# Unusual substitution to evaluate $\int_{0}^{\infty}\frac{\sin x}{x}\mathrm{d}x$

I was trying to calculate the Dirichlet integral using Complex integration and I was trying out a different substitution from the one that is usually employed to solve the problem, i.e. $$\operatorname{f}\left(z\right) = {\rm e}^{{\rm i}z}/z$$, that is I was trying to use the Residue theorem on $$\sin\left(z\right)/z$$ on a half circumference centered at the origin, extrema on real axis and radius $$R \to \infty$$.

• The integral over the entire curve vanishes because the function is analytic on the whole complex plane ( we can consider the analytic extension of the function to remove the singularity at zero ) so I am left calculating $$\int_{0}^{\pi}{\rm i} \sin\left(R{\rm e}^{{\rm i}\theta}\right) {\rm d}\theta$$ which is the contribution of the contour integral from the half circumference.
• Here is my problem: in the limit $$R\to\infty$$ the integrand seems to diverge because by writing $$\sin\left(z\right) = \frac{{\rm e}^{{\rm i}z} - {\rm e}^{-{\rm i}z}}{2{\rm i}}$$ and expanding the terms in the exponents I get a term $${\rm e}^{R\sin\left(\theta\right)}\times$$(terms of modulus $$1$$) from $${\rm e}^{-{\rm i}z}$$ whereas the first term contributes to zero in the limit.

Obviously the integral has to be finite, so what am I doing wrong ?.

• Hi, welcome to Math SE. Hint: since the integrand is even, write your integral as$$\frac12\Im\int_{-\infty}^\infty\frac{e^{\mathrm{i}z}-1}{z}\mathrm{d}z.$$
– J.G.
Commented Jul 27 at 22:52

You correctly conclude that the integrand $$i\sin(R e^{i \theta})$$ becomes infinitely large for $$R \rightarrow \infty$$ and $$0 < \theta < \pi$$. However, in that limit the integrand is oscillating infinitely fast and this gives the integral a chance of having a finite limit. And indeed you have shown that
$$\int_0^\infty \frac{\sin x}{x} \mathrm{d} x = -\frac{1}{2}\lim_{R \rightarrow \infty} \int_0^\pi i\sin(R e^{i \theta}) \mathrm{d}\theta = \frac{1}{4} \lim_{R \rightarrow \infty} \int_0^\pi \exp(-i R e^{i \theta}) \mathrm{d}\theta \, .$$