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The original quesion is $$\int_0^\infty\left(\frac{\sin x}{x^3}-\frac1{x^2}\right)\,dx$$ Can I divide them into two parts? Then using residue theorem?

Parcly Taxel
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Cristian YuuKi
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  • Are you open to solutions that don't use the residue theorem? – J.G. Oct 08 '20 at 09:50
  • Please show it. Thank you. – Cristian YuuKi Oct 08 '20 at 09:57
  • OK, see my answer. – J.G. Oct 08 '20 at 09:58
  • Both terms have a pole of order 2 at $0$, and so, their respective integrals over $(0,\infty)$ will diverge. You really need to make use of the fact that those singularities cancel out. – Sangchul Lee Oct 08 '20 at 10:03
  • Let $$f(z) = \frac {e^{i z} - e^{-i z}} {2 i z^3} - \frac 1 {z^2}, \quad g(z) = \frac {e^{i z}} {2 i z^3},$$ then $$\int_{-\infty}^\infty f(z) dz = \int_{-\infty - i0}^{\infty - i0} f(z) dz = \int_{-\infty - i0}^{\infty - i0} g(z) dz.$$ Or, in a slightly more roundabout way, $$\int_{-\infty}^\infty f(z) dz = \lim_{R \to \infty} \int_{C_R} f(z) dz = \lim_{R \to \infty} \int_{C_R} g(z) dz = \lim_{R \to \infty} \int_{|z| = R} g(z) dz,$$ where $C_R = { z: |z| = R \land \pi \leq \arg z \leq 2 \pi }$, traversed in the ccw direction. – Maxim Oct 09 '20 at 03:52

2 Answers2

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By a Schwinger parametrization, this integral is$$\begin{align}\frac{1}{4i}\int_0^\infty dx\int_0^\infty dy(e^{ix}-e^{-ix}-2ix)y^2e^{-xy}&=\frac{1}{4i}\int_0^\infty dy\left(\frac{1}{y-i}-\frac{1}{y+i}-\frac{2i}{y^2}\right)y^2\\&=-\frac12\int_0^\infty\frac{dy}{y^2+1}\\&=-\frac{\pi}{4}.\end{align}$$WA agrees.

J.G.
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If you like special functions $$I=\int\left(\frac{\sin x}{x^3}-\frac1{x^2}\right)\,dx=-\frac{\text{Si}(x)}{2}-\frac{\sin (x)}{2 x^2}+\frac{1}{x}-\frac{\cos (x)}{2 x}$$ $$J(p)=\int_0^p\left(\frac{\sin x}{x^3}-\frac1{x^2}\right)\,dx=-\frac{\text{Si}(p)}{2}-\frac{\sin (p)}{2 p^2}+\frac{1}{p}-\frac{\cos (p)}{2 p}$$ nad the limit of $\text{Si}(p)$ is $\frac \pi 2$.

Claude Leibovici
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