Proving: $(\ln{x})^n$ Integral from 0 to 1

[v0id19]

Homework Statement

If n is a positive integer, prove that \int_{0}^{1}(\ln{x})^ndx=(-1)^n\cdot n!

Homework Equations

The Attempt at a Solution

I'm assuming that since ln(0) is undef and \mathop{\lim}\limits_{x \to 0^+}\ln{x}=- \infty i need to rewrite the integral as \mathop{\lim}\limits_{t \to 0^+}\int_{t}^{1}(\ln{x})^ndx=(-1)^n\cdot n!. But I have no idea how to integrate that since n is a variable...

 

[Dick]

This looks like a job for induction. Can you show it's true for n=1? To do the induction step integrate by parts with u=(ln(x))^(n+1) and dv=dx.

 

[v0id19]

so i proved n=1, and assumed it's true for n=k
now for n=k+1, I get:
\mathop{\lim}\limits_{t \to 0^+}\int_{t}^{1}ln(x)^(k+1)dx which i need to prove equals (-1)^k^+^1\cdot (k+1)! Now i know that ln(x)^k+1=ln(x)^kln(x).

so if i integrate by parts using f=ln(x)^k and g'=ln(x), i go around in circles...

am i missing something?

 

[Dick]

I told you what parts to use. To integrate f*dg=(ln(x))^(k+1)*dx take f=(ln(x))^(k+1), dg=dx. So g=x. What's g*df?

 

[v0id19]

i'm not familiar with that notation for integration by parts, but if i understand you correctly g*df=\frac{\ln{x}^k \cdot (k+1)}{x} \cdot 1

 

[v0id19]

so then i get
ln(x)^k+1*x-∫ln(x)^k(k+1)dx

 

[Dick]

g=x. So I get x*(k+1)*ln(x)^k*(1/x) dx. Do you see how the x's cancel? I was trying to imitate your notation for integration by parts. Integral(f*dg)=f*g-Integral(g*df).

 

[Dick]

so then i get
ln(x)^k+1*x-∫ln(x)^k(k+1)dx

Ok then. Put in your induction hypothesis for the integral of ln(x)^k.

 

[v0id19]

can i do that with the (k+1) inside the integral?
or do i do parts again, and set
f'=ln(x)^k f=(-1)^kk!
g=k+1 g'=0

that gets me
ln(x)k+1*x-[((-1)^kk!)(k+1)]+∫0dx

 

[Count Iblis]

Physicists don't like proofs by induction.

\int_{0}^{1}x^{p}dx = \frac{1}{p+1}

Differentiate both sides n times w.r.t. the parameter p:

\int_{0}^{1}x^{p} \log^{n}(x)dx = (-1)^{n}\frac{n!}{(p+1)^{n+1}}

Put p = 0 to get the desired result.

 

[Dick]

(k+1) IS a CONSTANT.

 

[v0id19]

(k+1) IS a CONSTANT.

oh duh thanks

 

[v0id19]

so i have ln(x)^k+1x-(k+1)(-1)^kk!
which equals
ln(x)^k+1x-(-1)^k(k+1)!

so somehow ln(x)^k+1x needs to get my (-1)^k to (-1)^k+1

 

[Dick]

Physicists don't like proofs by induction.

\int_{0}^{1}x^{p}dx = \frac{1}{p+1}

Differentiate both sides n times w.r.t. the parameter p:

\int_{0}^{1}x^{p} \log^{n}(x)dx = (-1)^{n}\frac{n!}{(p+1)^{n+1}}

Put p = 0 to get the desired result.

I'm a physicist, and I don't have any particular problems with induction. But sure, you can do it that way. Note the problem says 'prove that'. Physicists also HATE proofs.

 

[Dick]

so i have ln(x)^k+1x-(k+1)(-1)^kk!
which equals
ln(x)^k+1x-(-1)^k(k+1)!

so somehow ln(x)^k+1x needs to get my (-1)^k to (-1)^k+1

(-1)*(-1)^k=(-1)^(k+1), doesn't it? The term x*ln(x)^(k+1) vanishes at x=1. And it's limit is 0 as x->0. You can use l'Hopital and another induction argument to show that, if you don't already know it.

 

[v0id19]

yeah if i just show the l'hopital for the t-->0 that should be sufficient.

THANKS! :D :D