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

Since you know that
$$\int_0^\infty \frac{\sin t}tdt=\frac\pi2$$
so it suffices to find
$$\int_0^\infty\frac{e^{-t}}t\sin tdt$$
so let
$$f(x)=\int_0^\infty\frac{e^{-t}}t\sin (xt)dt=\int_0^\infty h(x,t)dt$$
so using Leibniz theorem and since
$$\left|\frac{\partial h}{\partial x}(x,t)\right|\le e^{-t}\in L^1((0,\infty))
$$
so we have
$$f'(x)=\int_0^\infty\cos(xt)e^{-t}dt=\frac1{1+x^2}$$
Now since $f(0)=0$ then we find
$$f(x)=\arctan x$$
and we have the desired result by taking $x=1$.
