Why $ \int_{0}^{\infty} x^p \,dx $ does not converge for any $p$?

Question

We already know that

$$ \int_{0}^{\infty} x^p \,dx $$

is not convergent for any $p$. However I found that it actually converges for $p<-1$ . Here's my attempt

$$ \int_{0}^{\infty} x^p \,dx =\lim_{m \to \infty} \dfrac{m^{p+1}}{p+1}$$

We know that if $p>-1$ then the integral diverges ; however if $p<-1$ , the integral becomes $0$. So this means that this integral converges for $p<-1$. And this is very strange.

So what is wrong with my computation?

Answer 1

Your computation of this potentially doubly improper integral is incorrect: it should read $$ \int_{0}^{\infty} x^p \,dx =\lim_{m \to \infty} \dfrac{m^{p+1}}{p+1} - \lim_{\delta \to 0^+} \dfrac{\delta^{p+1}}{p+1}. $$ You seem to be assuming that the second term is always $0$, but that is not the case—it depends on the value of $p$.

Answer 2

In calculus course, we saw that we must split the improper integral into pieces so that each piece contains only one singularity assuming $\infty$ and $-\infty$ are singularities.

In this example, $x=0$ and $x=\infty$ are singularities of the function. So, first thing we do is to split, for example like this: $$\int_0^{\infty} x^p dx = \int_0^{1} x^p dx + \int_1^{\infty} x^p dx$$ Here, we have two integrals to consider. The second one is convergent for $p<-1$ and its value is $$\int_1^{\infty} x^p dx=\lim_{M\rightarrow\infty}\frac{1}{p+1}(M^{p+1}-1)=-\frac{1}{p+1}.$$ But, the first integral is not convergent for $p<-1$ since the limit $$\int_0^{1} x^p dx=\lim_{\delta\rightarrow 0^{+}}\frac{1}{p+1}(1-\delta^{p+1})$$ does not exist for $p<-1$. Informally, we say that it the limit is $\infty$.

An improper integral is convergent when each piece in such a splitting is convergent. Therefore, this integral is not convergent for $p<-1$.