Is this integral improper? If yes - why? $$ \int\limits^2_0 \,\frac{1}{x-1} dx $$

  • 1
    $\begingroup$ at x=1 function $\dfrac{1}{x-1}\;$ is undefined $\endgroup$ Jun 10, 2013 at 13:48
  • 2
    $\begingroup$ Hint: Check the definition of improper integral. $\endgroup$
    – Git Gud
    Jun 10, 2013 at 14:21

3 Answers 3



The integral $\int_a^b f(x)dx$ is called improper integral if:

  • $a=+\infty$ or $b=\infty$ or both.

  • $f(x)$ is unbounded at one or more points of $a\le x\le b$.

As @Git suggested verify which ones of above is satisfying the definition. You'll get the answer. ;-)

enter image description here

  • $\begingroup$ Very nicely argued (just the right nudges needed!) +1 $\endgroup$
    – amWhy
    Jun 11, 2013 at 0:05
  • $\begingroup$ @amWhy: Thanks my dear Amy. Yes indeed. :-) $\endgroup$
    – Mikasa
    Jun 11, 2013 at 0:07

It is improper because the function "blows up" between the end points. That is, the function approaches $\pm \infty$ because the denominator is 0.


An improper integral is the limit of a definite integral as an endpoint of the interval(s) of integration approaches either a specified real number or ∞ or −∞ or, in some cases, as both endpoints approach limits.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.