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# The Integration of the Function

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given that integral from 0 to infinity of e^(-x^2)dx=sqrt(pi)/2
show by integrating e^(-z^2) around a rectangle with vertices at z=0,p,p+ai and ai (with a positive and letting p go to infinity) that

integral from 0 to infinity of e^(-x^2)cos(2ax)dx= [e^(-a^2)]*sqrt(pi)/2
integral from 0 to infinity of e^(-x^2)sin(2ax)dx= [e^(-a^2)]*[integral from 0 to a of e^(t^2)dt]

##### Solution Summary

The integration of the function is performed in this case.

##### Solution Preview

Remark: In the problem statement, "cos(2ax)" needs to be changed to "cos(ax)" and "sin(2ax)" to "sin(ax)".

Problem:

given that integral from 0 to infinity of e^(-x^2)dx=sqrt(pi)/2
show by integrating e^(-z^2) around a rectangle with vertices at z=0,p,p+ai and ai (with a positive and letting p go to infinity) that

integral from 0 to infinity of e^(-x^2)cos(ax)dx= [e^(-a^2)]*sqrt(pi)/2
integral from 0 to infinity of e^(-x^2)sin(ax)dx= [e^(-a^2)]*[integral from 0 to a of e^(t^2)dt]

Solution:

Based on the integral
( 1)
it is ...

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