TITLE:  How can we evaluate $\int_{0}^{\infty}e^{-xt}\sin(t)dt$ without using Laplace transforms?
QUESTION [8 upvotes]: Without using Laplace transforms, how do I show that for every positive number $x$ the following equation is valid? $$\int_{0}^{\infty}e^{-xt}\sin(t)dt=\frac{1}{x^2+1}. $$

REPLY [9 votes]: 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}