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

\int_1^\infty1/x^pdx

I managed to integrate it, to \lim_(t\rarr\infty)[((t^(-p+1))/(-p+1))-((1^(-p+1))/(-p+1))] but now I am stuck :(

1 Answer
May 10, 2018

The integral converges for p>1

Explanation:

So, we'll take the integral:

int_1^oodx/x^p=lim_(t->oo)int_1^tdx/x^p

=lim_(t->oo)x^(1-p)/(1-p)|_1^t

=lim_(t->oo)t^(1-p)/(1-p)-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=-(1-p), so lim_(t->oo)t^-(1-p)/(1-p)=lim_(t->oo)1/(t^(1-p)(1-p))-1/(1-p)=-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 -1/(1-p).

Let's assume p<1.

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

lim_(t->oo)t^(1-p)/(1-p)-1/(1-p)=oo

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^oodx/x^p, we integrate

int_1^oodx/x=lim_(t->oo)int_1^tdx/x

=lim_(t->oo)ln(x)|_1^t

=lim_(t->oo)lnt=oo and we have divergence.

Thus, the integral only converges for p>1