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 March 12th, 2019, 04:01 PM #1 Newbie   Joined: Mar 2019 From: USA Posts: 4 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 ...
Attached Images AC4414CF-C3A4-4D67-980E-FEFDA60EA81D.jpg (13.6 KB, 5 views) 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. Tags integration, singularity Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Loren Number Theory 4 January 18th, 2017 02:56 PM Loren Number Theory 3 December 22nd, 2016 04:27 AM honzik Real Analysis 1 January 26th, 2016 12:00 PM aaron-math Complex Analysis 3 October 16th, 2013 04:51 PM mathbalarka Calculus 5 April 14th, 2012 01:48 PM

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