3
$\begingroup$

How comes this true?

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

$\endgroup$
  • $\begingroup$ Do you know the complex exponential form of $\sin$? $\endgroup$ – Simply Beautiful Art Sep 25 '16 at 16:50
1
$\begingroup$

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 $M\to\infty$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.