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

Integration by parts twice yields
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}a}\int_0^\infty\frac{1-e^{-at}}{t}\sin(t)\,\mathrm{d}t
&=\int_0^\infty e^{-at}\sin(t)\,\mathrm{d}t\\
&=\frac1a\int_0^\infty e^{-at}\cos(t)\,\mathrm{d}t\\
&=\frac1{a^2}-\frac1{a^2}\int_0^\infty e^{at}\sin(t)\,\mathrm{d}t\\
&=\frac1{1+a^2}
\end{align}
$$
Noting that when $a=0$ the original integral is $0$, we get
$$
\int_0^\infty\frac{1-e^{-at}}{t}\sin(t)\,\mathrm{d}t=\arctan(a)
$$
