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How can I find the values of $\mu$ such that:

  1. $\displaystyle\int_0^1\frac{1}{x\sqrt{1+x^\mu}}\,\mathrm{d}x$ is finite

  2. $\displaystyle\int_1^\infty\frac{\log(x)}{x^\mu}\,\mathrm{d}x$ is finite

The second one looks like a Taylor series, but my professor wants a "basic" argument. For the first one I have no idea.

Can you help me with some hint?

Riccardo.Alestra
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sinbadh
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  • (2) For any $\epsilon>0$ and $x\gg0$, you have $\log(x)<x^\epsilon$. – Kenta S Apr 22 '21 at 16:00
  • @KentaS that's usefull to see that $\mu>2$ is a part of the set of values. But how can i show that that is all the set of values? – sinbadh Apr 22 '21 at 16:03
  • Whenever $\mu>1$, say $\mu=1+\epsilon$, for $x$ large enough the integral can be bounded above by $\int_M^\infty x^{-1-\epsilon/2}dx$, which clearly converges. – Kenta S Apr 22 '21 at 16:07
  • @KentaS ok. But it means that if $\mu>1$ then the integral converges. But what about other values? With your idea we only know "a part" of the set of solutions – sinbadh Apr 22 '21 at 16:33
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    You could also evaluate $\int_1^{\infty}\frac{\ln(x)}{x^t}\mathrm{d}x$ via integration by parts... Set $$u=\ln(x),\mathrm{d}v=x^{-t}\mathrm{d}t$$ This approach would only work if $t\neq 1$, but it's not hard to see what's happening in this case. – Matthew H. Apr 22 '21 at 16:38
  • Because you did not put any restrictions on μ, I will consider μ=$log_x(log(x))$ for number 2 with a removable discontinuity at ln(0) and ln(1) where there are indeterminate forms. Also, for number 1, you can do $μ\in \Bbb Z^-$ or a negative integer. – Тyma Gaidash Apr 22 '21 at 17:04
  • @sinbadh if $\mu<1$, basically the same argument shows that it diverges. Otherwise, if $\mu=1$, integration by parts tells you that it diverges. I never meant my comment to be a full solution. It was just that---a comment. – Kenta S Apr 22 '21 at 21:22

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