So, I want to find the value of $\int\limits_{x=0}^{\infty} \frac{\sin x}{x} dx$.
Let $I(a)=\int\limits_{x=0}^{\infty} \frac{e^{ix}}{x} e^{ax} dx$. So $\Im(I(0))$ will be the answer. $$\frac{\mathrm{d}I(b)}{\mathrm{d}b}=\int\limits_{x=0}^{\infty} e^{ix} e^{ax} dx=\int\limits_{x=0}^{\infty} e^{x(i+a)} dx = \left[ \frac{1}{a+i} e^{x(a+i)} \right]_{x=0}^{\infty}$$ And here comes the limitation: $a$ must be less than 0 to be able to calculate it, because for $a>0$ it explodes, for $a=0$ it oscillates. $$\frac{\mathrm{d}I(b)}{\mathrm{d}b}= \left[ \frac{1}{a+i} e^{ax}e^{ix} \right]_{x=0}^{\infty}=\frac{-1}{a+i}=\frac{i-a}{a^2+1}$$ Now let's only integrate the imaginary part, since that's what we need: $$\Im(I(a))=\tan^{-1}{a}+C$$ $$\Im(I(0))=\tan^{-1}{0}+C=C$$ Because $\lim\limits_{a \to -\infty} \Im(I(a))=0=\lim\limits_{a \to -\infty} \tan^{-1}{a}+C=-\frac{\pi}{2}+C$, the original integral's value is $\frac{\pi}{2}$.
So, why could I use $a=0$?