Does the Integral of 1/(x^2 + a^2) from 0 to Infinity Exist?

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Homework Help Overview

The discussion revolves around the existence of the integral of the function \( \frac{1}{x^2 + a^2} \) from 0 to infinity, with participants exploring the implications of the variable \( a \) on the integral's convergence.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants question the placement of \( a^2 \) in the denominator and its effect on the integral's convergence. Some express confusion regarding the original problem statement and the conditions under which the integral exists.

Discussion Status

The discussion is active, with multiple interpretations of the integral being explored. Some participants have offered clarifications regarding the placement of \( a^2 \) and its necessity for convergence, while others have suggested that the original poster clarify their attempts to facilitate further assistance.

Contextual Notes

There are indications that the original problem may have been misinterpreted, leading to divergent views on the integral's setup. Additionally, forum rules emphasize the importance of showing prior attempts before receiving help.

king imran
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integratio 1/x2+a2.dx=1
intervel0 to +infinity
find value of 'a'
 
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Hi there imran,

I'm guessing that your integrand, is something along the lines of;

\int^{\infty}_{0} \frac{1}{x^2} + a^2 \;dx = 1

Please note, that the rules of this forum prohibit us to provide help to someone who has not attempted there homework question before hand (assuming of course this is homework). Perhaps, if you posted your attempt, we could help you further...
 
The a^2 factor must be on the denominator, otherwise the integral is divergent both horizontally at infinity and vertically at zero.

Just clearing up the interpretation. Since the original poster has access to the original problem anyway, this doesn't constitute helping him. ;)

--Stuart Anderson
 
The Integral \int \frac{1}{x^2 + a^2} dx looks very familiar to me :)

Maybe some trig will help you out?
 
mr_homm said:
The a^2 factor must be on the denominator, otherwise the integral is divergent both horizontally at infinity and vertically at zero.

Just clearing up the interpretation. Since the original poster has access to the original problem anyway, this doesn't constitute helping him. ;)

--Stuart Anderson
You are quite correct of course!
Gib Z said:
The Integral \int \frac{1}{x^2 + a^2} dx looks very familiar to me :)
And so it should :wink:
Gib Z said:
Maybe some trig will help you out?
Good suggestion :approve:
 
find volume of solid y=x2-1 ,x=2 ,y=0 is relvived about y-asis
 
If this is a new problem, start a new thread.

Also, since it is obviously homework, post it in the homework thread and show what you have tried!
 
please give me good sites to refer for integration,its applications
 
  • #10
\[<br /> \int {\frac{1}<br /> {{x^2 + a^2 }}dx} = \frac{1}<br /> {a}\tan ^{ - 1} \frac{x}<br /> {a} + c<br /> \]

Just in case.
 
  • #11
Hootenanny said:
\int^{\infty}_{0} \frac{1}{x^2} + a^2 \;dx = 1
does the following integral even exist?
\int^{\infty}_{0}\left(\frac{1}{x^2} + a^2\right) dx
 
  • #12
murshid_islam said:
does the following integral even exist?
\int^{\infty}_{0}\left(\frac{1}{x^2} + a^2\right) dx

Look at the above post:

mr_homm said:
The a^2 factor must be on the denominator, otherwise the integral is divergent both horizontally at infinity and vertically at zero.

Just clearing up the interpretation. Since the original poster has access to the original problem anyway, this doesn't constitute helping him. ;)

--Stuart Anderson
 
  • #13
murshid_islam said:
does the following integral even exist?
\int^{\infty}_{0}\left(\frac{1}{x^2} + a^2\right) dx

The question was actually with the a^2 also in the denominator, look at above posts. And that integral you stated only exists if a=0.
 

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