How comes this true?

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

  • $\begingroup$ Do you know the complex exponential form of $\sin$? $\endgroup$ – 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 $M\to\infty$

  • $\begingroup$ are you sure $\int_{-\infty}^\infty \frac{\cos(x)-1}{x}dx$ is well-defined ? $\endgroup$ – reuns Sep 25 '16 at 19:54
  • $\begingroup$ @user1952009 No problem at 0, integration by parts at infinity. $\endgroup$ – Aforest Sep 25 '16 at 20:00
  • $\begingroup$ @user1952009 Well...I think it cannot be well-defined, you are right. $\endgroup$ – Aforest Sep 25 '16 at 20:13
  • $\begingroup$ OP's formula is true in the $\int_{-\infty}^\infty = \lim_{A \to \infty} \int_{-A}^A$ sense $\endgroup$ – reuns Sep 25 '16 at 21:14

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