MHB Improper Integral Question (convergence & evaluation)

[Amad27]

Hello, Two questions will be posed here.

(1) Question about Convergence; quick way.

Hello, I am trying to learn this concept on my own. My major question here is that,

Is there a quick way, to tell if an integral converges or diverges?

Suppose $\int_{0}^{\infty} \frac{x^3}{(x^2 + 1)^2}\,dx$

This integral does not converge.

It took a while, a long while to actually evaluate this with the natural log, partial fraction decomposition etc...

The question is, Is there a way to tell the convergence or divergence of an improper integral, WITHOUT evaluating it?

Or a general easier method.

If you know of any, please let me know. Absolutely ANY TECHNIQUE?

Secondly, an actual integral for evaluation.

$\int_{0}^{\infty} \frac{4}{\sqrt{x}(x+6)}\,dx$

$(4) \cdot \int_{0}^{\infty} \frac{1}{x^{3/2} + 6x^{1/2}}\,dx$

First, I think it's a good idea to find the antiderivative. So the focus is, $\int \frac{1}{x^{1/2}(x + 6)}$

Some suggestions?

 

[Prove It]

Hello, Two questions will be posed here.

(1) Question about Convergence; quick way.

Hello, I am trying to learn this concept on my own. My major question here is that,

Is there a quick way, to tell if an integral converges or diverges?

Suppose $\int_{0}^{\infty} \frac{x^3}{(x^2 + 1)^2}\,dx$

This integral does not converge.

It took a while, a long while to actually evaluate this with the natural log, partial fraction decomposition etc...

The question is, Is there a way to tell the convergence or divergence of an improper integral, WITHOUT evaluating it?

Or a general easier method.

If you know of any, please let me know. Absolutely ANY TECHNIQUE?

Secondly, an actual integral for evaluation.

$\int_{0}^{\infty} \frac{4}{\sqrt{x}(x+6)}\,dx$

$(4) \cdot \int_{0}^{\infty} \frac{1}{x^{3/2} + 6x^{1/2}}\,dx$

First, I think it's a good idea to find the antiderivative. So the focus is, $\int \frac{1}{x^{1/2}(x + 6)}$

Some suggestions?

$\displaystyle \begin{align*} \int{ \frac{4}{\sqrt{x} \left( x + 6 \right) }\,\mathrm{d}x } &= 8 \int{ \frac{1}{\left( \sqrt{x} \right) ^2 + 6} \, \frac{1}{2\,\sqrt{x}}\,\mathrm{d}x } \\ &= 8 \int{ \frac{1}{u^2 + 6} \, \mathrm{d}u} \textrm{ after making the substitution } u = \sqrt{x} \implies \mathrm{d}u = \frac{1}{2\,\sqrt{x}}\,\mathrm{d}x \end{align*}$

You should be able to keep going...

 

[Amad27]

$\displaystyle \begin{align*} \int{ \frac{4}{\sqrt{x} \left( x + 6 \right) }\,\mathrm{d}x } &= 8 \int{ \frac{1}{\left( \sqrt{x} \right) ^2 + 6} \, \frac{1}{2\,\sqrt{x}}\,\mathrm{d}x } \\ &= 8 \int{ \frac{1}{u^2 + 6} \, \mathrm{d}u} \textrm{ after making the substitution } u = \sqrt{x} \implies \mathrm{d}u = \frac{1}{2\,\sqrt{x}}\,\mathrm{d}x \end{align*}$

You should be able to keep going...

Excellent trick,

How did you even think of that?

$= (8) (\frac{1}{\sqrt{6}})arctan(\frac{\sqrt{x}}{6}$

Keeping in mind that it is improper.

$\lim (8)[\frac{1}{\sqrt{6}arctan(\frac{\sqrt{t}}{6}} - \frac{1}{0}$
${{t}\to{\infty}} $

Uh oh, just noticed a major problem -_-

 

[jacobi1]

Excellent trick,

How did you even think of that?

$= (8) (\frac{1}{\sqrt{6}})arctan(\frac{\sqrt{x}}{6}$

Keeping in mind that it is improper.

$\lim (8)[\frac{1}{\sqrt{6}arctan(\frac{\sqrt{t}}{6}} - \frac{1}{0}$
$ $

How did it become [math]\frac{1}{\sqrt{6} \arctan (\frac{\sqrt{t}}{6})} - \frac{1}{0}[/math]?

It should be [math] \lim_{x \to \infty} \frac{1}{\sqrt 6 } \arctan \sqrt \frac{x}{6}[/math], because arctan(0)=0.
Can you proceed?

 

[Prove It]

Excellent trick,

How did you even think of that?

It's just from experience, I've developed a taste for when certain things will work.

In this case, because there was $\displaystyle \begin{align*} \sqrt{x} \end{align*}$ in the denominator, and because I know that $\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \left( \sqrt{x} \right) = \frac{1}{2\,\sqrt{x}} \end{align*}$, I realized that this $\displaystyle \begin{align*} \frac{1}{\sqrt{x}} \end{align*}$ factor could be used (after being scaled correctly) to change to a $\displaystyle \begin{align*} \mathrm{d}u \end{align*}$ if a substitution of $\displaystyle \begin{align*} u = \sqrt{x} \end{align*}$ was made everywhere else...