# Integration over a singularity

#### 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 â€œnoâ€ 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 â€œyesâ€ 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?

#### skeeter

Math Team
Wolfram says the integral does not converge ...

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#### SDK

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 â€œnoâ€ 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 â€œyesâ€ 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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