How do you determine if the improper integral converges or diverges #int ln(x)dx# from 0 to 2?

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Answer 1 · 2016-09-12T19:31:11

It converges to #2ln(2)-2#.

Note that #intln(x)dx=xln(x)-x+C#, which is solved through integration by parts here and I'm sure other places on Socratic.

So, applying this on the interval #[0,2]#:

#int_0^2ln(x)dx=[xln(x)-x]_0^2#

We cannot evaluate #ln(0)#, so take the limit of that portion:

#=(2ln(2)-2)-lim_(xrarr0)(xln(x)-x)#

The limit can be simplified in regard to the subtracted #x#:

#=(2ln(2)-2)-lim_(xrarr0)xln(x)#

This can be rewritten (rather sneakily) to prime for l'Hopital's rule:

#=(2ln(2)-2)-lim_(xrarr0)ln(x)/(1/x)#

The limit is in the indeterminate form #oo/oo#, so apply l'Hopital's by taking the derivative of the numerator and denominator:

#=(2ln(2)-2)-lim_(xrarr0)(d/dxln(x))/(d/dx(1/x))#

#=(2ln(2)-2)-lim_(xrarr0)(1/x)/(-1/x^2)#

#=(2ln(2)-2)-lim_(xrarr0)(-x)#

The limit can now be evaluated:

#=(2ln(2)-2)-0#

#=2ln(2)-2#