i was filling some gaps on a proof my teacher made in class, to compute $$\int_0^\infty \frac{\sin x}{x} \, \mathrm{d}x.$$ The first he made was saying that this integral is equal to $$\lim_{\varepsilon \to 0^+} \int_0^\infty \frac{\varepsilon \sin x}{e^{\varepsilon x}-1} \, \mathrm{d}x,$$ and he said "okay, you can check that at home". I tried to apply Dominated Convergence Theorem in $$\lim_{\varepsilon \to 0^+} \int_0^\infty \frac{\sin x}{x}-\frac{\varepsilon \sin x}{e^{\varepsilon x}-1} \, \mathrm{d}x,$$ and show that it's zero, because I already know that $\frac{\sin x}{x}$ is not Lebesgue-integrable. So, I tried to bound the function inside the integral sign, but I'm not getting very sharp bounds (essentialy, I tried to apply MVT a couple times and get $$\frac{\sin x}{x}\xi e^{-\tau},$$ with $0<\tau<\xi<\varepsilon x$, but then I take absolute value and get something like $\varepsilon$). Do you have any idea/hint to bound this with an integrable function?
Edit: As Delta-u pointed out, my approach can't work. Do you have any idea to prove the equality?