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I'm trying to evaluate the following integral:

$\int_{0}^{\infty}r^2\cdot\sin\big(\frac{b}{r^3}\big)\mathrm{d}r$

I had to solve a similar integral with cosine rather than sine and it was helpful to use a variable subsitution to obtain the format shown below. When it was cosine, this was evaluable using CPV. However, I don't think this is the case with sine and am searching for any insight into whether or not this integral is evaluable using CPV or otherwise.

$\int_{0}^{\infty}\frac{\sin(bx)}{x^2}\mathrm{d}x$

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    $\begingroup$ It is divergent due to the behaviour in a right neighbourhood of the origin, $\frac{b}{r}$. $\endgroup$ – Jack D'Aurizio Aug 14 '18 at 22:07
  • $\begingroup$ Even its principal value diverges. $\endgroup$ – Szeto Aug 14 '18 at 22:58
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You have $$I=\int\frac{\sin(bx)}{x^2}\,dx=b\, \text{Ci}(b x)-\frac{\sin (b x)}{x}$$ $$J=\int_\epsilon^\infty\frac{\sin(bx)}{x^2}\,dx=\frac{\sin (b \epsilon )}{\epsilon }-b\,\text{Ci}(b \epsilon )$$ Developed as series $$J=-b\, (\log (b)+\log (\epsilon )+\gamma -1)+\frac{b^3 \epsilon ^2}{12}+O\left(\epsilon ^4\right)$$ where you see the problem when $\epsilon \to 0$.

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