# How do integrals relate to converging and diverging series?

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

$$\int_0^1{\frac{1}{x^n}}dx$$

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.

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

• 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? – Addison Jan 15 at 20:16
• I added to my answer. I hope it's useful :) – Emilio Novati Jan 15 at 20:28
• 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. – Addison Jan 15 at 20:47
• Note: $-\frac{1}{1-n}=\frac{1}{n-1}$ and $\frac{1}{n-1}$ is a common factor. – Emilio Novati Jan 15 at 20:53
• 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? – 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

• I don't see how you were able to do this step: $\int_{0}^1 x^{-n} \$dx = $\int_1^{\infty} y^{n-2} \ dx$ – Addison Jan 15 at 19:30
• 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. – Doug M Jan 15 at 19:38