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Consider the following integral: $$\int _1 ^\infty \frac{1}{x\sqrt{x}-1} dx$$ So I can see that this integral is improper both at $\infty$ and at $1$. Using $\lim\limits_{x\to\infty} x^{3/2}\frac{1}{x\sqrt{x}-1}=1 \in (0,\infty),3/2>1 $ we get that the integral is convergent at $\infty$ . As for convergence at $1$, I tried to use $\lim\limits_{x\to\infty} (1-x) \frac{1}{x\sqrt{x}-1}$, but this is equal to $-\frac{2}{3}$ so I can't apply the criterion of convergence ( or divergence ) in this case because the limit is not in the interval $[0,\infty)$. I want to know how to solve the convergence or divergence of this integral without calculating the integral is that's possible.

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    $\begingroup$ You can see it is divergent by taking $ u = \sqrt{x}$. The integral then has the form $\int_1^{\infty} \frac{u}{u^3-1}du$. You can bound this below by $\int_1^{\infty} \frac{1}{u^2-\frac{1}{u^2}}du$ which diverges. $\endgroup$ – fGDu94 Dec 14 '19 at 21:03
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Note that as $x\to\infty$, $${1\over x\sqrt{x} - 1} \sim {1\over x\sqrt{x}},$$ so this integrates at $\infty$.

At 1, we have $${1\over x\sqrt{x} - 1} = {x\sqrt{x} + 1\over x^3 - 1} \sim {2\over x^3 -1} = {2\over (x-1)(x^2 + x + 1)}\sim {2\over 3(x-1)}$$ as $x \to 1$.

This does not integrate at 1.

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  • $\begingroup$ Nice , wanted to write the same :) $\endgroup$ – Vuk Stojiljkovic Dec 14 '19 at 22:14
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By $y=\sqrt x-1 \implies dy=\frac12\frac1{\sqrt x}dx$ we have

$$\int _1 ^\infty \frac{1}{x\sqrt{x}-1} dx=\int _0 ^\infty \frac{2(y+1)}{(y+1)^2(y+1)-1} dy=\int _0 ^\infty \frac{2(y+1)}{y^3+3y^2+3y} dy$$

which diverges at $0$ by limit comparison test with $\int_0^1 \frac1y dy$.

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$$I=\int_1^\infty\frac{1}{x\sqrt{x}-1}dx\overset{\sqrt{x}=y}{=}2\int_1^\infty\frac{y}{y^3-1}dy\overset{1/y=x}{=}2\int_0^1\frac{1}{1-x^3}dx$$

$$=\frac23\int_0^1\frac{dx}{1-x}+\frac23\int_0^1\frac{x+2}{x^2+x+1}dx$$

The first integral diverges and the second one converges, so $I$ diverges.

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