What is the integral of #int (lnx)^2/x dx# from 0 to 2?

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Answer 1 · 2016-02-18T14:33:08

That improper integral diverges.

The integrand is not defined at the lower limit of integration, so the integral is improper.

If you have not yet learned aout improper integrals, the best answer might be "Is not defined (yet)"

If you have learned about improper integrals, read on
(Or if you want to learn about them, read on.)

To attempt to evaluate the interval we use the limit:

# int_0^2 (lnx)^2 1/x dx = lim_(ararr0^+) int_a^2 (lnx)^2 1/x dx#

By substitution, we get #int (lnx)^2 1/x dx = 1/3 (lnx)^3 color(gray)(+C)#

So

#lim_(ararr0^+) int_a^2 (lnx)^2 1/x dx = lim_(ararr0^+) 1/3(lnx)^3]_a^2#

# = 1/3 lim_(ararr0^+) [(ln2)^3-(lna)^3]#

# = 1/3 [(ln2)^3 - lim_(ararr0^+) (lna)^3#}

But, as #ararr0^+#, we know that #lna rarr -oo#. So, the limit above fails to exist.

The integral also fail to exist. We say the integral diverges. (Some prefer to say, "The integral does not converge".)