Evaluate the integral using the types of improper integrals.

Question

∫(1 to ∞)([ln(x)]/x²)dx

I started off by using the method of improper integrals:

∫(1 to ∞)([ln(x)]/x²)dx = lim (t→∞) ∫(1 to t)([ln(x)]/x²)dx

I then used integration by parts with u=lnx and dv=1/x²dx, getting du=1/xdx and v=-1/x:

∫(1 to t)([ln(x)]/x²)dx=(-1/x)(lnx) |₁^t= ∫(1 to t) (-1/x)(1/xdx)

=-lnx/x |₁^t - ∫(1 to t) -1/x²dx

Evaluating the integral: ∫(1 to t) -1/x²dx = 1/x |₁^t

So lim (t→∞) (-lnt/t + ln1/1)-(1/t-1)

And since the limits of lnt/t and 1/t goes to 0 and ln(1)=0, I got convergent; 1 as my answer.

Answer 1

∫₁^∞lnx/x²dx = lim A→∫₁^Alnx/x²dx = lim A→∞ (- lnx/x₁^A+ ∫₁^A1/x²dx) = lim A→∞ (- lnx/x - 1/x)₁^A =

lim A→∞[(- lnA/A - 1/A) + (ln1/1 + 1/1)] = 1