Reduction formula for ∫(lnx)^n dx

[privyet]

Homework Statement

derive a reduction formula for ∫(lnx)ⁿ dx and use it to evaluate ∫₁^e (lnx)³dx

Homework Equations

The Attempt at a Solution

In other examples we've started by saying ∫(lnx)ⁿdx = ∫(lnx)(lnx)^n-1dx and using integration by parts. So let:
f = (lnx)^n-1
f' = (n-1)(lnx)^n-2*1/x
g = x(lnx - 1)
g' = lnx

then ∫(lnx)ⁿdx
= (lnx)^n-1x(lnx - 1) - ∫(n-1)(lnx)^n-2(lnx - 1) dx
= (lnx)^n-1x(lnx - 1) - (n-1)∫(lnx)^n-1 - (lnx)^n-2 dx
= (lnx)^n-1x(lnx - 1) - (n-1)∫(lnx)^n-1 + (n-1)∫(lnx)^n-2

at this point I get stuck because in all the examples I have seen we get minus the initial integral on the right and side and can add it to both sides to get n∫lnx)ⁿ dx = ... so ∫lnx)ⁿ dx =1/n[...]

Any help would be very appreciated.

 

[haruspex]

If $I_n=\int\ln(x)^n.dx$ then you have found the recurrence relation $I_n=\ln(x)^{n-1}x(\ln(x) - 1) -(n-1)I_{n-1}+(n-1)I_{n-2}$.
You are not asked to solve it, except in the special case cited.

 

[nuuskur]

If $I_n=\int\ln(x)^n.dx$ then you have found the recurrence relation $I_n=\ln(x)^{n-1}x(\ln(x) - 1) -(n-1)I_{n-1}+(n-1)I_{n-2}$.
You are not asked to solve it, except in the special case cited.

Careful. $\ln (x)^n \neq (\ln x)^n$.

@privyet
You have solved the problem for $n\geqslant 2$ where $I_0 :=x$ may be assumed. Now use this recursive relation to evaluate the integral you need.

 

[haruspex]

Careful. $\ln (x)^n \neq (\ln x)^n$.

Really? I did not write $\ln (x^n)$. I would have thought the parentheses defined what is the argument to the function.

 

[pasmith]

By setting $$\begin{split} I_n &= \int (\ln x)^n\,dx \\ &= \int x \frac{(\ln x)^n}{x}\,dx \\ &= \int x \frac{1}{n+1}\frac{d}{dx} (\ln x)^{n+1}\,dx \\ &= \frac{1}{n+1}\left[x(\ln x)^{n+1}\right] - \frac1{n+1}I_{n+1} \\ \end{split}$$ (which is a standard trick for dealing with powers of $\ln x$) I obtain the first-order recurrence $$I_{n} = x(\ln x)^n - nI_{n-1},\qquad I_0 = x.$$ This seems easier to work with than the second order recurrence given by @haruspex in #2 based on the OP's approach. That second order recurrence can be obtained from the above by subtracting $$I_{n-1} = x(\ln x)^{n-1} - (n-1)I_{n-2}$$ from both sides.