How can we evaluate $\int_{0}^{\infty}e^{-xt}\sin(t)dt$ without using Laplace transforms? Without using Laplace transforms, how do I show that for every positive number $x$ the following equation is valid? $$\int_{0}^{\infty}e^{-xt}\sin(t)dt=\frac{1}{x^2+1}. $$
 A: Yet another one (particularly useful when evaluating particular integral of ODE):
\begin{align}
\int_{0}^{\infty}e^{-xt}\sin t\ dt&=\Im\left[\int_0^\infty e^{-xt}e^{it}\ dt\right]\\
&=\Im\left[\int_0^\infty e^{-(x-i)t}\ dt\right]\\
&=\Im\left[\frac1{x-i}\right]\\
&=\Im\left[\frac1{x-i}\cdot\frac{x+i}{x+i}\right]\\
&=\frac1{x^2+1}.
\end{align}
A: $\rm\bf Hint$:
$$\large e^{it}=\cos(t)+i\sin(t)\implies \sin(t)=\frac{e^{it}-e^{-it}}{2i}.$$
A: Why the "complex analysis" tag? It's just a double integration:
$$
\int e^{-xt}\sin(t)dt=-\frac{1}{x}e^{-xt}\sin(t)+\frac{1}{x}\int e^{-xt}\cos(t)dt=
$$
$$
=-\frac{1}{x}e^{-xt}\sin(t)-\frac{1}{x^2}e^{-xt}\cos(t)-\frac{1}{x^2}\int e^{-xt}\sin(t)dt
$$
It follows that 
$$
\left(1+\frac{1}{x^2}\right)\int e^{-xt}\sin(t)dt= -\frac{1}{x}e^{-xt}\sin(t)-\frac{1}{x^2}e^{-xt}\cos(t)+
$$
It follows that
$$
\left(1+\frac{1}{x^2}\right)\int_0^\infty e^{-xt}\sin(t)dt= \frac{1}{x^2}
$$
$$
\left(\frac{x^2+1}{x^2}\right)\int_0^\infty e^{-xt}\sin(t)dt= \frac{1}{x^2}
$$
$$
 \int_0^\infty e^{-xt}\sin(t)dt= \frac{1}{1+x^2}
$$
It follows that the integral is equal to $\displaystyle \frac{1}{x^2+1}$, as you wrote.
A: If you integrate by parts twice, using $u=\exp(-xt), dv=\sin(t)dt$ then $dv= \cos(t)dt$ you get the same integral back and can solve for it.
