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

  • Thread starter v0id19
  • Start date
  • Tags
    Integral
In summary, physicists don't like proofs by induction, but if you use l'Hopital and another induction argument, you'll get the desired result.
  • #1
v0id19
49
0

Homework Statement


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

Homework Equations


The Attempt at a Solution


I'm assuming that since ln(0) is undef and [tex]\mathop{\lim}\limits_{x \to 0^+}\ln{x}=- \infty[/tex] i need to rewrite the integral as [tex]\mathop{\lim}\limits_{t \to 0^+}\int_{t}^{1}(\ln{x})^ndx=(-1)^n\cdot n![/tex]. But I have no idea how to integrate that since n is a variable...
 
Physics news on Phys.org
  • #2
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.
 
  • #3
so i proved n=1, and assumed it's true for n=k
now for n=k+1, I get:
[tex]\mathop{\lim}\limits_{t \to 0^+}\int_{t}^{1}[/tex]ln(x)(k+1)dx which i need to prove equals [tex](-1)^k^+^1\cdot (k+1)![/tex] 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?
 
  • #4
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?
 
  • #5
i'm not familiar with that notation for integration by parts, but if i understand you correctly g*df=[tex]\frac{\ln{x}^k \cdot (k+1)}{x} \cdot 1[/tex]
 
  • #6
so then i get
ln(x)k+1*x-∫ln(x)k(k+1)dx
 
  • #7
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).
 
  • #8
v0id19 said:
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.
 
  • #9
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
 
  • #10
Physicists don't like proofs by induction.

[tex]\int_{0}^{1}x^{p}dx = \frac{1}{p+1}[/tex]

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

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

Put p = 0 to get the desired result.
 
  • #11
(k+1) IS a CONSTANT.
 
  • #12
Dick said:
(k+1) IS a CONSTANT.

oh duh thanks
 
  • #13
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
 
  • #14
Count Iblis said:
Physicists don't like proofs by induction.

[tex]\int_{0}^{1}x^{p}dx = \frac{1}{p+1}[/tex]

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

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

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.
 
  • #15
v0id19 said:
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.
 
  • #16
yeah if i just show the l'hopital for the t-->0 that should be sufficient.

THANKS! :D :D
 

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

What is the purpose of proving the integral of $(\ln{x})^n$ from 0 to 1?

The purpose of proving the integral of $(\ln{x})^n$ from 0 to 1 is to determine the area under the curve of the function $(\ln{x})^n$ between the limits of 0 and 1. This is important in calculus and other areas of mathematics.

What is the method used to prove the integral of $(\ln{x})^n$ from 0 to 1?

The method used to prove the integral of $(\ln{x})^n$ from 0 to 1 is typically the substitution method or integration by parts. These are common techniques used in calculus to evaluate integrals.

Why is it important to prove the integral of $(\ln{x})^n$ from 0 to 1?

Proving the integral of $(\ln{x})^n$ from 0 to 1 is important because it allows us to accurately calculate the area under the curve of the function. This can have practical applications in various fields, such as physics and engineering.

What are the key steps in proving the integral of $(\ln{x})^n$ from 0 to 1?

The key steps in proving the integral of $(\ln{x})^n$ from 0 to 1 typically involve using the appropriate integration technique, simplifying the integrand, and evaluating the integral at the given limits. Depending on the method used, there may be additional steps involved.

Are there any special cases or exceptions when proving the integral of $(\ln{x})^n$ from 0 to 1?

Yes, there may be special cases or exceptions when proving the integral of $(\ln{x})^n$ from 0 to 1. For example, if the value of n is negative or zero, the integral may not exist. It is important to carefully consider the function and limits before attempting to evaluate the integral.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
496
  • Calculus and Beyond Homework Help
Replies
4
Views
576
  • Calculus and Beyond Homework Help
Replies
6
Views
599
  • Calculus and Beyond Homework Help
Replies
34
Views
2K
  • Calculus and Beyond Homework Help
Replies
14
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
547
  • Calculus and Beyond Homework Help
Replies
3
Views
6K
  • Calculus and Beyond Homework Help
Replies
8
Views
848
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
343
Back
Top