Determining if integral converges or diverges

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Discussion Overview

The discussion revolves around determining the convergence or divergence of the improper integral $\int_{3}^{\infty} \frac{1}{\sqrt{x} - 1} \,dx$. Participants explore various methods, including u-substitution and transformations, to analyze the integral's behavior as the upper limit approaches infinity.

Discussion Character

  • Mathematical reasoning, Homework-related

Main Points Raised

  • One participant initiates the discussion by proposing u-substitution with $u = \sqrt{x} - 1$, expressing uncertainty about the next steps.
  • Another participant encourages the first by suggesting to continue with the substitution, providing the transformation for $dx$ in terms of $u$.
  • A subsequent reply outlines the continuation of the integration process, reformulating the integral and introducing limits as $t$ approaches infinity.
  • Another participant presents a different approach using a substitution $t^2 = x$, leading to a similar transformation of the integral and suggesting further simplifications.

Areas of Agreement / Disagreement

Participants are generally aligned in their approach to using substitutions to analyze the integral, but there are multiple methods proposed without consensus on which is the most effective or correct.

Contextual Notes

Some steps in the transformations and limits are not fully resolved, and there may be dependencies on the definitions of convergence that are not explicitly stated.

tmt1
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$\int_{3}^{\infty} \frac{1}{\sqrt{x} - 1} \,dx$

I need to find if this converges or diverges.

I'm trying u-substitution, so $u = \sqrt{x} - 1$.

Therefore, $du = \frac{1}{2\sqrt{x}} dx$.

I'm not sure how to proceed from here.
 
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tmt said:
$\int_{3}^{\infty} \frac{1}{\sqrt{x} - 1} \,dx$

I need to find if this converges or diverges.

I'm trying u-substitution, so $u = \sqrt{x} - 1$.

Therefore, $du = \frac{1}{2\sqrt{x}} dx$.

I'm not sure how to proceed from here.
Good so far. Just keep going:
[math]\sqrt{x} = u + 1[/math]

Thus
[math]dx = 2(u + 1) ~ du[/math]

etc.

-Dan
 
I would continue as follows:

$$dx=2\sqrt{x}\,du=(2u+2)\,du$$

And so the integral becomes:

$$I=\int_{\sqrt{3}-1}^{\infty}\frac{2u+2}{u}\,du=2\int_{\sqrt{3}-1}^{\infty} 1+\frac{1}{u}\,du=2\lim_{t\to\infty}\left(\int_{\sqrt{3}-1}^{t} 1+\frac{1}{u}\,du\right)$$
 
$$\int_3^\infty\frac{1}{\sqrt{x}-1}\text{ d}x$$

$$t^2=x,\quad2t\text{ d}t=\text{d}x$$

$$2\int_\sqrt{3}^\infty\frac{t}{t-1}\text{ d}t=2\int_\sqrt{3}^\infty1+\frac{1}{t-1}\text{ d}t$$

etc.
 

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