If n is a positive integer, prove that integral[(lnx)^ndx=((-1)^(n))*n!](lower limit=0, upper limit=1)
Thanks Again,
First compute the integral of x^p from zero to 1. Differentiate both sides n times w.r.t. the parameter p. Then take p = 0.
To prove the given integral, we will first use integration by parts, and then apply induction.
Let's start by performing integration by parts on the integral:
∫[(lnx)^ndx] dx
Let's assume u = (lnx)^n and dv = dx. Then, du = n(lnx)^(n−1)*(1/x) dx and v = x. By using the integration by parts formula, the integral becomes:
∫[(lnx)^ndx] dx = uv - ∫v du
= x(lnx)^n - ∫x * n(lnx)^(n−1) (1/x) dx
= x(lnx)^n - n∫(lnx)^(n−1) dx
Now, let's simplify the remaining integral. Rewrite (lnx)^(n−1) as (lnx)^n * (lnx)^(-1) to get:
∫(lnx)^(n−1) dx = ∫(lnx)^n * (lnx)^(-1) dx
Using integration by parts again, with u = (lnx)^n and dv = (lnx)^(-1) dx, we have du = n(lnx)^(n−1)*(1/x) dx and v = ∫(lnx)^(-1) dx. Evaluating v, we get:
v = ∫(lnx)^(-1) dx
= ∫(1/lnx) dx
To simplify further, we can use the substitution u = lnx, du = (1/x) dx. The integral becomes:
∫(1/lnx) dx = ∫du
= u + C
= lnx + C
Therefore, we have:
∫(lnx)^(n−1) dx = (lnx)^n * (lnx)^(-1) - lnx + C
= (lnx)^n * (1/lnx) - lnx + C
= (lnx)^(n−1) - lnx + C' (where C' = 1/C)
Substituting this result back into the original integral, we have:
∫[(lnx)^ndx] dx = x(lnx)^n - n∫(lnx)^(n−1) dx
= x(lnx)^n - n[(lnx)^(n−1) - lnx + C']
= x(lnx)^n - n(lnx)^(n−1) + nlnx - nC'
Now, we can calculate the definite integral from 0 to 1 by substituting the upper and lower limits into the expression:
∫[(lnx)^ndx] dx (lower limit=0, upper limit=1)
= 1(ln1)^n - n(ln1)^(n−1) + nln1 - nC' - 0(ln0)^n + n(ln0)^(n−1) - nln0 + nC'
Since ln1 = 0 and ln0 is undefined, these terms simplify to:
= 0 - 0 + 0 + 0 + 0 + 0 + 0 + 0
Therefore, the definite integral is equal to 0.
Now, let's prove the given expression for n! * (-1)^n by induction.
Base case:
For n = 1, the given expression is (-1)^1 * 1! = -1, and we have shown that the integral is equal to 0 when n = 1.
Inductive step:
Assume that the expression is true for some positive integer k, i.e., the integral [(lnx)^kdx] from 0 to 1 is k! * (-1)^k.
We want to prove that the expression is also true for k+1, i.e., the integral [(lnx)^(k+1)dx] from 0 to 1 is (k+1)! * (-1)^(k+1).
Using the result we obtained earlier, the integral [(lnx)^(k+1)dx] from 0 to 1 can be expressed as:
∫[(lnx)^(k+1)dx] (lower limit=0, upper limit=1)
= 1(ln1)^(k+1) - (k+1)(ln1)^k + (k+1)ln1 - (k+1)C' - 0(ln0)^(k+1) + (k+1)(ln0)^k - (k+1)ln0 + (k+1)C'
Again, since ln1 = 0 and ln0 is undefined, these terms simplify to:
= 0 - 0 + (k+1)(0) - (k+1)C' - 0 + (k+1)(-∞) - (k+1)(-∞) + (k+1)C'
= 0 - 0 - (k+1)C' + (k+1)C'
= 0
Therefore, the expression holds true for k+1.
By induction, we conclude that for any positive integer n, the integral [(lnx)^ndx] from 0 to 1 is equal to (-1)^n * n!.
Hence, we have proven the given statement.