reduction formula: J_n = [integral of] x^m * (lnx)^n * dx

[dts5044]

Let J[sub:2aplqv9g]n[/sub:2aplqv9g] = [integral of] x[sup:2aplqv9g]m[/sup:2aplqv9g] * (lnx)[sup:2aplqv9g]n[/sup:2aplqv9g] * dx where m is a real number not equal to 1 and n is a positive integer. Establish a reduction formula for J[sub:2aplqv9g]n[/sub:2aplqv9g], then deduce a formula for J[sub:2aplqv9g]n[/sub:2aplqv9g]

I used by parts:

u = (lnx)[sup:2aplqv9g]n[/sup:2aplqv9g], du = n * (lnx)[sup:2aplqv9g]n-1[/sup:2aplqv9g] * (1/x) * dx
dv = x[sup:2aplqv9g]m[/sup:2aplqv9g] * dx, v = x[sup:2aplqv9g]m+1[/sup:2aplqv9g]/(m+1)

so the integral is equal to: v*u - [integral of] v*du
= x[sup:2aplqv9g]m+1[/sup:2aplqv9g]/(m+1) * (lnx)[sup:2aplqv9g]n[/sup:2aplqv9g] - n/(m+1) [integral of] x[sup:2aplqv9g]m+1[/sup:2aplqv9g] * (lnx)[sup:2aplqv9g]n-1[/sup:2aplqv9g] * (1/x) * dx

= x[sup:2aplqv9g]m+1[/sup:2aplqv9g]/(m+1) * (lnx)[sup:2aplqv9g]n[/sup:2aplqv9g] - n/(m+1) [integral of] x[sup:2aplqv9g]m[/sup:2aplqv9g] * (lnx)[sup:2aplqv9g]n-1[/sup:2aplqv9g] * dx

= x[sup:2aplqv9g]m+1[/sup:2aplqv9g]/(m+1) * (lnx)[sup:2aplqv9g]n[/sup:2aplqv9g] - n/(m+1) * J[sub:2aplqv9g]n-1[/sub:2aplqv9g]

I think this is the reduction formula for J[sub:2aplqv9g]n[/sub:2aplqv9g], but can't figure out how to deduce a formula for J[sub:2aplqv9g]n[/sub:2aplqv9g]!

 

[dts5044]

no one has an answer?

 

[Dr. Flim-Flam]

J_n = integral {x^m*[ln(x)]^n}dx. Use integration by parts.

integral(u)dv = uv - integral(v)du.

Let u = (lnx)^n, then du ={[ n(lnx)^(n-1)]/x}dx, let dv =(x^m)dx, then v = x^(m+1)/(m+1)

Hence J_n = [(lnx)^n*x^(m+1)]/(m+1) -[n/(m+1)]integral [x^m*(lnx)^(n-1)]dx