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I want to investigate the convergence of integral

$$\int_0^1 \frac{1}{\sqrt[3]{- x^{4} + 1}}\, dx$$

How shall I deal with it? I thought about taking $\frac{1}{x^p}$ and finding $p$ by $\lim\limits_{x \to 1} \frac{f(x)}{g(x)}$. Now I decided to use $\frac{1}{(1-x)^p}$, but how shall I do with finding $p$?

And others: $\int\limits_1^2 \frac{1}{\ln x}\, dx$ $\int\limits_{n/2}^\infty \sin x\frac{1}{x^2}\, dx$

How should I deal with them, just an idea please

Michael Hardy
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TaumCkBep
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  • The first apparently takes into use the Gamma function – John Lou Apr 10 '18 at 15:00
  • The second diverges, and I would use a comparison to $1/x$, perhaps change it to $\int_0^1 \frac{1}{\ln(x+1)}dx$, and then use taylor series to show how it is always greater than $1/x$, and $1/x$ diverges in that range – John Lou Apr 10 '18 at 15:04
  • Not 100% sure what you mean for the third – John Lou Apr 10 '18 at 15:05
  • For the third, I would suggest using a u-sub for $1/x = u$, which will give a function that I believe will always numerically converge (given non-infinite values of n) – John Lou Apr 10 '18 at 15:19
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    @JohnLou, thank you, man, i really appreciate your help. – TaumCkBep Apr 10 '18 at 15:56

1 Answers1

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Let x=$\frac1y\implies dx=-\frac1{y^2}dy$ then

$$\int_{0}^{1} \frac{1}{\sqrt[3]{- x^{4} + 1}}\, dx=\int_{1}^{+\infty} \frac{\sqrt[3]y}{y\sqrt[3]{y^{4} - 1}}\, dy$$

and thus since

$$ \frac{\sqrt[3]y}{y\sqrt[3]{y^{4} - 1}}\sim \frac1{y^2}$$

the integral converges by limit comparison test with $\int \frac1{y^2}$.

user
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