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

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

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Definition:

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. ;-)

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

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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.

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