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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
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