# Determine the values of p for which the integral below is convergent?

## $\setminus {\int}_{1}^{\setminus} \infty \frac{1}{x} ^ p \mathrm{dx}$ I managed to integrate it, to $\setminus {\lim}_{t \setminus \rightarrow \setminus \infty} \left[\left(\frac{{t}^{- p + 1}}{- p + 1}\right) - \left(\frac{{1}^{- p + 1}}{- p + 1}\right)\right]$ but now I am stuck :(

May 10, 2018

The integral converges for $p > 1$

#### Explanation:

So, we'll take the integral:

${\int}_{1}^{\infty} \frac{\mathrm{dx}}{x} ^ p = {\lim}_{t \to \infty} {\int}_{1}^{t} \frac{\mathrm{dx}}{x} ^ p$

$= {\lim}_{t \to \infty} {x}^{1 - p} / \left(1 - p\right) {|}_{1}^{t}$

$= {\lim}_{t \to \infty} {t}^{1 - p} / \left(1 - p\right) - \frac{1}{1 - p} , p \ne 1$ (this constraint is very important to take note of, as we'll see later)

Let's assume $p > 1.$

In that case, $1 - p$ in the numerator is negative, IE, $1 - p = - \left(1 - p\right)$, so ${\lim}_{t \to \infty} {t}^{-} \frac{1 - p}{1 - p} = {\lim}_{t \to \infty} \frac{1}{{t}^{1 - p} \left(1 - p\right)} - \frac{1}{1 - p} = - \frac{1}{1 - p}$

and we have convergence as the exponent $1 - p$ that is now moved to the denominator is positive, so sending the base $t$ to infinity gives a denominator of infinity and an overall value of $- \frac{1}{1 - p} .$

Let's assume $p < 1.$

In this case, $1 - p > 0 = 1 - p ,$ so

${\lim}_{t \to \infty} {t}^{1 - p} / \left(1 - p\right) - \frac{1}{1 - p} = \infty$

and we have divergence as the exponent $1 - p$ in the numerator is now positive and sending the base $t$ to infinity yields a numerator of infinity and an overall value of infinity.

However, the limit we're evaluating requires that $p \ne 1$. If we let $p = 1$ in ${\int}_{1}^{\infty} \frac{\mathrm{dx}}{x} ^ p$, we integrate

${\int}_{1}^{\infty} \frac{\mathrm{dx}}{x} = {\lim}_{t \to \infty} {\int}_{1}^{t} \frac{\mathrm{dx}}{x}$

$= {\lim}_{t \to \infty} \ln \left(x\right) {|}_{1}^{t}$

$= {\lim}_{t \to \infty} \ln t = \infty$ and we have divergence.

Thus, the integral only converges for $p > 1$