MHB What is the Integration Formula for x and p in Maxima?

[karush]

$\Large{§8.8.15} \\
\tiny\text {Leeward 206 Integration to Infinity}$
$$\displaystyle
\int_{e^{2}}^{\infty} \frac{dx}{x\ln^p\left({x}\right)}\,dx \,, p>1$$
$\text{not sure how to deal with this} $
$\text{since there are two variables x and p} $

$\text{answer by maxima is:'} $
$$\displaystyle \dfrac{\ln^{1-p}\left(x\right)}{1-p}$$$\tiny\text{ Surf the Nations math study group}$

 

[MarkFL]

Okay, the first thing I would do is set the definite integral up as a limit because of the infinite upper bound:

$$I=\lim_{t\to\infty}\left(\int_{e^2}^t \frac{1}{x\ln^p(x)}\,dx\right)$$

Next, I would use the following $u$-substitution:

$$u=\ln(x)\,\therefore\,du=\frac{1}{x}dx$$

And for out limits of integration

$$u\left(e^2\right)=\ln\left(e^2\right)=2$$

$$\lim_{x\to\infty}\left(u(x)\right)=\lim_{x\to\infty}\left(\ln(x)\right)=\infty$$

So, now we have:

$$I=\lim_{t\to\infty}\left(\int_{2}^t u^{-p}\,du\right)$$

Can you proceed? :D

 

[karush]

$\text{are you sugesting} $
$$\displaystyle
I=\left[\frac{-\ln^{1-p} \left({x}\right)}{p-1}\right]_2 ^\infty$$

$\text{the book answer was} $
$$\displaystyle
\frac{1}{\left(p-1\right){2}^{p-1}}$$

 

[MarkFL]

$\text{are you sugesting} $
$$\displaystyle
I=\left[\frac{-\ln^{1-p} \left({x}\right)}{p-1}\right]_2 ^\infty$$

More like:

$$I=\frac{1}{1-p}\lim_{t\to\infty}\left(\left.u^{1-p}\right|_2^t\right)$$

 

[karush]

Sorry it's simple I know: but I don't see it?

 

[MarkFL]

Sorry it's simple I know: but I don't see it?

Nothing is "simple" when you are first learning it...I learned things like this more than 20 years ago, so I've likely done it a few more times than you have. :D

To finish it up, I would write:

$$I=\frac{1}{1-p}\lim_{t\to\infty}\left(\left.u^{1-p}\right|_2^t\right)=\frac{1}{1-p}\lim_{t\to\infty}\left(t^{1-p}-2^{1-p}\right)=\frac{2^{1-p}}{p-1}$$

The term $t^{1-p}$ goes to zero as $t$ grows without bound, because $1<p$...does this all make sense?

 

[karush]

OK got what's happening
Sure appreciate the help