Evaluating $\int_{0}^{\infty}\frac{1-e^{-t}}{t}\sin{t}\operatorname d\!t$ Find this integral
$$I=\int_{0}^{\infty}\dfrac{1-e^{-t}}{t}\sin{t}\operatorname d\!t$$
I know this
$$\int_{0}^{\infty}\dfrac{\sin{t}}{t}\operatorname d\!t=\dfrac{\pi}{2}$$But I can't find this value,Thank you

By Frullani's theorem
$$ \int_{0}^{+\infty}\frac{e^{it}-e^{(i-a)t}}{t}\,dt =\log(1+ia) $$
hence by taking the imaginary part:

$$ \int_{0}^{+\infty}\frac{1-e^{-at}}{t}\,\sin(t)\,dt = \arctan(a).$$

As an alternative, by using the Laplace transform we have:
$$ \mathcal{L}(\sin x)=\frac{1}{1+s^2},\qquad\mathcal{L}^{-1}\left(\frac{1-e^{-at}}{t}\right)= I_{s\leq a}(s) $$
hence:
$$ I = \int_{0}^{a}\frac{ds}{1+s^2}=\arctan(a).$$
