# $\int^\infty_0\frac{\sin x}{x} \, dx = \frac{1}{2i}\int^\infty_{-\infty} \frac{e^{ix}-1}{x} \, dx$, why?

How comes this true?

$$\int^\infty_0\frac{\sin x} x \, dx = \frac{1}{2i}\int^\infty_{-\infty} \frac{e^{ix}-1} x \, dx$$

• Do you know the complex exponential form of $\sin$? – Simply Beautiful Art Sep 25 '16 at 16:50

Note that $\int^\infty_{-\infty} \frac{e^{ix}-1}{x} \, dx$ is not well-defined, in fact $$\frac{e^{ix}-1} x=\frac{\cos(x)-1} x+\frac{i\sin(x)} x$$ $x\mapsto\frac{\sin(x)} x$ is an even function, therefore $$\frac{1}{2i}\int^\infty_{-\infty} \frac{i\sin(x)} x \, dx=\int^\infty_0\frac{\sin x} x \, dx$$ $x\mapsto\frac{\cos(x)-1} x$ is an odd function, but its integral is divergent, because $$\int^M_{0} \frac{\cos(x)-1} x \, dx=[\frac{\sin(x)-x} x]_{0}^{M}+\int^M_{0} \frac{\sin(x)-x} {x^2} \, dx=A-\int^M_{1}\frac{1} {x}dx$$ where A is finite but $\int^M_{1}\frac{1} {x}dx$ is divergent when $M\to\infty$
• are you sure $\int_{-\infty}^\infty \frac{\cos(x)-1}{x}dx$ is well-defined ? – reuns Sep 25 '16 at 19:54
• OP's formula is true in the $\int_{-\infty}^\infty = \lim_{A \to \infty} \int_{-A}^A$ sense – reuns Sep 25 '16 at 21:14