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Homework Statement
Integrate exp(-z^2) over the rectangle with vertices at 0, R, R + ia, and ia.Homework Equations
int(0, inf)(exp(-x^2)) = sqrt(pi/2)
This is what I would have thought, but I'm supposed to be using the integral of e^(-z^2) to evaluate the real integral int(0,inf)((e^(-x^2))*cos(2ax)), which is apparently equal toideasrule said:Isn't the contour integral equal to 0 if there are no poles?
A contour integral is a type of line integral in complex analysis. It involves integrating a complex-valued function along a specific path or contour in the complex plane.
The purpose of a contour integral is to calculate the value of a complex integral by breaking it down into simpler integrals along a specific path in the complex plane. It is a useful tool in many areas of mathematics and physics, including complex analysis, number theory, and quantum mechanics.
A contour integral can be used to calculate the value of the Gaussian function exp(-z^2) along a specific path in the complex plane. This is known as the Gaussian integral and has important applications in probability, statistics, and physics.
The contour integral of exp(-z^2) is a special case of the error function and has a number of important properties. These include the fact that it is an even function, its value at z=0 is the square root of pi, and it can be used to calculate the probability of a standard normal distribution.
The contour integral of exp(-z^2) can be calculated using various techniques, including the Cauchy integral theorem, Cauchy's integral formula, and the residue theorem. These methods involve evaluating the function along the chosen contour and using complex analysis techniques to simplify the integral.