When $n$ is an integer greater than $1$,


The answer key says the answer is $\infty$.

Now I have experience with integration, finding convergence/divergence of series, etc. But, I'm not seeing the connection with series. $\frac{1}{x^n}$ is clearly the form of the power series, but how do I connect it with this integral?

This leads to a more general question about the relationship between integrals and series, which I have searched for. I found some decent answers here, but I am certainly open to any thoughts about that topic.

Anyways, let me know what you think about solving that integral at the top.

  • $\begingroup$ Do you know the power rule for integrals? $\endgroup$ – Eric Towers Jan 15 at 19:26
  • $\begingroup$ @EricTowers yes, and when I did that I got $\frac{x^{-n+1}}{-n+1}$ from 1 to 0 $\endgroup$ – Addison Jan 15 at 19:28

$\int_0^1{\frac{1}{x^n}}dx$ is an improper integral, because for $x \to 0$ we have $\frac{1}{x^n} \to \infty$.

So, by definition, the value of the integral for $n>1$ is given by:

$$ \int_0^1{\frac{1}{x^n}}dx= $$ $$ = \frac{1}{1-n}(1)^{1-n}-\lim_{x \to 0}\frac{1}{1-n}x^{1-n}=\frac{1}{n-1}\left[\lim_{x \to 0}\;(x^{1-n})-1 \right] $$ that, with the substitution $y=1/x$ becomes: $$ =\frac{1}{n-1}\left[\lim_{y \to \infty}\;(y^{n-1})-1 \right] $$ Note that here we have a limit that is the limit of a sequence, not the sum of a series, and, for $n>1$, this sequence is clearly divergent.

The last step comes from: $$ \lim_{x \to 0}(x^{1-n}) $$ $$ =\lim_{x \to 0}\frac{1}{x^{n-1}} $$ $$ =\lim_{x \to 0}\left(\frac{1}{x}\right)^{n-1}= $$ and the substitution: $$ \frac{1}{x}=y $$ for which we have that if $ x\to 0$ than $y \to \infty$.

  • $\begingroup$ So I see how in the first step you simply integrated and plugged in the limits. I see you set the second term, as a limit, and that's fair enough. However, I don't see how you were able to get rid of the first term as you went to the next step. Also why does the denominator turn from 1-n to n-1? $\endgroup$ – Addison Jan 15 at 20:16
  • $\begingroup$ I added to my answer. I hope it's useful :) $\endgroup$ – Emilio Novati Jan 15 at 20:28
  • $\begingroup$ sorry, I'm still having trouble getting this. So I see that you integrated and put it into two terms: $\frac{1}{1-n}(1)^{1-n}-\lim_{x \to 0}\frac{1}{1-n}x^{1-n}$. I'm guessing the reason the denominator of $\lim_{x \to 0}\frac{1}{1-n}x^{1-n}$ becomes n-1 is because of the negative in front of the limit, making things negative. But then how did you get the -1 in $\frac{1}{n-1}\left[\lim_{x \to 0}\;(x^{1-n})-1 \right]$. The term $\frac{1}{1-n}(1)^{1-n}$ doesn't simplify to -1. $\endgroup$ – Addison Jan 15 at 20:47
  • $\begingroup$ Note: $-\frac{1}{1-n}=\frac{1}{n-1}$ and $\frac{1}{n-1}$ is a common factor. $\endgroup$ – Emilio Novati Jan 15 at 20:53
  • $\begingroup$ So what happens to the exponent of 1-n ? Doesn't that make the sign change? How can we factor out $\frac{1}{n-1}$ if 1 is alternating? $\endgroup$ – Addison Jan 15 at 20:55

let $y = x^{-1}$

$dy = -x^{-2} \ dx\\ dx = - y^{-2} dy$

$\int_{0}^1 x^{-n} \ dx = \int_1^{\infty} y^{n-2} \ dx $

If $n \ge 2$ then the integral is clearly divergent as $\lim_\limits {y\to \infty} y^{n-2}\ne 0$

If $n < 2$ then $y^{n-2}$ is montonic and decreasing.

$\sum_\limits {k=1}^{\infty} (k^{n-2}) \le\int_1^{\infty} y^{n-2} \ dy\le \sum_\limits {k=2}^{\infty} (k^{n-2})$

if $n\ge1$

$\sum_\limits {k=1}^{\infty} (k^{n-2})$ diverges

if $n < 1$

$\sum_\limits {k=2}^{\infty} (k^{n-2})$ converges

  • $\begingroup$ I don't see how you were able to do this step: $\int_{0}^1 x^{-n} \ $dx = $\int_1^{\infty} y^{n-2} \ dx$ $\endgroup$ – Addison Jan 15 at 19:30
  • $\begingroup$ The limits of integration, $\lim_\limits {x\to 0^+} y(x) = \infty, y(1) = 1.$ Plugging all of these we get $\int_{\infty}^1 (y^n)(-y^{-2}) \ dy. $ We can flip the limits of integration if we also flip the sign. $\endgroup$ – Doug M Jan 15 at 19:38

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