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?
On
Consider the integral $$\int_0^\infty \frac{1-\cos t}{t} e^{-st} dt. $$ If we define $f(t)=1-\cos t$, this integral is the Laplace transform of $f(t)/t$. Moreover, we have the following identity for Laplace transforms:
$$\mathcal{L} \left\{ \frac{f(t)}{t} \right\}(s)=\int_s^\infty F(\sigma)d \sigma $$ where $F(\sigma)$ is the Laplace transform of $f(t)$, which in our case is known to be $$F(\sigma)=\frac{1}{\sigma}-\frac{\sigma}{\sigma^2+1}. $$ Thus we have $$\int_0^\infty \frac{1-\cos t}{t} e^{-st}=\int_s^\infty \left( \frac{1}{\sigma}-\frac{\sigma}{\sigma^2+1} \right) d \sigma= \log \sigma-\frac{1}{2} \log (\sigma^2+1) \big]^{\sigma=\infty}_{\sigma=s} .$$ Plugging in $s=1$ gives $$\int_0^\infty \frac{1-\cos t}{t} e^{-t} dt=\frac{1}{2} \log 2- \log 1=\frac{1}{2} \log 2 .$$
On
Another approach
Consider Laplace transform $$ \mathcal{L}\left[f(t)\right]=F(s)=\int_0^\infty f(t)\ e^{-st}\ dt $$ and property of the unilateral Laplace transform $$ \mathcal{L}\left[\frac{f(t)}{t}\right]=\int_s^\infty F(\omega)\ d\omega, $$ where $F(\omega)$ is Laplace transform of $f(t)$. We choose $f(t)=(1-\cos at)$ and it is easy to show that $$ \mathcal{L}\left[1-\cos at\right]=F(s)=\int_0^\infty (1-\cos at)\ e^{-st}\ dt=\frac{a^2}{s(s^2+a^2)}, $$ then $$\eqalign { \mathcal{L}\left[\frac{1-\cos at}{t}\right]&=\int_1^\infty \frac{a^2}{s(s^2+a^2)}\ ds\\ &=\int_1^\infty \left[\frac{1}{s}-\frac{s}{s^2+a^2}\right]\ ds\\ &=\left.\frac12\left[\ln s^2-\ln(s^2+a^2)\right]\right|_1^\infty\\ &=\left.\frac12\ln\left(\frac{s^2}{s^2+a^2}\right)\right|_1^\infty\\ &=\frac12\ln\left(1+a^2\right). } $$ Taking $a=1$ yields $$ \int_0^\infty \left[\frac{1-\cos t}{t}\right]\ e^{-t}\ dt=\large\color{blue}{\frac{\ln 2}{2}}. $$
On
Assume that $a>1$.
Using the Maclaurin series of the cosine function, we get
$$\begin{align} \int_{0}^{\infty} \frac{1-\cos t}{t} \, e^{-at} \, dt &= -\int_{0}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^{n}t^{2n}}{(2n)!} \frac{e^{-at}}{t} \, dt \\ &= \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{(2n)!} \int_{0}^{\infty}t^{2n-1}e^{-at} \, dt \\ &=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{(2n)!} \frac{(2n-1)!}{a^{2n}} \\ &= \frac{1}{2}\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \left( \frac{1}{a^{2}} \right)^{n} \\ &=\frac{1}{2} \,\ln \left(1+ \frac{1}{a^{2}} \right) \, , \end{align}$$ where interchanging the order of summation and integration is justified by Fubini's theorem.
Letting $a \to 1^{+}$, we get $$\int_{0}^{\infty} \frac{1-\cos t}{t} \, e^{-t} \, dt = \frac{\ln 2}{2}. $$
(See this question about justification for moving the limit inside the integral.)
On
Here's a solution using Frullani's integral.
$$\int_0^{\infty} \frac{1-\cos t}{t}e^{-t}\,dt=\Re\left(\int_0^{\infty} \frac{1-e^{it}}{t}e^{-t}\,dt\right)=\Re\left(\int_0^{\infty} \frac{e^{-t}-e^{-(1-i)t}}{t}\,dt\right)$$ $$=\Re\left(\ln\left(\frac{1-i}{1}\right)\right)=\Re\left(\ln\left(\sqrt{2}e^{i(-\pi/4+2k\pi)}\right)\right)=\boxed{\dfrac{\ln 2}{2}}$$
On
$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\infty}{1 - \cos\pars{t} \over t} \expo{-t}\,\dd t =\ {\large ?}}$
\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}
On
By double integral $$ \begin{aligned} \int_0^{\infty} \frac{1-\cos t}{t} e^{-t} d t = & \int_0^{\infty}\left(\int_0^1 \sin (t x) d x\right) e^{-t} d t \\ = & \int_0^1 \int_0^{\infty} e^{-t} \sin (t x) d t d x \\ = & \int_0^1 \frac{x}{x^2+1} d x \\ = & \frac{1}{2}\left[\ln \left(x^2+1\right)\right]_0^1 \\ = & \frac{1}{2} \ln 2 \end{aligned} $$
Consider $$\mathcal{I}(a)=\int_0^{\infty}\frac{1-\cos at}{t}e^{-t}dt.$$ Differentiating w.r.t. $a$, we find $$\mathcal{I}'(a)=\int_0^{\infty}\sin at\;e^{-t}dt=\frac{a}{1+a^2}.$$ Now integrating back and using that $\mathcal{I}(0)=0$ yields $$\mathcal{I}(a)=\int_0^a\frac{a\,da}{1+a^2}=\frac12\ln\left(1+a^2\right).$$ It remains to set $a=1$.