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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!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 shouldThe Integral [tex]\int \frac{1}{x^2 + a^2} dx[/tex] looks very familiar to me :)
Good suggestionMaybe 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]
Look at the above post:does the following integral even exist?
[tex]\int^{\infty}_{0}\left(\frac{1}{x^2} + a^2\right) dx[/tex]
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 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.does the following integral even exist?
[tex]\int^{\infty}_{0}\left(\frac{1}{x^2} + a^2\right) dx[/tex]