What is the integral from 0 to 4 of lnx dx?

1 Answer
May 1, 2015

For #int_0^4lnx*dx#, it converges by a sum of #4ln4-4#.

Let's introduce the idea of improper integrals. Remember that integrals are based on the sums of the individual terms as shown in the Sigma notation below:

#int_a^bf(x)dx = lim_(n to oo)sum_(i=0)^nf(x_i)Deltax_i#, for #Deltax =(b-a)/n#,
(From http://www.math.wvu.edu/~hjlai/Teaching/Tip-Pdf/Tip1-29.pdf)

where #a# is the first term, #b# is the last term, and #n# is the number of "parts." Think of it as going from a to b by splitting individual rectangles of area and adding them up.

According to the function #lnx#, if you insert the first term #0# for x, you get no solution since the line on the graph below continues going down to #-oo#, making #x<=0# undefined:
graph{lnx [-7.87, 12.13, -4.13, 5.875]}
Using the Type 2 definition of improper integrals, let's instead make #0# as a #t# limit variable for the integral as #t to 0# from the right (#0^+#):

(1) #lim_(t to 0^+)[int_t^4lnx*dx]#

To integrate #lnx#, integration by parts comes in handy if you set it as #1*lnx# using the equation:

(2) #intuv'=uv-intvu'# (the prime indicates the derivative).

Let #u=lnx#, and #v'=1*dx#. By differentiating the #u# and integrating the #v'#,

#u'=1/x*dx# and #v=x#.

Substituting the #u# and #v# variables for Equation 2,

#int (lnx*1)dx=xlnx-int 1*dx#.

This allows us to easily integrate #lnx#, only now we need to evaluate its definite integral from #t to 4#:

(3) #int_t^4lnx*dx=[lnx(x)~|_t^4-int_t^4 1*dx=(4ln4-tlnt)-(4-t)#

Now we can do the math using Eqs. 1 and 3 and split the limit:

#lim_(t to 0^+)[int_t^4lnx*dx] = lim_(t to 0^+)[4ln4-tlnt-4+t]#

=#lim_(t to 0^+)(-tlnt)+lim_(t to 0^+)(4ln4-4+t)#

=#lim_(t to 0^+)(-tlnt)+4ln4-4#

Notice though that #lim_(t to 0^+)(-tlnt)# gives us a #-0*oo# indeterminate form.
Rewriting it would allow us to use l'Hôpital's Rule to find the limit:

#lim_(t to 0^+)(-lnt/(1/t))=>-oo/oo=>#indeterminate form. So by LHR,
#lim_(t to 0^+)(-lnt/(1/t))=lim_(t to 0^+)((-1/cancel(t))/(-1/t^cancel(2)))=lim_(t to 0^+)(t)=0#

Thus, #int_0^4lnx*dx=4ln4-4#.

Hopefully it makes sense with such a long text!