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Kinda tough one – been thinking about this integral for a while.

$$\int_0^{\infty} \left| x^{-2s} \right| \text{d}x$$

for any complex $s$ and $x \in (0;\infty)$ obv.

Intuition 'tells' me, this shouldn't converge, but I cannot find any way to prove it – any inequality or hint will be helpful :-)

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    $\begingroup$ I get it’s “obv.” as you put it, but it took me a second there. The $dx$ is kinda mandatory when you write your integrals. $\endgroup$ – Chase Ryan Taylor Nov 3 '18 at 0:46
  • $\begingroup$ @ChaseRyanTaylor I added $dx$ term. Thanks. $\endgroup$ – user464980 Nov 3 '18 at 10:43
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Let $s=a+ib$ so $$x^{-2s}=x^{-2a}x^{-2ib}$$ Writing $x^{-2ib}$ as $$x^{-2ib}=e^{-2ib\log x}$$ So, $$ |x^{-2s}|=|x^{-2a}e^{-2ib\log x}|=|x^{-2a}||e^{-2ib\log x}|$$ The first factor is a positive real number and the second one is a complex number with modulus equal to 1,so $$|x^{-2s}|=|x^{-2a}|=x^{-2a}$$ Now you can see that for any value of $a\in R$ the integral diverges. Hope this helps.

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Since the integrand is absolute value, the imaginary part of $s$ can be ignored. Assume $s$ is real, then for $s\le 0$ the integral obviously diverges at the upper limit. For positive $s$ there are two cases to consider. For $2s\le 1$, the integral diverges at the upper limit. For $2s\ge 1$, the integral diverges at the lower limit. Conclusion: the integral is divergent for all values of $s$.

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