Is my integral-convergence contradiction proof valid?

I have to brush up on my proof making. I am a little rusty. I was not sure if the following really held up.

I wanted to prove the following is convergent/divergent (working with improper fractions) $\beta \in $ Real numbers:

$$\int_{0}^1 \frac {1}{x\sqrt {1+x^\beta}}dx$$

The following is my thought process.

The 'improper' aspect happens at 0, thus the root is of no concern which means $\beta$ can be any real number. If I am to solve this, I would use the Ratio convergence test. $\frac 1x$ is a similar function at $x\to0^+$.

I recognize, the following sounded iffy to me.

I will attempt to prove it by contradiction. In the next steps I will use the ratio test.

Let's suppose $\int_{0^+}^1 \frac {1}{x\sqrt {1+x^\beta}}dx$ converges.

If $\lim_{x\to0^+} \frac{\frac1x}{\frac1{x\sqrt {1+x^\beta}}} $=L where L$\in$Reals excluding $0$ Then $\int_{o^+}^1 \frac1x dx $ converges since their ratio is the same.

But wait! Therein lies a contradiction, $\int_{o^+}^1 \frac1x dx $ does not converge! My original supposition fails. And if the original function is not convergent, then it is divergent by definition.

I welcome other ways to do it, I really am just curious and looking to better myself.

  • $\begingroup$ $\displaystyle\int\frac{dx}{x\sqrt{1+x^k}} ~=~ -\frac2k~\text{arctanh}~\sqrt{1+x^k}.$ $\endgroup$ – Lucian Aug 14 '15 at 0:49

Your proof is not correct.

The integrand is continuous on $(0,1]$, thus potential problems are as $x \to 0^+$.

I would distinguish three cases.

  • $\beta=0$. As $x \to 0^+$, $$ \frac {1}{x\sqrt {1+x^\beta}} \sim\frac {1}{\sqrt {2}} \frac1x$$ the initial integral is divergent in this case.

  • $\beta>0$. As $x \to 0^+$, $$ \frac {1}{x \sqrt{1+x^\beta}} \sim\frac1x$$ the initial integral is divergent in this case.

  • $\beta<0$. As $x \to 0^+$, $$ \frac {1}{x\sqrt {1+x^\beta}}=\frac {1}{x^{1-|\beta|/2}\sqrt {1+x^{|\beta|}}} \sim\frac1{x^{1-|\beta|/2}}$$ the initial integral is convergent in this case.

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