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.

**First, consider if p=1.**
$$\int_{2}^{\infty}\frac{1}{x\ln\left(x\right)}dx$$
$$=\lim_{T\to\infty}\int_{2}^{T}\frac{1}{x\ln\left(x\right)}dx$$

$$\lim_{T\to\infty} ln(ln(x))dx$$ evaluated from 2 to T gives us

$$\lim_{T\to\infty} [ln(ln(T))]-[ln(ln(2))]dx$$

When the value for T increases, the output value also increases towards infinity.

**Now, consider if p is not equal 1**

$$\int_{2}^{\infty}\frac{1}{x\ln\left(x\right)^p}dx$$
$$=\lim_{T\to\infty} \frac{1}{1-p} ln(x)^{1-p}$$

$$=\lim_{T\to\infty} [\frac{1}{1-p} ln(T)^{1-p}]-[\frac{1}{1-p} ln(2)^{1-p}]$$

When the value for T increases, the output value will be infinite or finite depending on if p is greater than 1 or p is less than or equal to one.

**Conclusion**
Therefore, when p is greater than 1, it converges because as we put in bigger values for p, the exponent (1-p) will make the output get smaller and smaller because it will have a large negative exponent.
When p is less than or equal to one, it diverges because the exponent gets bigger making our output value larger towards infinity.

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