Why does the integral $\int_{1}^{\infty} x^p \,dx$ diverge for values of $p ≤ 1$?

Question

$$ \int_{1}^{\infty} x^p \,dx $$

For the above integral, it was defined in class that this diverges for $$p \leq 1$$ but if $p$ is $0$, then wouldn't that mean $1/1$ which is $1$ and thus a number, so the function would actually converge in the $$p \leq 1$$ interval. Why does it diverge?

Answer 1

Let’s see why with a simple power rule:

$$\int_1^\infty x^p dx=\frac{x^{p+1}}{p+1}\bigg|_0^\infty=\frac{1}{p+1}\lim_{x\to\infty} x^{p+1}-\frac{1}{p+1}$$

The limit here matters most:

This immediately implies that $p\ne 1$. Let’s do each case:

$p<-1$:

$$\frac{1}{p+1}\lim_{x\to\infty} x^{p+1}=\frac{0}{p+1}=0$$

$p>-1$:

$$\frac{1}{p+1}\lim_{x\to\infty} x^{p+1}=\frac{\infty}{p+1}=\infty$$

Now let’s try $p=\frac12$:

$$\frac 23\lim_{x\to\infty} x^{\frac32>1}=\infty$$

p=0:

$$1\lim_{x\to\infty} x^{1}=\infty$$

$p=i$:

$$\frac{1}{i+1}\lim_{x\to\infty} x^{i+1}=(1+i)\infty$$

$p=i-1$:

$$-i\lim_{x\to\infty} x^{i-1+1}\in [-1,1](1+i)$$

$p=i-2$:

$$\frac{1}{i-1}\lim_{x\to\infty} x^{p+1}=0$$

Ignoring other possible values of $p$, our final answer is:

$$\int_1^\infty x^p dx=-\frac{1}{p+1},\text{Re}(p)<-1$$

If $p=-1$, then see this proof for

$$\int x^{-1} dx=\ln(x)+C$$

Please correct me and give me feedback!