0
$\begingroup$

I tried to calculate the $\int_{-\infty}^{\infty}\frac{\sin{x}}{x}\mathrm{d}x$ by the complex integral $\oint_{\gamma}\frac{\sin{z}}{z}\mathrm{d}z$, where $\gamma$ is the half-circle with $y > 0$.

$$\begin{align*} \int_{-\infty}^{\infty}\frac{\sin{x}}{x}\mathrm{d}x &= \Im\left \{\oint_{\gamma}\frac{\mathrm{e}^{iz}}{z}\mathrm{d}z \right \}=\Im\left \{ 2\pi i Res\left [ \frac{\mathrm{e}^{iz}}{z}, 0 \right ] \right \}=\Im\left \{ 2\pi i \lim_{z\to0} \left ( z \frac{\mathrm{e}^{iz}}{z} \right ) \right \} \\ &= \Im\left \{ 2\pi i \lim_{z\to0} \mathrm{e}^{iz} \right \}=\Im\left \{ 2\pi i {e}^{i0} \right \}=\Im\left \{ 2\pi i \cdot 1 \right \}=2\pi \end{align*}$$ Now, I know the integral $\int_{-\infty}^{\infty}\frac{\sin{x}}{x}\mathrm{d}x = \pi$ by many other theorems. So, what did I do wrong?

If you know, can you please write the correct answer? Thanks!

$\endgroup$
  • $\begingroup$ Drop that annoying $\;\Im\;$ , do the integral and just write down $\;\Im\;$ at the end. $\endgroup$ – DonAntonio Apr 9 '18 at 11:43
  • $\begingroup$ Your question has no meaning at all if you don't specify clearly what the integration path $\;\gamma\;$ is... $\endgroup$ – DonAntonio Apr 9 '18 at 11:44
  • $\begingroup$ @DonAntonio it's doesn't metter, It will still be a $2\pi$, and the $\gamma$ is a standart letter for the half circle of the xy plane where y > 0. $\endgroup$ – Ziv Sdeor Apr 9 '18 at 11:46
  • 2
    $\begingroup$ No, of course not. It won't still be $\;2\pi\;$ at all. Read my answer. And again: everything depends on what path you choose! Without specifying that a complex integral has no meaning at all. $\endgroup$ – DonAntonio Apr 9 '18 at 11:55
  • 6
    $\begingroup$ "the $\gamma$ is a standart letter for the half circle of the xy plane where y > 0." You may be overgeneralizing from a not-very-large set of examples. The letter $\gamma$ is widely used to name any curve in the plane. So while it's used for the half-circle you mention, it's also used for everything else. $\endgroup$ – John Hughes Apr 9 '18 at 12:17
0
$\begingroup$

Define for real $\;r>0\;$ :

$$\gamma_r^{\pm}:=\left\{\,z\in\Bbb C\;|\;\;|z|=r\;,\;\;\text{Im}>0\,\right\}$$

where the sign is $\;+\;$ if we take that half circle in the positive direction, and $\;-\;$ otherwise.

Now take the closed, simple path

$$\gamma:=[-R,-\epsilon]\cup\gamma_\epsilon^-\cup[\epsilon, R]\cup\gamma_R^+\;\;,\;\;\;0<R\in\Bbb R$$

Use now the corollary of this lemma to get for $\;f(z)=\cfrac{e^{iz}}z\;$ ,

$$\int_{\gamma_\epsilon^-}f(z)\,dz\xrightarrow[\epsilon\to0]{}-i\pi\cdot1=-\pi i$$

and then

$$0=\lim_{R\to\infty,\,\epsilon\to0}\int_\gamma f(z)\,dz=\int_{-\infty}^\infty f(x)\,dx-\pi i+0$$

and now just equate imaginary parts and we're done. Observe the integral over $\;\gamma_R\;$ vanishes at the limit because of Jordan's Lemma...or directly also by Cauchy's Estimate.

$\endgroup$
  • $\begingroup$ Please accept my edit. You are equating $0$ and $-\pi i$ in your last equation. $\endgroup$ – Szeto Apr 9 '18 at 13:22
  • $\begingroup$ @Szeto You are completely right, thanks. That equality sign was an ugly typo, yet someone else already rejected your edit. Anyway, I already edited that. Thanks again. $\endgroup$ – DonAntonio Apr 9 '18 at 13:34

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.