Integral_c tan(z/2) / (z-x_0)^2 dz ~ it's answer has a sec^2 in it how?

[laura_a]

Homework Statement

I have this question in my text and the answer has a sec^2 in it, and according to what I know, there is no where that you need to use derivatives, but maybe I don't understand the formula? Up until now all the answers have just been regular functions, not derivatives.

ANyways, here is what I've done

Question is

C is a positively oriented boundary of the square whose sides lie on lines x & y = +/- 2 so eval

Int_c tan(z/2) / (z-x_0)^2 dz

The answer in the text is i*pi*sec^2(x_0)/2

Homework Equations

The Cauchy Integral formula I'm using is as follows
f(z_0) = 1/ (2 * pi * i) ⌠_c f(x) / (z-z_0) dz

The Attempt at a Solution

modified the intergral so I could use the above formula

⌠_c tan(z/2) / (z-x_0)(z+x_0) dz

so f(x) = tan(z/2) / (z+x_0)

and z_0 = x_0
so f(z_0) = tan(x_0/2) / (2x_0)

Using Cauchy Integral I get
tan(x_0/2) / (2x_0) * (2 * pi * i)

so the 2's cancel and I get

p * i * tan(x_0 / 2) / x_0

What don't I know about sec^2 (x) that makes me not be able to get

the answer?

 

[Gib Z]

Perhaps the sec^2 came up naturally and not precisely as the derivative of tan. Or perhaps they did some integration by parts.

 

[HallsofIvy]

Cauchy's integral formula is more general than that:
$$f^{(n)}= \frac{n!}{2\pi i}\int \frac{f(z)dz}{(z- z_0)^{n+1}}$$
where the integral is around a closed curve with z₀ in its interior.

Since your integral has a square in the denominator, look at the first derivative of tan(z) evaluated at z₀.