TITLE: Frullani 's theorem in a complex context.
QUESTION [21 upvotes]: 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.

REPLY [4 votes]: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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}