Determine whether the improper integral is convergent or divergent. If it is convergent, evaluate it.?

Question

1
∫ ln(x) dx
0

Answer 1

D(lnx)=(0, +inf)

∫ln(x)·dx= I. Part[ u=ln(x)--> du=1/x·dx ; dv=dx --> v=x] =
x·ln(x) - ∫dx = x ·ln(x) - x + K = x · [ ln(x) -1 ] + K

lim_{x-->1} x · [ ln(x) -1 ] = 1·(0-1)= -1 =F(1)

lim_{x-->0+} x · [ ln(x) -1 ] = (0·Inf) Ind =
lim_{x-->0+} [ ln(x) -1 ] / (1/x) = (inf/inf) L'Hopital =
lim_{x-->0+} [1/x]/[-1/x^2] =
lim_{x-->0+} (-x) = 0 =F(0)

lim_{x-->+inf} x · [ ln(x) -1 ] = +inf


--> the improper integral is convergent in [0, K] with K>0

1
∫ ln(x) dx = F(1)-F(0)= -1
0

Saludos.

Answer 2

Int ln(x) dx = xln x -x
At 1 the value is -1 and the limit as x ==>0 is 0
as xlnx ==>0
You can see this by L´Hôpital
lim lnx/(1/x) = lim (1/x<) /(-1/X^2=-lim x=0
So the value of this improper integral is -1