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I'm asked to investigate this integral,

$$\int_0^3 \frac{1}{(1-x)^2}\,\mathrm{d}x$$

I get -3/2, but the integrand has a discontinuity at $x=1$, so I can't just integrate it. What is the answer? And why do things go wrong when integrating over a discontinuity?

Also what about $$\int_{-1}^{1} \frac{1}{x}\,\mathrm{d}x$$

When evaluated it gives 0, but this is correct since the areas cancel.

user85798
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  • For the first one you probably used the second part of the fundamental theorem of calculus. It requires that the function is riemann integrable over the interval. Yours aren't riemann integrable. – Git Gud Dec 01 '13 at 15:54
  • If you use the Riemann integral, a function needs to be bounded on the interval in order to be integrable over that interval. If you want to integrate over a discontinuity, you have to take limits of integrals (i.e. split the integral from -1 to epsilon and epsilon to 1 and take the limit epsilon to 0 seperately per integral). This limit doesn't exist for the second integral, so the improper Riemann integral doesn't exist. – Pol van Hoften Dec 01 '13 at 15:54
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    Seeing the negative result $-\frac{3}{2}$ for an integral of a function that is positive (everywhere where it is defined at all) is enough to say that something is very wrong. As was pointed out in Git Gud's comment, the function isn't integrable. When understood as an improper integral, the integral diverges to infinity. – Dan Shved Dec 01 '13 at 15:59

1 Answers1

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It is not so much the discontinuity of the integrand that is of issue as the discontinuity of the naïvely computed antiderivative, i.e. $F(x)=1/(1-x)$. The theorem that $$\int_a^b F'(x)\,dx=F(b)-F(a)$$ requires at the very least that $F$ be continuous on $[a,b]$, and that is not the case here.

The second integral you mention can be fruitfully considered to be zero by its Cauchy principal value. Whether that is appropriate or not depends on the intended use for the integral.

Harald Hanche-Olsen
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