# Integral with complex perimeter – convergence.

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

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 :-)

• 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. – Chase Ryan Taylor Nov 3 '18 at 0:46
• @ChaseRyanTaylor I added $dx$ term. Thanks. – user464980 Nov 3 '18 at 10:43

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.
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 $$s$$.