If I(n)=-n * I(n-1) deduce that I(n) = (-1)^n * n!?

UPDATE:

Hemant: To be honest, I don't see that you have added any content beyond what I had originally posted; which is below:

============================================

============================================

Let's first look at the original problem:

L(n) = ∫ (ln(x) )^n dx , [x = 0,1]

Integrating by parts,

L(n) = x*(ln(x))^n - ∫ x * (1/x)(n*(ln(x))^(n-1) dx , [x = 0, 1]

...... = x*(ln(x))^n - n * ∫(ln(x))^(n-1) dx , [x = 0, 1]

...... = (0 - 0) - n*L(n-1)

because x dominate ln(x) when x=> 0.

Therefore, as you stated:

L(n) = - n * L(n-1)

Before we can deduce what L(n) is, we have to have a starting point. In particular, does L(1) exist?

L(1) = ∫ ln(x) dx , [x = 0,1]

...... = x*ln(x) - x , [x = 0, 1]

...... = 1*0 - 0 - 1 = -1

So L(1) exists and = -1; and this matches the formula, because (-1)^1 * 1! = -1.

Now it is straightforward:

L(1) = -1

L(2) = -2*L(1) = +2

L(3) = -3*L(2) = -3*2

L(4) = -4*L(3) = +4*3*2

Obviously, it works. More formally, we would prove it by induction:

If L(n) = (-1)^n * n! for n = m, and

L(m+1) = -(m+1)*L(m), then

L(m+1) = -(m+1)*(-1)^m * m!

........... = (-1)^(m+1) * (m+1)*(m!)

........... = (-1)^(m+1) * (m+1)!

So if it works for m, it works for m+1.

Since we know it works for m = 1, that implies that it works for all integers 1 and above.

QED

UPDATE :

Nealjkin... frankly, i haven't read your solution.

I had just cast a casual glance at your solution and

those m and m+1 gave me the impression that you were

using the method of mathematical induction. So I shot off

my own solution.

...........................................................................................

Let : I(n) = - n. I(n-1) ... (1)

Then,

I(n-1) = -(n-1). I(n-2)

I(n-2) = -(n-2). I(n-3)

... ... ... ... ... ... ... ...

... ... ... ... ... ... ... ...

I(3) = -3. I(2)

I(2) =- -2. I(1)

I(1) = -1. I(0) = - 0∫1 ( ln x )° dx = - 0∫1 (1)dx

....................= -[x] on [0,1]

....................= - [ 1 - 0 ]

....................= -1.

Hence, combining all these equations backwards, i.e.,

using I(1) in I(2), ... I(2) in I(3), ... , I(n-1) in I(n),

I(n) = (-n)[-(n-1)].[-(n-2)].[-(n-3)]. ... (-3).(-2).(-1)

...... = [ (-1)(-1)(-1) ... n times ].[ 1.2.3. ... (n-1).n ]

...... = (-1)ⁿ. n! ................ Q.E.D.

...............................................................................................................................

Happy To Help !

..........................................................................................................................................