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

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In summary, the conversation revolved around a question about an integral with a variable 'a' and how to find its value. The participants offered suggestions and advice on approaching the problem, and also discussed the existence of the integral depending on the placement of the a^2 term. The conversation also touched on the use of trigonometric functions to solve the integral.
  • #1
king imran
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integratio 1/x2+a2.dx=1
intervel0 to +infinity
find value of 'a'
 
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  • #2
Help Me Inthis Question
 
  • #3
Hi there imran,

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

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

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...
 
  • #4
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
 
  • #5
The Integral [tex]\int \frac{1}{x^2 + a^2} dx[/tex] looks very familiar to me :)

Maybe some trig will help you out?
 
  • #6
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 [tex]\int \frac{1}{x^2 + a^2} dx[/tex] looks very familiar to me :)
And so it should :wink:
Gib Z said:
Maybe some trig will help you out?
Good suggestion :approve:
 
  • #7
find volume of solid y=x2-1 ,x=2 ,y=0 is relvived about y-asis
 
  • #8
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!
 
  • #9
please give me good sites to refer for integration,its applications
 
  • #10
[tex]\[
\int {\frac{1}
{{x^2 + a^2 }}dx} = \frac{1}
{a}\tan ^{ - 1} \frac{x}
{a} + c
\][/tex]

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

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?
[tex]\int^{\infty}_{0}\left(\frac{1}{x^2} + a^2\right) dx[/tex]

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.
 

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

What is the solution to the integral 1/x^2+a^2.dx=1?

The solution to this integral is 1/a * arctan(x/a) + C, where C is the constant of integration.

How do you solve this integral using substitution?

To solve this integral using substitution, let u = x/a and du = 1/a * dx. Then, the integral becomes ∫ 1/(a^2 * u^2 + a^2) * du.

What is the domain of this integral?

The domain of this integral is all real numbers except for where the denominator, a^2 * u^2 + a^2, is equal to 0. This occurs when u = 0, so the domain is (-∞, 0) U (0, ∞).

Can this integral be solved using partial fractions?

Yes, this integral can be solved using partial fractions by rewriting the integrand as 1/(a^2 * (x^2 + 1)). We can then use the formula for the integral of 1/(x^2 + a^2) to solve it.

How can this integral be applied in real life?

This integral can be used in various fields of science and engineering, such as physics and electrical engineering, to solve problems involving inverse square relationships. For example, it can be used to calculate the electric potential due to a point charge, or the gravitational potential due to a planet.

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