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 $\displaystyle u=1x$. 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 
Math Team Joined: Jul 2011 From: Texas Posts: 2,838 Thanks: 1481 
Wolfram says the integral does not converge ...

March 12th, 2019, 06:38 PM  #3  
Senior Member Joined: Sep 2016 From: USA Posts: 578 Thanks: 345 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
You are also correct that a change of variables can't turn a divergent integral into a convergent one. If you change to $u = 1x$ then you just get \[ \int_{1}^1 \frac{u1}{u^3} \ du \] which is, as you noticed, still divergent. Last edited by skipjack; March 12th, 2019 at 09:12 PM.  

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