Because #ln(0)# is not defined (does not exist), #0# is not in the domain of the integrand.
This is an improper integral, so we use:
#int _0^1 x ln(x) dx = lim_(a rarr 0) int _a^1 x ln(x) dx#
Use integration by parts (with #u = lnx# and #dv = x#) to get:
#lim_(a rarr 0) int _a^1 x ln(x) dx = lim_(a rarr 0) (1/2 x^2 lnx - 1/4 x^2]_a^1) #
Recalling that #ln(1)=0#, we get:
#int _0^1 x ln(x) dx = lim_(a rarr 0) ((0-(1/4))- a^2/2 lna - 1/4 a^2)#
#lim_(a rarr 0) 1/4 a^2 = 0#,
But, #lim_(a rarr 0) ( a^2/2 lna )# has indeterminate form: #0 * -oo#.
Rewriting as: #lna/(2/a^2)# Which has indeterminate form: #(-oo)/oo#.
Applying l'Hopital's Rule gives us:
#lim_(a rarr 0) ( a^2/2 lna ) = lim_(a rarr 0) lna/(2/a^2) = lim_(a rarr 0) (1/a)/((-4)/a^3) = lim_(a rarr 0)(a^2 / -4) = 0#
So, we conclude:
#int _0^1 x ln(x) dx = -1/4 - 0 - 0 = -1/4#