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By using this method we can evaluate $\displaystyle\int^\infty_0\dfrac{x-\sin x}{x^3}\,dz=\dfrac{\pi}{4}$ and I intended to solve $\displaystyle\int^\infty_0 \dfrac{\sin x}{x^3}\,dx=\displaystyle\int^\infty_0\dfrac{1}{x^2}\,dx-\dfrac{\pi}{4}$. But how can I integrate $\displaystyle\int^\infty_0\dfrac{1}{x^2}\,dx$ when the integral does not converge?

Alvin
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    $\int^\infty_0 \dfrac{\sin x}{x^3},dx$ does not exist. – Kavi Rama Murthy Apr 29 '22 at 06:11
  • My Sage agrees with Kavi - "Integral is divergent." – Gareth Ma Apr 29 '22 at 06:13
  • Does that mean $\displaystyle\int^\infty_0\dfrac{a\sin x}{x^3},dx$ is also divergent for any complex number $a$? – Alvin Apr 29 '22 at 06:23
  • For $\ x\ge0\ $, $$ \frac{\sin x}{x^3}\ge\frac{1}{x^3}-\frac{1}{2x}\ , $$ so $$ \int_b^c\frac{\sin x}{x^3},dx\ge\int_b^c\left(\frac{1}{x^3}-\frac{1}{2x}\right),dx\ . $$ for $\ 0\le b\le c\ $. What happens to the integral on the right of this inequality as $\ b\rightarrow0^+\ $? What happens to the one on the left of the inequality if you multiply it by a complex number $\ a\ $ and take its limit as $\ b\rightarrow0^+\ $? – lonza leggiera Apr 29 '22 at 06:30
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    $x^{-3} \sin x \sim x^{-2}$ as $x \to 0$. – Christophe Leuridan Apr 29 '22 at 07:06
  • @Alvin You can factor out $a$ into the front of the integral and you are left with the case $a=1$. – Gary Apr 29 '22 at 08:04

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You must keep in mind that an equality like $$ \int_A (f(x)+g(x))\,dx = \int_A f(x)\, dx + \int_A g(x)\,dx $$

only holds if all three integrals exist. In your example, as it was pointed out in the comments, this is not the case.

PierreCarre
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