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

Question

Ezekiel Alexander · Accepted Answer

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#

Scarlett Allison · Answer

undefined To determine if the improper integral ∫ ln(x) dx from 0 to 2 converges or diverges, we evaluate the integral and check if the result is finite or infinite.

$$∫ ln(x) dx$$ from 0 to 2

$$= \lim_{a \to 0^+} ∫_{a}^{2} ln(x) dx$$

$$= \lim_{a \to 0^+} [x ln(x) - x]_{a}^{2}$$

$$= \lim_{a \to 0^+} [(2 ln(2) - 2) - (a ln(a) - a)]$$

$$= (2 ln(2) - 2) - \lim_{a \to 0^+} (a ln(a) - a)$$

$$= (2 ln(2) - 2) - \lim_{a \to 0^+} a(ln(a) - 1)$$

As $a$ approaches 0 from the right, $ln(a)$ approaches negative infinity, so $a(ln(a) - 1)$ approaches negative infinity. Therefore, the limit of the integral as $a$ approaches 0 from the right is negative infinity.

Since the integral evaluates to negative infinity, the improper integral ∫ ln(x) dx from 0 to 2 diverges.