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: 
Just integrate now, $\displaystyle\int e^{ax}\,dx = a^{-1}e^{ax}+C$, so apply this and get a common denominator to see

> 
> $$\int\_0^\infty {1\over 2i}\left({e^{ai-t}\over ai-t}+{e^{-ai-t}\over ai+t}\right|\_0^\infty = {1\over 2i}\left({1\over ai-t}+{1\over ai+t}\right)$$
>  $$={a\over t^2+a^2}$$
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**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.