What is the value of the integral$\int_{0}^{+\infty} \frac{1-\cos t}{t} \, e^{-t} \, \mathrm{d}t$?

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 $a=1$.

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.)

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 $$\int_0^\infty \frac{1-\cos t}{t} e^{-t} dt=\frac{1}{2} \log 2- \log 1=\frac{1}{2} \log 2 .$$