How do you evaluate the integral #int ln(x-1)# from 1 to 2?

1 Answer
Jim H
Dec 8, 2016

Please see the explanation section below.

Explanation:

#int_1^2 ln(x-1) dx# is an improper integral because the integrand is not defined at one of the limits of integration. (#ln(x-1)# is not defined for #x=1#.)

Before attempting to evaluate this improper integral, let's change the variable of integration. (Let's do a substitution.)

Let #t = x-1#, so that #dt = dx# and when #x=1#, the #t = 0# and when #x=2#, then #t=1#

Our integral becomes

#int_0^1 lnt dt = lim_(ararr0) int_a^1 lnt dt# #" "# if the limit exists.

We need #int lnt dt# which may be found by integration by parts. with #u = lnt# and #vd = dt#

#int lnt dt = tlnt-t +C#

Returning to the main question,

#int_0^1 lnu du = lim_(ararr0^+) int_a^1 lnt dt#

# = lim_(ararr0^+) [t ln t -t]_a^1#

# = lim_(ararr0^+) [(0-1) - (a ln a -a)]#

# = -1 - lim_(ararr0^+) (a ln a-a)#

# = -1 - lim_(ararr0^+) (a ln a)#

To evaluate #lim_(ararr0^+) (a ln a)#, we'll use l"Hospital's Rule.

#lim_(ararr0^+) (a ln a)# has initial form #0*-oo#, so we need to rewrite

# = lim_(ararr0^+) (ln a/(1/a))# has form #(-oo)/oo# so we try

# = lim_(ararr0^+) ((1/a)/(-1/a^2)) = lim_(ararr0^+) (-a) = 0 #.

So, #int_0^1 lnu du = -1 - lim_(ararr0^+) (a ln a) = -1#

And finally,

#int_1^2 ln(x-1) dx = -1# .

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