Approximation question -- How does this approximation simplify the integral?

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The discussion focuses on how a specific approximation simplifies an integral by treating the integrand as constant and evaluating it at the midpoint, z'=0. This approach reduces complexity by allowing for straightforward multiplication with the range of z'. The conditions l<<lambda and k_z<<pi ensure that the exponential function's argument remains small, maintaining the validity of the approximation. The participants highlight that the smallness of k_z l/r compared to 1 is crucial for the approximation's accuracy. Overall, the approximation streamlines the integral while adhering to specific parameter constraints.
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how does this aproximation reduse the integral
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Where are lambda and k_z in the integral?

The approximation simply assumes the integrand to be constant, evaluates it at the midpoint (z'=0) and multiplies it with the range z' takes.
 
so its because r'=0 ??
then why we need
l<<lambda kz<<pi
?
 
The exponential function should not change its argument too much, so kzl/r should be small compared to 1.
 

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