10. $[\mathbf{8}]$ Compute
$$
\int_{0}^{\infty} \frac{e^{-x} \sin (x)}{x} d x
$$
Answer: $\frac{\pi}{4}$. We can compute the integral by introducing a parameter and exchanging the order of integration:
$$
\begin{aligned}
\int_{0}^{\infty} e^{-x}\left(\frac{\sin (x)}{x}\right) \mathrm{d} x & =\int_{0}^{\infty} e^{-x}\left(\int_{0}^{1} \cos (a x) \mathrm{d} a\right) \mathrm{d} x=\int_{0}^{1}\left(\int_{0}^{\infty} e^{-x} \cos (a x) \mathrm{d} x\right) \mathrm{d} a \\
& =\int_{0}^{1} \operatorname{Re}\left[\int_{0}^{\infty} e^{(-1+a i) x} \mathrm{~d} x\right] \mathrm{d} a=\int_{0}^{1} \operatorname{Re}\left[\left.\frac{e^{(-1+a i) x}}{-1+a i}\right|_{x=0} ^{\infty}\right] \mathrm{d} a \\
& =\int_{0}^{1} \operatorname{Re}\left[\frac{1}{1-a i}\right] \mathrm{d} a=\int_{0}^{1} \operatorname{Re}\left[\frac{1+a i}{1+a^{2}}\right] \mathrm{d} a \\
& =\int_{0}^{1} \frac{1}{1+a^{2}} \mathrm{~d} a=\left.\tan ^{-1}(a)\right|_{a=0} ^{1}=\frac{\pi}{4}
\end{aligned}
$$