Improper integral comparison test

Question

Having this integral $$\int_1^{\infty}\frac{3x^2+2x +1}{x^3+6x^2+x+4}$$

In order to do the comparison test at some point it gets like

$$\frac{3x^2+2x +1}{x^3+6x^2+x+4}\geq \frac{1}{4x}$$

How is $\frac{1}{4x}$ found ? It doesnt seem obvious to me.

EDIT
Also how is this found

$$0\leq\frac{3x +1}{\sqrt{x^5+4x^3+5x}}\leq \frac{4}{x^{3/2}}$$

Where does 4 came from here ?Minimum value ? 3x1 +1 ? Why.

André Nicolas · Accepted Answer · 2014-06-06T16:27:21.563

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For positive $x$, the top is $\ge 3x^2$. For $x\ge 1$, the bottom is $\le x^3+6x^3+x^3+4x^3$.

Edit: For your added question, the range of values of $x$ is not specified. However, if $x\ge 1$, then the top is $\le 4x$. The bottom is $\ge \sqrt{x^5}$. So for $x\ge 1$, the whole thing is $\le \frac{4x}{x^{5/2}}$, which simplifies to $\frac{4}{x^{3/2}}$.

edited Jun 06 '14 at 16:27

answered Jun 06 '14 at 16:03

André Nicolas

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why we dont keep 3 too then ? 3/4x ? Because it has lower power ? – GorillaApe Jun 06 '14 at 16:14
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I kept the $3$ in $3x^2$ on top. The bottom is $\le 12x^3$. Then the $3$ cancels part of the $12$. – André Nicolas Jun 06 '14 at 16:19
For your added question, you do not specify the range of values of $x$, so it is not possible to give an answer. However, for $x\ge 1$, the term $3x+1$ is $\le 4x$. – André Nicolas Jun 06 '14 at 16:22
they are the same with the first. Integral from 1 to infinity – GorillaApe Jun 06 '14 at 16:24
if bottom $x^5$ was like $10x^5$ would it matter too ? – GorillaApe Jun 06 '14 at 16:26
The $x^{5/2}$ (see my addition to the answer) could then be replaced by $(10)^{1/2} x^{5/2}$. However, one can still accurately say that the bottom is $\ge x^{5/2}$, which is enough to show convergence. – André Nicolas Jun 06 '14 at 16:30
Where can I find texts about these tricks ? – GorillaApe Jun 06 '14 at 16:30
Calculus/analysis books usually spend little time on this business of knowing what to "give away." But after doing a smallish number of examples you will get reasonably good control over the ideas. – André Nicolas Jun 06 '14 at 16:34

score 1 · Answer 2 · answered Jun 06 '14 at 16:04

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For example, and taking into account the fact that for large values $\;x\;$ only the summands with the largest exponent of $\;x\;$ are important (in this sense), we get

$$\frac{3x^2+2x +1}{x^3+6x^2+x+4}\ge\frac{3x^2}{x^3+6x^3+x^3+4x^3}=\frac1{4x}$$

answered Jun 06 '14 at 16:04

DonAntonio

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Improper integral comparison test

2 Answers2