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## Homework Statement

## The Attempt at a Solution

**ap1)**

0, since there are no singularities for the function inside contour

**ap2)**

Provided that INT{-infinite,infinite} e^-z^2dz = sqrt(pi) -gaussian integral-,

INT{0,infinite} e^-z^2dz = sqrt(pi)/2

**ap3)**

show that int(arc)-> 0 as R->infinite. Using parametrization of z = R*e^(i*B) where 0<=B<=alpha/2, dz = i* R*e^(i*B) dB we can substitute z and dz :

INT{0,alpha/2} e^-(R*e^(i*B))^2 * i*R*e^(i*B) dB =

= INT{0,alpha/2} i*R*e^(i*B) / e^(R^2 * e^(2*i*B)) dB

hence, we have arrived to something of the form R / e^(R^2),

and if R->infinite, the integral -> zero

(Note: I think that this may not be valid because we need an ML inequality

to demonstrate this. However I don't know how to do this in this problem )

**ap4)**

Using parametrization: z = r*e^(i*alpa/2) with r from R to 0,

dz = e^(i*alpha/2) dr,

and e^(ai) = cos a + i * sin a,

I want to transform INT(III) 1/e^z^2 dz so that we can have integrals A and B inside.

Is this way of proceeding correct? I have tried transforming the integral but haven't been

able to get to A and/or B.

Any kind of help or directions will be greatly appreciated.

Thanks in advance.