Convergence or divergence for $\int_1^\infty \frac{1}{x\sqrt{x}-1} dx$

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.

• 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. – fGDu94 Dec 14 '19 at 21:03

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.

• Nice , wanted to write the same :) – Vuk Stojiljkovic Dec 14 '19 at 22:14

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 $$\int_0^1 \frac1y dy$$.

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

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

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