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$$\int_0^\infty \dfrac{\sin x}x\,\mathrm dx = \mathrm{Im}\left(\int \dfrac{e^{ix}}x\,\mathrm dx\right)$$ Considering $f(z) = \dfrac{e^{iz}}z$, it has a simple pole at $z = 0$. Residue of $f(z)$ at $z = 0$ is $1$. This imples that

$$\mathrm{Im}\left(\int_0^\infty \dfrac{e^{iz}}z\,\mathrm dz\right) = \dfrac12\int_{-\infty}^\infty\dfrac{e^{iz}}z\,\mathrm dz = \dfrac12(2\pi\cdot1) = \pi$$

I found out that the answer is $\dfrac12\pi$. Why??

an4s
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ENEJO SUNDAY OCHE
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  • In order to get answers, you should format your question in MathJax: https://math.stackexchange.com/help/notation – Miksu Jun 09 '18 at 18:08
  • He, welcome on the MathSE! Somebody already started to beautify your question, but if you remain here, I suggest to start to learn MathJax. Btw, all capital nicks don't look very well on the first spot, I suggest to become simply "Enejo Sunday Oche". Good luck on the site! – peterh Jun 09 '18 at 18:16
  • @ENEJOSUNDAYOCHE I assume you are relating the residue to the integral using a semicircular contour. Integrating over the semicircle would give an extra factor of 1/2. – Tyberius Jun 09 '18 at 18:21
  • Your first equation is not right. The right hand side does not exist. What you want is a principal value. And then as Tyberius's comments, one way of computing this involves going around the pole at the origin in a semicircle, which only gives you half the residue – spaceisdarkgreen Jun 09 '18 at 18:25

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