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!