Question: > 
> Let $t > 0$ and $a$ is a constant. Show that $$ \int\_{0}^{\infty} e^{-tx} \sin(ax) dx = \frac{a}{t^2 + a^2} $$
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The hint was to use the exponential formula, and I get that

$\sin(ax) = \frac{e^{iax} - e^{-iax}}{2i}$

and I replace it within the integral to get:

$$I = \frac{1}{2i} \cdot \int\_{0}^{\infty} e^{-tx}(e^{iax} - e^{-iax}) dx$$

But I cannot go further

Answer: 
An alternative approach. The sine function is a solution of the differential equation
$$ f''+f = 0 $$
and by applying a [Laplace transform](https://en.wikipedia.org/wiki/Laplace_transform) to both sides we get
$$ (\mathcal{L}f)(s) + s^2 (\mathcal{L} f)(s) = f'(0)+s\,f(0) $$
so that
$$ (\mathcal{L} f)(s) = \int\_{0}^{+\infty}\sin(x) e^{-sx}\,dx = \frac{1}{s^2+1} $$
and the claim follows from a straightforward substitution.
**Hint:**

It's simpler to use the $\mathrm e^{-tx}\sin ax=\operatorname{Im}\mathrm e^{(-t+ia)x}$. The antiderivative of this function (w.r.t. $x$) is
$$\frac1{-t+ia}e^{(-t+ia)x}=-\frac{t+ia}{t^2+a^2}\,e^{-tx}e^{iax},$$
and all you have to do is to finish extracting its imaginary part.