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This question already has an answer here:

I need help solving this integral:

$$\int_0 ^\infty \frac{\sin(x)}{x} dx$$

I have a help that says that try to calculate the integral of $$\frac{e^{iz}}{z}$$ for a "proper path"... but I don't know how to use that, help.

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marked as duplicate by Felix Marin, user61527, user147263, mathematics2x2life, user91500 Jun 28 '14 at 4:16

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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$(1)$ $$f(-z) = \frac{\sin(z)}{z} \Rightarrow f \ \ \text{is even}.$$

$(2)$ $$\int_{-\infty}^{\infty} f(z) \ dz = 2 \int_{0}^{\infty} f(z)\ dz$$

$(3)$ $$\int_{-R}^{R} \frac{e^{iz}}{z} \ dz = Im\{ \pi i (\sum Res f|_{z_k})\}, \text{for}\ \ z_k (\text{being the singularities})$$

$(4)$ $$\int_{-R}^{R} \frac{e^{iz}}{z} \ dz =Im\{ \pi i (e^{i \cdot 0}) \} = \pi$$

$(5)$ $$\int_{0}^{\infty} \frac{\sin(z)}{z}\ dz = \frac{\pi}{2}$$ \

$\textbf{Comment}$: Of course a massive amount of details were left out.

$$\int_{-R}^{R} f(z) dz = \int_{C} f(z) - \int_{C_r} f(z) dz$$

$\cdot$ The latter goes to zero by Jordan's Lemma.

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  • $\begingroup$ Can you explain a little more the equality number 3? $\endgroup$ – Dimitri Jun 28 '14 at 3:33
  • $\begingroup$ en.wikipedia.org/wiki/Cauchy_principal_value $\endgroup$ – Mr.Fry Jun 28 '14 at 3:36
  • $\begingroup$ I think you make a mistake, on the left side of 3 shouldnt be the integral of f, and on the right side the residues of $\frac{e^z}{z}$? Also i dont understand the path that you use to integrate, because you can integrate trough -R to R because the singularity at $z=0$ $\endgroup$ – Dimitri Jun 28 '14 at 3:49

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