Integration over a singularity

Mar 2019
4
0
USA
On a recent assignment, my engineering professor gave us a list of functions to treat as integrands and corresponding ranges to integrate them over. We were asked to answer yes or no as to whether they each had a numerical answer. One of the integrands was \(\displaystyle y = x / (1 - x)^3\) with the range \(\displaystyle x=0\) to \(\displaystyle x=2\). My response was “no” because the integrated function has a singularity at \(\displaystyle x=1\). He later posted the solutions for the assignment and it said that the answer was “yes” because you can use the substitution \(\displaystyle u=1-x\). I was under the impression that making such a substitution would not change the fact that the integral diverges in the given range; is my thinking correct or is the teacher correct in saying it has a numerical answer?
 

SDK

Sep 2016
804
545
USA
On a recent assignment, my engineering professor gave us a list of functions to treat as integrands and corresponding ranges to integrate them over. We were asked to answer yes or no as to whether they each had a numerical answer. One of the integrands was \(\displaystyle y = x / (1 - x)^3\) with the range \(\displaystyle x=0\) to \(\displaystyle x=2\). My response was “no” because the integrated function has a singularity at \(\displaystyle x=1\). He later posted the solutions for the assignment and it said that the answer was “yes” because you can use the substitution \(\displaystyle u=1-x\). I was under the impression that making such a substitution would not change the fact that the integral diverges in the given range; is my thinking correct or is the teacher correct in saying it has a numerical answer?
You are correct that this integral diverges and your professor is wrong. It should be pointed out, though, that the integrand having a singularity doesn't automatically mean the integral doesn't converge. It's more subtle than that.


You are also correct that a change of variables can't turn a divergent integral into a convergent one. If you change to $u = 1-x$ then you just get
\[ \int_{-1}^1 \frac{u-1}{u^3} \ du \]
which is, as you noticed, still divergent.
 
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