Technical question on integrals.

  • Thread starter Thread starter MathematicalPhysicist
  • Start date Start date
  • Tags Tags
    Integrals
Click For Summary

Homework Help Overview

The discussion revolves around two integral expressions involving binomial coefficients and polynomial integrals. The first expression requires demonstrating a relationship involving the integral of a polynomial raised to a power, while the second expression relates a binomial coefficient to an integral representation.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply the binomial theorem to the first integral but finds it unhelpful. They express uncertainty about how to approach the second integral, questioning the applicability of integration techniques like integration by parts or substitution.

Discussion Status

Some participants suggest integrating term by term after applying the binomial theorem, while others propose using integration by parts for the second expression. There is ongoing exploration of methods without a clear consensus on the best approach.

Contextual Notes

Participants are navigating the complexities of integrating polynomial expressions and are considering various mathematical techniques, with some expressing doubts about the effectiveness of certain methods in this context.

MathematicalPhysicist
Science Advisor
Gold Member
Messages
4,662
Reaction score
372
i have to show that:
1)[tex](-1)^n\int_{-1}^{1}(x^2-1)^ndx=2^{2n+1}(n!)^2/(2n+1)![/tex]
2) [tex]\binom{n}{k}=[(n+1)\int_{0}^{1}x^k(1-x)^{n-k}dx]^{-1}[/tex]

for the first part i thought to use Newton's binomial, i.e:
[tex](1-x^2)^n=\sum_{k=0}^{n}\binom{n}{k}(-x^2)^k[/tex]
but it didn't get me far.
for the second part i don't have a clue, i don't think you can integrate the integral by parts or substitution can you?
 
Physics news on Phys.org
Where'd you get stuck after using the binomial theorem? Just integrate term by term.
 
this is what i got:
2-2n/3+2n(n-1)/10+...+2(-1)^n/(2n+1)
i don't know how to procceed from here.
 
yeah, i think number 2 can be worked out using integral by parts... the tabular method.
all except the last term of the resulting series is zero when the bounds are substituted in.
 

Similar threads

  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K