Estimate a integral in the complex

[handiman]

Homework Statement

Let γ denote the portion of the circle lzl = √(ln(3)) which lies in the left-half plane.

Prove l(∫_γ(e^(iz²)/(z-1) dz l ≤ (3pi√(ln(3))/(ln(3)+1)

Note: the l l around the integral represents modulus.

Homework Equations

In the past we used this inequality.

l∫_λf(z)/g(z) dz l ≤ max lf(z)/g(z) l * length(lzl)

The Attempt at a Solution

So far I found the length of lzl to be 2pi√(ln(3)) and g(z) ≥ √(3) - 1.

I am not sure about my work on g(z) and for f(z) = le^(iz²) I am very confused as to how to proceed.

 

[mighty2000]

Thank you for your post. In order to prove the given inequality, we will need to use the Cauchy Integral Formula and the triangle inequality.

First, let's rewrite the given integral as follows:

l∫γ(e^(iz2)/(z-1) dz l = l∫γ(e^(iz2)/((z-1)(z-1)) dz l

We can do this because the function e^(iz2) is analytic in the region enclosed by γ, and the point z = 1 is not contained within this region.

Next, we can use the Cauchy Integral Formula to write the above integral as:

l∫γ(e^(iz2)/((z-1)(z-1)) dz l = l(2πi * e^(i*1^2))/(1-1)^2 l

= l(2πi * e^(i))/(0)^2 l

= l(2πi * e^(i))/(0) l

= 0

Now, we can use the triangle inequality to write the modulus of the integral as:

l∫γ(e^(iz2)/(z-1) dz l = l∫γ(e^(iz2)/(z-1) dz l ≤ 2π * max l(e^(iz2)/(z-1) l * length(γ)

= 2π * max l(e^(iz2) l/(z-1) l * length(γ)

= 2π * max l(e^(iz2) l * l 1/(z-1) l * length(γ)

= 2π * max l(e^(iz2) l * l 1/(z-1) l * 2π√(ln(3))

= 4π^2 * max l(e^(iz2) l * l 1/(z-1) l * √(ln(3))

Now, we can use the given inequality from the past work to write:

l∫γ(e^(iz2)/(z-1) dz l ≤ 4π^2 * max l(e^(iz2) l * l 1/(z-1) l * √(ln(3))

≤ 4π^2 * max le^(iz2) / √(3) - 1 l * √(ln(3))

= 4