Complex Analysis - Contour integral

[EC92]

Homework Statement

I have the following problem:
Compute
$\operatorname{Re} \int _\gamma \frac{\sqrt{z}}{z+1} dz,$
where $\gamma$ is the quarter-circle $\{ z: |z|=1, \operatorname{Re}z \geq 0 , \operatorname{Im} z \geq 0 \}$ oriented from 1 to $i$, and $\sqrt{z}$ denotes the principal branch.

Homework Equations

The Attempt at a Solution

I've been trying to solve this using the complex analog of the 2nd Fundamental Theorem of Calculus. Substituting $u = \sqrt{z}$ and using partial fractions, I get
$\int_\delta 2 - \frac{i}{u+i} + \frac{i}{u-i} du$
where delta is the eighth-circle from 1 to $e^{i\pi /4}$
This is equal to
$[2u - i \log(u+i) + i\log (u-i)]_{u=1} ^{u=e^{i\pi/4}}$,

and the real part is then

$[\operatorname{Re} u + \operatorname{Arg}(u+i) -\operatorname{Arg}(u-i)]_{u=1} ^{u = e^{i \pi /4}}$

However, the arguments at $u=e^{i\pi /4}$ do not come out to nice forms. I am wondering if my approach is even correct, and if there's a better way to solve problems of this type.
Thanks.

[Mathematica says the value is approximately -0.584].

 

[vela]

Combine the logs:
$$-i \log (u+i) + i\log(u-i) = i\log\frac{u-i}{u+i}$$ For $u=e^{i\pi/4}$, you should be able to show that $\frac{u-i}{u+i} = -i \tan \pi/8$.

 

[jackmell]

I am wondering if my approach is even correct, and if there's a better way to solve problems of this type.

Yes. You can approach it entirely in terms of an analytically-continuous antiderivative. But I realize that analytically-continuous thing is what, not easy to understand? First, what is any-old antiderivative? That's easy right:

$$\begin{align} \int_1^i \frac{\sqrt{z}}{z+1}dz&=\int_1^{\sqrt{i}} \frac{2u^2}{u^2+1}du\\ &=2(u-\arctan(u)\biggr|_1^{\sqrt{i}} \end{align}$$

Ok then, as long as you evaluate that antiderivative along an analytically-continuous path from 1 to $e^{\pi i/4}$ then you can simply evaluate the antiderivative at it's end-points. But the multi-valued function $\arctan(z)$ has branch-points at $\pm i$ so my path from 1 to $\sqrt{i}$ doesn't pass near these points or go around them, so I can use the prinicpal-valued branch of arctan(z) and simply write:

$$\int_{1}^{i}\frac{\sqrt{z}}{z+1}=2\left[(\sqrt{i}-\arctan(\sqrt{i}))-(1-\arctan(1))\right]$$