king imran
- 6
- 0
integratio 1/x2+a2.dx=1
intervel0 to +infinity
find value of 'a'
intervel0 to +infinity
find value of 'a'
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.
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.
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.
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 \int \frac{1}{x^2 + a^2} dx looks very familiar to me :)
Good suggestionGib Z said:Maybe some trig will help you out?

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