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title: Math 182 Miniproject 6 Another $p$-test

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Math 182 Miniproject 6 Another $p$-test.md
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Math 182 Miniproject 6 Another $p$-test 
===

**Overview:** In this project we develop a $p$-test to determine whether a certain type of integral converges or diverges.

**Prerequisites:** Section 6.5 of _Active Calculus_

In class we learned the $p$-test for integrals of the flavor 
$$
\int_1^\infty\frac{1}{x^p}dx.
$$
__The $p$-test:__ $\int_1^\infty\frac{1}{x^p}dx$ converges if and only if $p>1$.

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Your task is to identify conditions on $p$ that let us know when the integral 
$$
\int_2^\infty\frac{1}{x(\ln(x))^p}dx
$$

converges. You may want to break your exploration into separate cases. Include all of your work below. 

Being that $x(ln(x))^p$ is a composite function I will use u-substitution to simplify the integral such that
$$u=ln(x)$$
$$du= \frac{1}{x}dx$$
Substituting we get 
$$
\int_2^\infty\frac{1}{u^p}du
$$
Integrating we find that it equals
$$=lim_{a\to \infty} \left[\frac{1}{1-p}u^{1-p}\right]_.6931471806^a$$
$$=lim_{a\to \infty} \left[\frac{1}{1-p}a^{1-p}-\frac{1}{1-p}.6931471806^{1-p}\right]$$
in which $p \neq 1$.


Now we will consider the  cases when $P >0$, $P<0$, and $P=1$
Case 1: $p<0$
Since p is less than 0, p is a negative number. That means we get a positive number for ever $1-p$ since it would be $1-(-n)$ for some integer n. That being said
$$lim_{a\to \infty} \frac{1}{1-p}a^{1-p}= \text{a really big number(inifnity)}$$
and 
$$\frac{1}{1-p}.6931471806^{1-p}= \text{approaches zero from the right}$$
that is 
$$\text{infinity - 0= infinity}$$
**Therefore when $p<0$ it diverges**.  

Case 2: $p >0$ but $p \neq 1$
Since p is greater than 0, p is a positive number. That means we get a negative number for ever $1-p$ since it would be $1-(n)$ for some integer n. That being said
$$lim_{a\to \infty} \frac{1}{1-p}a^{1-p}= \text{approaches zero from the left}$$
and 
$$\frac{1}{1-p}.6931471806^{1-p}= \text{a negative value value}$$
that is 
$$\text{0 - -value= value}$$
**Therefore when $p>0$ but $p \neq 1$ it converges**. 

**Case 3: p=1**
In this case even using L'hopitals rule we can not reach a set up where we can conduct an integral and evaluate at x=1 without reaching an undefined answer. 
**Therefore when $p = 1$ it is undefined**.


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