integration - Frullani 's theorem in a complex context.

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.

Answer

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

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

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{\text{Re}(a\bar b)-|a|^2}{\text{Im}(a\bar b)}\right)-\arctan\left(\frac{|b|^2-\text{Re}(a\bar b)}{\text{Im}(a\bar 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}$$