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
 A: Hints:


*

*One has $\displaystyle\frac{1-\mathrm e^{-t}}t=\int_0^1\mathrm e^{-xt}\mathrm dx$

*For every real numbers $x$ and $t$, one has $\mathrm e^{-xt}\sin t=\Im(\mathrm e^{-(x-\mathrm i)t})$

*For every complex number $z$ such that $\Re z\gt0$, one has $\displaystyle\int_0^\infty\mathrm e^{-zt}\mathrm dt=\frac1z$

*For every real number $x$, one has $\displaystyle\frac1{x-\mathrm i}=\frac{x+\mathrm i}{x^2+1}$ hence $\displaystyle\Im\left(\frac1{x-\mathrm i}\right)=\frac1{x^2+1}$

*And finally, one has $\displaystyle\int_0^1\frac1{x^2+1}\mathrm dx=\frac\pi4$


Extension/Consequence: for every nonnegative $a$,
$$
\int_{0}^{\infty}\dfrac{1-\mathrm e^{-at}}{t}\sin{t}\,\mathrm dt=\arctan a$$
A: 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 $x=1$.
A: 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
$$
\int_0^\infty\frac{1-e^{-at}}{t}\sin(t)\,\mathrm{d}t=\arctan(a)
$$
A: 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:
$$ I = \int_{0}^{a}\frac{ds}{1+s^2}=\arctan(a).$$
