TITLE: What is the value of the integral$\int_{0}^{+\infty} \frac{1-\cos t}{t} \, e^{-t} \, \mathrm{d}t$?
QUESTION [10 upvotes]: I have a question about evaluating

$$\int_{0}^{\infty} \frac{1-\cos t}{t}  \, e^{-t} \, \mathrm{d} t$$


Since $\lim_{t \to 0} \frac{1-\cos(t)}{t}  =0$, we know that the integrand is integrable near $t=0$.
But when evaluating the integral, how do you deal with the $t$ in the denominator of the integrand?

REPLY [5 votes]: $\newcommand{\+}{^{\dagger}}
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$\ds{\int_{0}^{\infty}{1 - \cos\pars{t} \over t} \expo{-t}\,\dd t =\ {\large ?}}$

\begin{align}
&\color{#44f}{\large\int_{0}^{\infty}{1 - \cos\pars{t} \over t} \expo{-t}\,\dd t}
=\Re\int_{0}^{\infty}\bracks{%
\expo{-t} - \expo{-\pars{1 + \ic}t}}\int_{0}^{\infty}\expo{-t\xi}\,\dd\xi\,\dd t
\\[3mm]&=\Re\int_{0}^{\infty}\int_{0}^{\infty}\bracks{%
\expo{-\pars{1 + \xi}t} - \expo{-\pars{\xi + 1 + \ic}t}}\,\dd t\,\dd\xi
=\Re\int_{0}^{\infty}\bracks{{1 \over \xi + 1} - {1 \over \xi + 1 + \ic}}\,\dd\xi
\\[3mm]&=\Re\ln\pars{1 + \ic} = \color{#44f}{\large\half\,\ln\pars{2}}
\end{align}