- #1
king imran
- 6
- 0
integratio 1/x2+a2.dx=1
intervel0 to +infinity
find value of 'a'
intervel0 to +infinity
find value of 'a'
You are quite correct of course!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
And so it shouldGib Z said:The Integral [tex]\int \frac{1}{x^2 + a^2} dx[/tex] looks very familiar to me :)
Good suggestionGib Z said:Maybe some trig will help you out?
does the following integral even exist?Hootenanny said:[tex]\int^{\infty}_{0} \frac{1}{x^2} + a^2 \;dx = 1[/tex]
murshid_islam said:does the following integral even exist?
[tex]\int^{\infty}_{0}\left(\frac{1}{x^2} + a^2\right) dx[/tex]
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
murshid_islam said:does the following integral even exist?
[tex]\int^{\infty}_{0}\left(\frac{1}{x^2} + a^2\right) dx[/tex]
The solution to this integral is 1/a * arctan(x/a) + C, where C is the constant of integration.
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.
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, ∞).
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.
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.