Frullani 's theorem in a complex context.

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.

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

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

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!