- #1

- 90

- 2

## Homework Statement

Integrate [tex] \int_{-\infty}^{\infty} dx \ x^{-3}(sin x -x )[/tex]

## Homework Equations

Residual theorem, Jordan's lemma

## The Attempt at a Solution

[tex] \int_{-\infty}^{\infty} dx \ x^{-3}(sin x -x )

= Im \ (PV\int_{-\infty}^{\infty} dx \ x^{-3}(e^{ix} -ix ))[/tex]

In the contour, if we let R goes to infinity and ρ goes to zero, then we have

[tex]\oint_{Total}^{ } dz \ z^{-3}(e^{iz} -iz) = \int_{C_{1}}^{ } dz \ z^{-3}(e^{iz} -iz) + \int_{C_{2}}^{ } dz \ z^{-3}(e^{iz} -iz) + PV \int_{-\infty}^{\infty} dx \ x^{-3}(e^{ix} -ix) [/tex]

Because the only singularity is 0, not inside the contour. By residual theorem (or Cauchy's theorem),

[tex]\oint_{Total}^{ } dz \ z^{-3}(e^{iz} -iz) = 0[/tex]

By Jordan's lemma, as R goes to infinity,

[tex] \int_{C_{2}}^{ } dz \ z^{-3}(e^{iz} -iz) =0[/tex]

For the C1 contour

[tex] \int_{C_{1}}^{ } dz \ z^{-3}(e^{iz} -iz)= \int_{\pi}^{0 } d(\rho e^{i\theta }) \frac{e^{i\rho e^{i\theta }} -i\rho e^{i\theta }}{(\rho e^{i\theta })^{3}}= i \int_{\pi}^{0 }\frac{e^{i\rho e^{i\theta }}-i\rho e^{i\theta}}{\rho ^{2}e^{2i\theta }} d \theta[/tex]

But as ρ goes to 0, [tex]\frac{e^{i\rho e^{i\theta }}-i\rho e^{i\theta}}{\rho ^{2}e^{2i\theta }} [/tex]

goes to infinity. So I am stuck.

The final answer, according to WolframAlpha, is -π/2