integral: (lnx)^n * dx from 0 to 1

[dts5044]

8. Let I[sub:3qzcjcp0]n[/sub:3qzcjcp0] = [integral from 0 to 1] (lnx)[sup:3qzcjcp0]n[/sup:3qzcjcp0] * dx, where n is a positive integer. Establish a reduction formula for I[sub:3qzcjcp0]n[/sub:3qzcjcp0], then deduce a formula for I[sub:3qzcjcp0]n[/sub:3qzcjcp0] and prove it by induction.

I think I got it but if someone could check my work that'd be great

first rewrite the integral as [integral from s to 1] (lnx)[sup:3qzcjcp0]n[/sup:3qzcjcp0] * dx
when s is approaching 0[sup:3qzcjcp0]+[/sup:3qzcjcp0]


taking the integral by parts:
u = (lnx)[sup:3qzcjcp0]n[/sup:3qzcjcp0], du = n(lnx)[sup:3qzcjcp0]n-1[/sup:3qzcjcp0] * 1/x *dx
dv = dx, v = x

so it is equal to x(lnx)[sup:3qzcjcp0]n[/sup:3qzcjcp0] {value at 1 minus value at s} - n[integral from s to 1] (lnx)[sup:3qzcjcp0]n-1[/sup:3qzcjcp0] * dx

x(lnx)[sup:3qzcjcp0]n[/sup:3qzcjcp0] {value at 1 minus value at s} approaches 0-0 as s approaches 0[sup:3qzcjcp0]+[/sup:3qzcjcp0] (I did the work and know this to be correct)
so we are left with (-n) * I[sub:3qzcjcp0]n-1[/sub:3qzcjcp0] which is the reduction formula

so:
I[sub:3qzcjcp0]n[/sub:3qzcjcp0] = -(n) * I[sub:3qzcjcp0]n-1[/sub:3qzcjcp0]
I[sub:3qzcjcp0]n-1[/sub:3qzcjcp0] = -(n-1) * I[sub:3qzcjcp0]n-2[/sub:3qzcjcp0]
I[sub:3qzcjcp0]n-2[/sub:3qzcjcp0] = -(n-2) * I[sub:3qzcjcp0]n-3[/sub:3qzcjcp0]

so I[sub:3qzcjcp0]n[/sub:3qzcjcp0]= -(n) * -(n-1) * -(n-2)... * -(n-(n+1)) * I[sub:3qzcjcp0]0[/sub:3qzcjcp0]
= (-1)[sup:3qzcjcp0]n[/sup:3qzcjcp0] * n! (I[sub:3qzcjcp0]0[/sub:3qzcjcp0] = 1) this is the formula for I[sub:3qzcjcp0]n[/sub:3qzcjcp0]

by induction: for n = 1 I[sub:3qzcjcp0]n[/sub:3qzcjcp0] = [integral from s to 1] lnx * dx = -1 (after going through by parts) = (-1)[sup:3qzcjcp0]1[/sup:3qzcjcp0] * 1!

assume I[sub:3qzcjcp0]n[/sub:3qzcjcp0] = (-1)[sup:3qzcjcp0]n[/sup:3qzcjcp0] * n!
for n+1: I[sub:3qzcjcp0]n+1[/sub:3qzcjcp0] = -(n+1) * I[sub:3qzcjcp0]n[/sub:3qzcjcp0] (by the reduction formula)
so I[sub:3qzcjcp0]n+1[/sub:3qzcjcp0] = -(n+1) * (-1)[sup:3qzcjcp0]n[/sup:3qzcjcp0] * n!
= (-1)[sup:3qzcjcp0]n+1[/sup:3qzcjcp0] *(n+1)!

so the statement I[sub:3qzcjcp0]n[/sub:3qzcjcp0] = (-1)[sup:3qzcjcp0]n[/sup:3qzcjcp0] * n! is proven for all integers n >= 1

Sorry, I don't know how to type in a more easily readable format. Can someone double-check my work quickly?

 

[tkhunny]

approaches 0-0

What does that mean?

know this to be correct

Why do you not know this of the entire problem?

so we are left with (-n) * I[sub:1vgsdxev]n-1[/sub:1vgsdxev] which is the reduction formula

Where did the integral go?

so I[sub:1vgsdxev]n[/sub:1vgsdxev]= -(n) * -(n-1) * -(n-2)... * -(n-(n+1)) * I[sub:1vgsdxev]0[/sub:1vgsdxev]
= (-1)[sup:1vgsdxev]n[/sup:1vgsdxev] * n! (I[sub:1vgsdxev]0[/sub:1vgsdxev] = 1) this is the formula for I[sub:1vgsdxev]n[/sub:1vgsdxev]

Once you get it down to \(\displaystyle \int ln(x) dx\), or n = 1, does it still work?

 

[dts5044]

1. x(lnx)[sup:19nls277]n[/sup:19nls277] approaches 0 - 0 when it is evaluated at s then subtracted from the value at 1

1*(ln1)[sup:19nls277]n[/sup:19nls277] = 0

s * (lns)[sup:19nls277]n[/sup:19nls277] = 0

so it is 0 - 0 = 0

2. I know s (lns)[sup:19nls277]n[/sup:19nls277] as s approaches 0[sup:19nls277]+[/sup:19nls277] approaches 0 because I am comfortable with doing limits and did the work to figure this out. (You rewrite as (lns)[sup:19nls277]n[/sup:19nls277]/ (1/s) and apply L'Hospital's rule n times...) I do not know my whole problem is correct because it is my first attempt at a problem involving reduction formulas, whereas I have had two years working with limits and can do them easily.

3. by my definition of I[sub:19nls277]n[/sub:19nls277] ( I[sub:19nls277]n[/sub:19nls277] = [integral from 0 to 1] (lnx)[sup:19nls277]n[/sup:19nls277] * dx)
I[sub:19nls277]n-1[/sub:19nls277] = [integral from 0 to 1] (lnx)[sup:19nls277]n-1[/sup:19nls277] *dx
so the integral is still there, as the term I[sub:19nls277]n-1[/sub:19nls277]

4. In the section where I did the work by induction the first thing I did was state the [integral from 0 to 1] (lnx) *dx = -1 = (-1)[sup:19nls277]1[/sup:19nls277] * 1!

not to complain, but these seem like nitpicky questions that have no relevance to my initial question of whether the problem is correct. Pointing out errors is one thing, but asking irrelevant questions or questions I clearly answered in my post is entirely something else. I appreciate you volunteering your time to help me and others like me but all I really need is someone who knows very well what they're doing to check this and tell me if it's right or I messed up somewhere.

 

[tkhunny]

1) Clear notation is important. "0-0" is not a clear notation. My first impression was an attempt at infinity. Thank you for clearing that up.

2) Okay. A reasonable reply.

3) I'll give you this one. I was lost in the array of new notation. I realize it is hard to write in-line when one is used to hand-written communication. See note #1.

4) I'll give you this one, too. Good work. The notation is really difficult to read. Try to remember that your audience can't see into your head.

NonComplaint) Learning to nit-pick is an important skill. Try getting a Masters Degree without it. Normally, when you seek assistance, you might want to park your opinion just a bit and listen to the qualified responses - particularly concerning what is or is not relevant. Look what you have done. You have defended your work quite nicely. Is this of no value? Answer the remaining objections and let's learn some mathematics.