My Math Forum Integration over a singularity

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 March 12th, 2019, 04:01 PM #1 Newbie   Joined: Mar 2019 From: USA Posts: 1 Thanks: 0 Integration over a singularity 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?
March 12th, 2019, 04:23 PM   #2
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Wolfram says the integral does not converge ...
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March 12th, 2019, 06:38 PM   #3
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Quote:
 Originally Posted by asaiz 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.

Last edited by skipjack; March 12th, 2019 at 09:12 PM.

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