What is the standard convention for computing something like

$$\int_0^3 \frac{1}{\sqrt{1-x}}\,\mathrm{d}x? $$

Is it equivalent, since the integrand is not defined on $(1,\infty)$, to $$\int_0^1 \frac{1}{\sqrt{1-x}} \, \mathrm{d}x$$ where one can use the standard technique for an improper integral, or does it simply not exist?

  • 1
    $\begingroup$ Hint: You can't integrate a nonexistent function. So the second one is what you want. We may take the first as a clumsy notation for the second. $\endgroup$
    – Allawonder
    Feb 1, 2018 at 6:49
  • 1
    $\begingroup$ I think that people just don't compute such things. I don't think that they would encounter such integral on the reals. If you encounter it on the reals then all kinds of bells must ring. There is something messed up and you need to find out what. $\endgroup$
    – Shashi
    Feb 1, 2018 at 7:05

1 Answer 1


For $x>1$, the term $ \frac{1}{\sqrt{1-x}}$ is not defined. Hence $\int_0^3 \frac{1}{\sqrt{1-x}}\,\mathrm{d}x$ is nonsense.


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