Question

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 :(

Answer 1

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`