# Thread: Help integrating complex function!

1. ## Help integrating complex function!

I'm stuck... its been awhile since I've done integrals.

So the integral is $\int\frac{dz}{z}$ along the straight line segment from 1 to i. I parameterized the curve as $\gamma:[0,1]\rightarrow C$, defined as $\gamma(t)=(-t+1)+it$ and then used the property that dividing 1 by z is the equivalent of $z^{-1}$. So I have (with the constant (-1+i) omitted):

$)\int(\frac{-t+1}{(-t+1)^2+t^2})-i\frac{t}{(-t+1)^2+t^2}dt\\=\int(\frac{-t+1}{(-t+1)^2+t^2})dt-i\int\frac{t}{(-t+1)^2+t^2}dt$

I have no clue how to integrate these... is there some trick I'm forgetting? any help?

2. Originally Posted by platinumpimp68plus1
I'm stuck... its been awhile since I've done integrals.

So the integral is $\int\frac{dz}{z}$ along the straight line segment from 1 to i. I parameterized the curve as $\gamma:[0,1]\rightarrow C$, defined as $\gamma(t)=(-t+1)+it$ and then used the property that dividing 1 by z is the equivalent of $z^{-1}$. So I have:

$\int(\frac{-t+1}{(-t+1)^2+t^2})-i\frac{t}{(-t+1)^2+t^2}dt\\=\int(\frac{-t+1}{(-t+1)^2+t^2})dt-i\int\frac{t}{(-t+1)^2+t^2}dt$

I have no clue how to integrate these... is there some trick I'm forgetting? any help?
$\int\frac{-t+1}{(-t+1)^2+t^2}dt=\int\frac{-t+1}{t^2-2t+1+t^2}dt=\int\frac{-t+1}{2t^2-2t+1}dt$. Let $z=2t^2-2t+1\implies dz=2t-2$..

3. Originally Posted by Drexel28
$\int\frac{-t+1}{(-t+1)^2+t^2}dt=\int\frac{-t+1}{t^2-2t+1+t^2}dt=\int\frac{-t+1}{2t^2-2t+1}dt$. Let $z=2t^2-2t+1\implies dz=2t-2$..
I got kind of excited until I noticed its actually $dz=4t-2$... but that lead me on this train of thought:

(Integrating the real term):

$z=2t^2-2t+1 \implies dz=4t-2$
$-t+1= \frac{-1}{4} (4t-4)=\frac{-1}{4} (4t-2-2)=\frac{-1}{4} (4t-2)+\frac{1}{2}$

So:

$=\frac{-1}{4}\int\frac{4t-2}{2t^2-2t+1}dt+\frac{1}{2}\int
\frac{1}{2t^2-2t+1}dt$

Which should be...

$=-\frac{1}{4}\ln(2t^2-2t+1)+\frac{1}{2}\tan^-1(2x-1)$

yay/nay?