- #1
kexue
- 196
- 2
When replacing x with 1/kx then
[tex]
\int_{1/k}^\infty {\left[ {\left( {x^2 - 1} \right)\left( {k^2 x^2 - 1} \right)} \right]} ^{ - 1/2} dx = \int\limits_0^1 {\left[ {\left( {\frac{1}{{k^2 x^2 }} - 1} \right)\left( {\frac{1}{{x^2 }} - 1} \right)} \right]} ^{ - 1/2} \frac{{dx}}{{kx^2 }}
[/tex]
I do not see how. Why ranges the LHS integral over infinity, whereas the RHS from 0 to 1?
Any help and hints very much appreciated.
thanks
[tex]
\int_{1/k}^\infty {\left[ {\left( {x^2 - 1} \right)\left( {k^2 x^2 - 1} \right)} \right]} ^{ - 1/2} dx = \int\limits_0^1 {\left[ {\left( {\frac{1}{{k^2 x^2 }} - 1} \right)\left( {\frac{1}{{x^2 }} - 1} \right)} \right]} ^{ - 1/2} \frac{{dx}}{{kx^2 }}
[/tex]
I do not see how. Why ranges the LHS integral over infinity, whereas the RHS from 0 to 1?
Any help and hints very much appreciated.
thanks