# Differing (divergent) limits in the improper integral $\int_{-1}^{3}x^{-3}dx$

In a typical Calculus class, the improper integral $$\int_{-1}^{3}x^{-3}dx$$ is to be split up as $$\lim_{b \rightarrow 0^{-}}\int_{-1}^{b} x^{-3} dx + \lim_{b \rightarrow 0^{-}} \int_{b}^{3} x^{-3} dx,$$ and we arrive at the sum-of-limits $$\lim_{b \rightarrow 0^{-}}\big( -\frac{1}{2b^{2}} + \frac{1}{2} \big) + \lim_{b \rightarrow 0^{+}}\big( -\frac{1}{18} + \frac{1}{2b^{2}} \big).$$ Now, in each of these limits, it seems to me that the "$+$" and "$-$" don't matter on the $0$ since $b^{2}$ is a square. So we can write these limits as $\lim_{b \rightarrow 0}$ instead, and "pulling out the limit", we get $$\lim_{b \rightarrow 0}\big( -\frac{1}{2b^{2}} + \frac{1}{2} - \frac{1}{18} + \frac{1}{2b^{2}} \big) = \lim_{b \rightarrow 0} \frac{4}{9} = \frac{4}{9}.$$

My instructor says that the integrals have to be convergent independently in order for the whole improper integral to converge, and so this whole integral diverges. But the problem I have is that one limit is $\infty$ while the other is $-\infty$. So we get something like $\infty - \infty$, which I guess doesn't make much sense. And the fact that we have a $b^{2}$ in both limits makes me think that we don't have to worry about which side the limit is coming from.

Am I missing anything?

• No, you are correct. The integral does not converge. – Biggs Mar 10 '17 at 4:24

The fact that $\infty-\infty$ doesn't make sense is precisely why we say that the integral is divergent.

Note that you should use different variable names for both limits. You should write $$\lim_{a \rightarrow 0^{-}}\bigg(-\frac{1}{2a^{2}} + \frac{1}{2} \bigg) + \lim_{b \rightarrow 0^{+}}\bigg( -\frac{1}{18} + \frac{1}{2b^{2}} \bigg)$$ and now it is clear that you can't "pull out the limit".

What you are calculating has a name though. It is the Cauchy principal value.

The improper integral $\int_{-1}^3 x^{-3}\,dx$ fails to exist. Note that

$$\int_{-1}^{-\epsilon_1}x^{-3}\,dx=\frac12-\frac{1}{2\epsilon_1^2} \tag 1$$

and

$$\int_{\epsilon_2}^3x^{-3}\,dx=\frac{1}{2\epsilon_2^2}-\frac1{18}\tag 2$$

and neither the integral in $(1)$ not the one in $(2)$ converges as $\epsilon_1\to 0^+$ and $\epsilon_2\to 0^+$, respectively.

However, the Cauchy Principal Value integral,

$$\text{PV}\int_{-1}^3x^{-3}\,dx=\lim_{\epsilon\to 0^+}\left(\int_{-1}^{-\epsilon}x^{-3}\,dx+\int_{\epsilon}^3x^{-3}\,dx\right)=\frac12-\frac{1}{18}=\frac49$$

exists.

• Mike, please let me know how I can improve this answer too. As always, I really want to give you the best answer I can. If the answer was not useful, I am happy to delete it. -Mark – Mark Viola Mar 29 '17 at 19:57
• Mike, please let me know how I can improve this answer too. As always, I really want to give you the best answer I can. If the answer was not useful, I am happy to delete it. -Mark – Mark Viola Apr 12 '17 at 17:10