A Definite integral where solution. involves infinity - infinity

Click For Summary

Homework Help Overview

The discussion revolves around evaluating a definite integral that involves limits approaching infinity. The original poster expresses confusion regarding the solution provided in their textbook, which states the answer is pi/4.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using partial fractions and limits, with some suggesting combining terms that approach infinity. There are questions about alternative methods to evaluate the integral without using L'Hôpital's rule.

Discussion Status

Several participants have shared their approaches and expressions for the integral. There is acknowledgment of similar integrals among contributors, but no consensus has been reached on a definitive method or solution. Some participants are focused on clarifying algebraic expressions and ensuring proper notation.

Contextual Notes

Participants mention restrictions on using certain calculus techniques, such as L'Hôpital's rule, which may affect their approaches to the problem.

Physics news on Phys.org
After using partial fractions decomposition, I get this:
[tex] \int_0^B \frac{-1/2}{x + 1} + (1/2)\frac{x + 1}{x^2 + 1}dx[/tex]

I'm not sure this is what you got. After integrating, the antiderivative will be in terms of B. Take the limit of this expression as B approaches infinity.
 
Combine the parts that go to infinity into one term. ln((x+1)^(1/2)/(x^2+1)^(1/4)). You want to figure out the limit of that as x->infinity. Can you see where to go from there?
 
Thank you very much Mark44 and Dick.
I got the same integral as Mark44.
I understand what Dick is saying but I'm not allowed to use L hospital's rule at school. I can't figure out any other way.
Is there an alternate way to evaluate the above integral?
 
cpashok said:
Thank you very much Mark44 and Dick.
I got the same integral as Mark44.
I understand what Dick is saying but I'm not allowed to use L hospital's rule at school. I can't figure out any other way.
Is there an alternate way to evaluate the above integral?

Factor an x^(1/2) out of the numerator and denominator of (x+1)^(1/2)/(x^2+1)^(1/4).
 
Correct me if i am wrong.
ln [x^(1/2) { 1+ 1/ x^(1/2) } / x^(1/2) { 1+ 1/ x^(1/2) }]

I would get ln 1.

which is equal to zero and hence the required answer.
Thanks a lot Mark 44 and Dick.
 
The conclusion is right, but your algebra leaves a lot to be desired. If you mean
(x+1)^(1/2)=(x)^(1/2)*(1+1/x)^(1/2), that's ok. If you mean (x+1)^(1/2)=x^(1/2)+1/(x^(1/2)) or x^(1/2)+(1/x)^(1/2), both of those are wrong. And what happened with the 1/4 power in the denominator? Try to use enough parentheses to make it clear what you mean.
 
oops!
{x^(1/2){1+1/x}^(1/2)} / x ^(1/2){1+1/(x^2)}^(1/4) .
i got this in the notebook, forgot to use paranthesis and forgot to edit the second term that i copy pasted from the first.
 
cpashok said:
oops!
{x^(1/2){1+1/x}^(1/2)} / x ^(1/2){1+1/(x^2)}^(1/4) .
i got this in the notebook, forgot to use paranthesis and forgot to edit the second term that i copy pasted from the first.

Perfect.
 

Similar threads

Replies
3
Views
9K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K