 2 '16 at 3:24
  • $\begingroup$ It is really impressive! Thank you both!!! $\endgroup$ Jun 2 '16 at 7:04

I think you may simply consider $$ f(\alpha) = \int_{0}^{+\infty}\frac{e^{-\alpha x}-e^{-x}}{x}\,dx $$ as a complex variable function with the assumption $\text{Re}(\alpha)>0$. Then: $$ f'(\alpha) = -\int_{0}^{+\infty}e^{-\alpha x}\,dx =-\frac{1}{\alpha} $$ and $f(1)=0$, so

$$ f(\alpha) = -\int_{1}^{\alpha}\frac{dz}{z}.$$

Since $\text{Re}(\alpha)>0$, the last complex integral is well defined, and you may define $\text{Re}\,f(\alpha)$ over $\left\{\text{Re}(z)\geq 0\right\}\setminus 2\pi i \mathbb{Z}$ by analytic continuation, since $\text{Re}\log\alpha = \log\|\alpha\|$. We also have $f(\alpha)=f(\bar{\alpha})$ by the Schwarz' reflection principle and $$ f(\alpha)=-f\left(\frac{1}{\alpha}\right) $$ by the obvious substitution. Another chance is given by the well-known lemma $$ \int_{0}^{+\infty}f(x)\frac{dx}{x} = \int_{0}^{+\infty}\mathcal{L}(f)(s)\,ds, $$ but we have to be careful with that, since in our case we are considering a Laplace transform on the boundary of its convergence domain.

The Cantarini-Frullani's theorem has just born :D

  • 2
    $\begingroup$ (+1) Nice, thanks. Even if I'm not interested for this particular integral, but when this type of identity holds. For example if we take $f(x)=\arctan(x)$ the mechanism seems to work. Maybe it follows from what you write. I'm tired now so even if someone write to me that it works because this is the will of the witch Baba Yaga I would have difficult to starting a discussion. Thank you again. $\endgroup$ May 31 '16 at 21:08
  • $\begingroup$ Jack, it is even easier than this, I believe - a simple application of Cauchy's Integral Theorem as in the answer I posted. -Mark $\endgroup$
    – Mark Viola
    May 31 '16 at 21:20

Note that from Cauchy's Integral Theorem

$$\oint_C \frac{e^{-iz}}{z}\,dz=0 \tag 1$$

where $C$ is the closed contour comprised of (i) the line segment from $\epsilon>0$ to $R$, (ii) the quarter circle of radius $R$ centered at the origin from $R$ to $-iR$, (iii) the line segment from $-iR$ to $-i\epsilon$, and (iv) the quarter circle of radius $\epsilon$ centered at the origin from $-i\epsilon$ to $\epsilon$.

We can write $(2)$ as

$$\begin{align}\oint_C \frac{e^{-iz}}{z}\,dz&=\int_\epsilon^R \frac{e^{-ix}}{x}\,dx+\int_R^\epsilon \frac{e^{-y}}{-iy}\,(-i)\,dy\\\\ &+\int_0^{-\pi/2}\frac{e^{iRe^{i\phi}}}{Re^{i\phi}}\,iRe^{i\phi}\,d\phi\\\\ &+\int_{-\pi/2}^0\frac{e^{i\epsilon e^{i\phi}}}{\epsilon e^{i\phi}}\,i\epsilon e^{i\phi}\,d\phi\tag 2 \end{align}$$

As $R\to \infty$, the third integral on the right-hand side of $(2)$ approaches zero. As $\epsilon \to 0$, the fourth integral on the right-hand side of $(2)$ approaches $i\pi/2$. Thus, we see that

$$\int_0^\infty \frac{e^{-ix}-e^{-x}}{x}\,dx=-i\pi/2 \tag 3$$

as was to be shown!

  • $\begingroup$ @JackD'Aurizio The imaginary part of the LHS is $$-\int_0^\infty \frac{\sin(x)}{x}\,dx$$which of course does converge to $\pi/2$. ;-)) $\endgroup$
    – Mark Viola
    May 31 '16 at 22:15
  • $\begingroup$ Yes, I am an idiot :D $\endgroup$ May 31 '16 at 22:21
  • 2
    $\begingroup$ @JackD'Aurizio Jack, if you're an idiot, then what does that say about the average human? I think I'm an idiot - and might be - especially when I know there were/are people like Euler, Gauss, Dirac, Einstein, et. al. whose intellects dwarf most others. You're one of the most creative folks on this site! -Mark $\endgroup$
    – Mark Viola
    May 31 '16 at 22:38
  • $\begingroup$ (+1) Thank you, but as I wrote to Jack I'm not interested to this particular integral. I already know how to find this closed form. Probably it is my fault, I could write the question in a better way. I would know when we can use the Frullani's theorem with complex parameters instead of positive parameters. For example if we take $f(x)=\arctan(x)$ the theorem holds. So it seems that if the integral converges and if exists $f(0)$ and $f(\infty)$ then the theorem works also for complex parameters. $\endgroup$ Jun 1 '16 at 6:01
  • $\begingroup$ Marco, I've added another answer that might be closer to the generalization that you seek. -Mark $\endgroup$
    – Mark Viola
    Jun 1 '16 at 20:35

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\int_{0}^{\infty}{\expo{-\ic x} - \expo{-x} \over x}\,\dd x = \int_{0}^{\infty}\pars{\expo{-\ic x} - \expo{-x}} \int_{0}^{\infty}\expo{-xt}\,\dd t\,\dd x \\[5mm] = &\ \int_{0}^{\infty}\int_{0}^{\infty} \bracks{\expo{-\pars{t + \ic}x} - \expo{-\pars{t + 1}x}}\dd x\,\dd t = \int_{0}^{\infty} \pars{{1 \over t + \ic} - {1 \over t + 1}}\dd t = \left.\ln\pars{t + \ic \over t + 1}\right\vert_{\ t\ =\ 0}^{\ t\ \to\ \infty} \\[5mm] = &\ -\ln\pars{\ic} =\ \bbox[#ffe,10px,border:1px dotted navy]{\ds{-\,{\pi \over 2}\,\ic}} \end{align}

  • $\begingroup$ Excuse me sir. Sorry for commenting here after a long time but what were you using or what theorem is this in which you were able to turn a single integral into a double one? $\endgroup$
    – user635988
    Oct 11 '21 at 0:55

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