Shouldn’t I find a function whose integral over the interval does not converges either and is always less than $\;\sqrt{1+\frac{1}{x^4}}\;$ when close to zero so that the first integral is also divergent? I’m afraid I’m only familiarized with this approach.
How do I show that this integral $\int _0^1 x\sqrt {1+\frac {1} {x^4}} dx$ does not converge?
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calculus
algebra-precalculus
functions
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0Set $x^2=\cot y$ – 2017-02-09
2 Answers
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Hint. One may observe that, as $x \to 0^+$, $$ x\sqrt {1+\frac {1} {x^4}}= \frac1{x}\sqrt {x^4+1}>\frac1x $$ thus the given integral is divergent.
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Away from zero, there is no promlem.
As you approach zero, the function behaves like $x/x^2 =1/x$ and the integral of this (being $\ln(x)$) diverges as $x \to 0$.