Integration Formula: Proving the Formula

Click For Summary

Homework Help Overview

The discussion revolves around the derivation of a specific integration formula involving the integral of the form \(\int\frac{du}{(u^2+\alpha^2)^m}\). Participants are exploring methods to prove this formula as presented in a textbook, which lacks a proof.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts integration by parts as suggested by the textbook but expresses difficulty in deriving the formula. Some participants suggest considering the homework section for a more appropriate context. Others propose using trigonometric or hyperbolic substitutions to simplify the integral and derive recursion formulas.

Discussion Status

The discussion is ongoing, with various approaches being suggested. There is no explicit consensus on the best method to derive the formula, but some participants are exploring alternative substitutions that may facilitate the process.

Contextual Notes

Participants note that the original post may not fit traditional homework criteria, leading to discussions about the appropriateness of the thread's placement in the forum.

uman
Messages
348
Reaction score
1
Hello all,

My textbook states the formula \int\frac{du}{(u^2+\alpha^2)^m}=\frac{1}{2\alpha^2(m-1)}\frac{u}{(u^2+\alpha^2)^{m-1}}+\frac{2m-3}{2\alpha^2(m-1)}\int\frac{du}{(u^2+\alpha^2)^{m-1}} but does not provide a proof of this formula. Anyone want to show me how it's derived? I tried integration by parts, which the book gives as the method for deriving the formula, but I couldn't figure it out :-(. Any help?
 
Physics news on Phys.org
You may want to try the homework section.
 
Why?

Also any mods can move this if necessary but it isn't homework for a class...
 
A "homework-type" question might get a faster response in the HW section.
 
You have a point. I don't know how to move it or if that's even possible. If a mod sees this please move it.
 
An obvious trig substitution converts this integral into a power of cosine integral, which has well known recursion formulae, or otherwise, easier to derive recursion formulae than the original integral.

EDIT: The hyperbolic substitution converts it into a power of hyperbolic cosine integral, which is even easier to deal with.
 
Last edited:

Similar threads

Compute sum (possibly using Parseval's formula)
  • · Replies 1 ·
Replies
1
Views
2K
Center of mass of spherical shell inside of cone (Apostol Problem).
  • · Replies 3 ·
Replies
3
Views
2K
How Does Substitution Affect Double Integration and Differentiation?
  • · Replies 3 ·
Replies
3
Views
2K
Prove that the integral is equal to ##\pi^2/8##
  • · Replies 105 ·
4
Replies
105
Views
11K
Solve the given problem that involves integration
  • · Replies 4 ·
Replies
4
Views
2K
Find the value of the definite integral
  • · Replies 8 ·
Replies
8
Views
2K
How should I show that all solutions of this equation are bounded?
  • · Replies 4 ·
Replies
4
Views
2K
Integrals that keep me up at night
  • · Replies 22 ·
Replies
22
Views
3K
How Do You Solve the Differential Equation dy/dx = 1 - y^2?
  • · Replies 12 ·
Replies
12
Views
3K
Is the method used to evaluate the given integral correct?
  • · Replies 7 ·
Replies
7
Views
2K