Computing $\int_{0}^{\infty}\frac{1}{x^{p}}\,dx$

Question

I got stuck at the following integral

Let $p>1$, then $$ \int_{0}^{\infty}\dfrac{1}{x^{p}}dx=\frac{p}{(p-1)}. $$

Could you suggest a helpful idea?

Answer 1

You can do that since $p>1$:

$$\int_0^\infty \frac 1{x^p}\mathrm d x=\left[\frac 1{-p+1}x^{-p+1}\right]_0^\infty=0-(-\infty)=+\infty.$$

So the integral is divergent and the result you gave is false.

Or you can use a comparaison to Riemann example (for $n\geqslant 0$):

$$\int_0^1\frac 1{x^n} \mathrm d x$$ is convergent if, and only if, $n<1$, and

$$\int_1^\infty\frac 1{x^n} \mathrm d x$$ is convergent if, and only if, $n>1$.

Answer 2

Although some of the other posts so far have the correct answer, none of them (as of the time I started this) are doing it the "right" way, i.e., the way you'll most likely need to show work if you don't want to lose any points (I'm assuming this is at or around the Calc II level, but even beyond that it's necessary — friends of mine lost points for not doing this in a graduate-level real analysis class). This is an improper integral, so it's important to be both rigorous and very careful here.

The "right" way is to do it with limits. We'll actually need two limits since both limits of integration cause the integral to be improper. In the following, $A \to 0^+$ means $A$ approaches $0$ only from the right.

\begin{align*} \int_0^{+\infty} \frac1{x^p} \, dx &= \lim_{A \to 0^+} \lim_{B \to +\infty} \int_A^B\frac1{x^p} \, dx \\[0.3cm] &= \lim_{A \to 0^+} \lim_{B \to +\infty} \int_A^B x^{-p} \, dx\\[0.3cm] &= \lim_{A \to 0^+} \lim_{B \to +\infty} \frac1{1-p} x^{1-p} \bigg|_{x=A}^{x=B} \end{align*} Note that $p$ is just a constant, so we can pull it (and expressions involving only it and other constants) out of both limits. Doing this and continuing further, we get \begin{align*} \lim_{A \to 0^+} \lim_{B \to +\infty} \frac1{1-p} x^{1-p} \bigg|_{x=A}^{x=B} &= \frac1{1-p} \lim_{A \to 0^+} \lim_{B \to +\infty} \left(B^{1-p} - A^{1-p}\right)\\[0.3cm] &= \frac1{1-p} \left(\lim_{B\to+\infty} B^{1-p} - \lim_{A \to 0^+}A^{1-p}\right) \end{align*} Now, since $p > 1$, then $1 - p < 0$ and so $B^{1-p} \to 0$ as $B \to +\infty$. Also, since $A \to 0^+$, then $A$ is always strictly positive and so $A^{1-p}$ is well defined for all $p > 1$. And because $1-p < 0$ then we have $A^{1-p} \to +\infty$ as $A \to 0^+$ $$ \frac1{1-p} \left(\lim_{B\to+\infty} B^{1-p} - \lim_{A \to 0^+}A^{1-p}\right) = +\infty $$ So the integral diverges and the result you wanted to show is not true.

Answer 3

The primitive of $\frac{1}{x^p} = x^{-p}$ for $p > 1$ can be found with the "usual formula" : $$ \int x^n = \frac{x^{n+1}}{n+1} $$ Therefore $$ \int \frac{1}{x^p} = \int x^{-p} = \frac{x^{-p+1}}{-p+1} = \frac{1}{1-p} \frac{1}{x^{p-1}} $$ Adding the bounds in we get that (Noting that $1-p$ < 0 by assumption) $$ \left [ \frac{1}{1-p} \frac{1}{x^{p-1}} \right]_{0^+}^\infty = \frac{1}{1-p} \left [ \frac{1}{x^{p-1}} \right]_{0^+}^\infty = 0 - (-\infty) = \infty $$

Answer 4

It looks like fake $$p>1\implies\int \frac{1}{x^p} =\int x^{-p}=\frac{x^{1-p}}{1-p}$$ Then, as $1-p<0$$$\lim_{x\rightarrow\infty}\frac{x^{1-p}}{1-p}=0$$ So $$\int_1^{\infty} \frac{1}{x^p}\mathrm{d}x=\frac{1}{p-1}$$ Therefore, we'd expect $$\int_0^1 \frac{1}{x^p}\mathrm{d}x=1$$ which is false because $\frac{1}{1^p}=1$ and $\frac{1}{x^p}>1,\forall x\in (0,1)$