Pn = integral (lnx)^n and Pn = x(lnx)^n - n(Pn-1). Hence find P3.?

Question

n> or equal to 1
x>0

Answer 1

n=1: x*(log(x) - 1)

n=2: x*log(x)^2 - 2*x(log(x)-1) = x*(log(x)^2 - 2*log(x) + 2)

n=3: x*log(x)^3 - 3*(x*(log(x)^2 - 2*log(x) + 2))
= x*(log(x)^3 -3*log(x)^2 + 6*log(x) - 6)

in general = x * a polynomial in log(x)

Answer 2

This is called a "reduction formula". It is proved by a simple application of integration by parts. But don't we need some limits of integration? I suggest that you need [1,x] for limits.

Then P3 = x(lnx)^3 - 3P2,
P2 = x(lnx)^2 - 2P1
P1 = x(lnx) - P0.

What is P0?
P0 = integral[lnx]^0 = x -1

and work back to P3.

The notation is misleading; if I am right it is better to define

Pn = integral[1,x] {lnt}^n .dt