# Thread: Evaluate the path integral in C

1. ## Evaluate the path integral in C

Hello,

I'm trying to evaluate the following path integral $$\int_{\alpha} \frac{dz}{z}$$ where $\alpha(t)=e^{i(\pi/4 +4\pi t)}r(t)$ with $0\leq t \leq 1$ and $r(t)>0$, $r(0)=\sqrt{2}$, and $r(1)=2\sqrt{2}$.

We have a definition $\int_\alpha f(z) dz = \int_a^b f(z(t))z'(t) dt$. This should give $$\int_\alpha \frac{dz}{z}=\int_0^1 \frac{r'(t)e^{i(\pi/4+4\pi t)} +i4\pi r(t) e^{i(\pi/4+4\pi t)} }{e^{i(\pi/4+4\pi t)} r(t) } dt.$$ Then after canceling the $e^{i(\pi/4+4\pi t)}$s you have $$\int_0^1 \frac{r'(t)}{r(t)} dt + \int_0^1 i4\pi dt.$$

The rightmost term I have no problem with. The left term though I'm not sure what to do. I tried integrating by parts with $g'=r'(t) \implies g=r(t)$, $f=1/r(t) \implies f'=-r(t)^{-2}r'(t)$. This gives $$\frac{r(t)}{r(t)}+\int r(t)\frac{r'(t)}{r(t)^2}dt=\int \frac{r'(t)}{r(t)}dt$$. But this implies 1=0...

2. ## Re: Evaluate the path integral in C

we can assume that $r(t)= \sqrt{2}(t+1)$

3. ## Re: Evaluate the path integral in C

Originally Posted by Idea
we can assume that $r(t)= \sqrt{2}(t+1)$
That would give $r(0)=\sqrt{2}$ and $r(1)=2\sqrt{2}$.

How do I justify it though? Something about integrals over paths being independent of parameterization though?

Anyways, $$\int_0^1 \frac{r'(t)}{r(t)}dt=\int_0^1 \frac{\sqrt{2}}{\sqrt{2}(t+1)}dt= \log(t+1)\Big|_{t=0}^{t=1}=\log(2)-\log(1).$$

There may be some problems with using logarithms. I don't know.

But then the integral would evaluate to $\log(2)+i4\pi$.

4. ## Re: Evaluate the path integral in C

using the Cauchy integral theorem

this is not about two parameterizations of the same path but rather

a line integral over two different paths connecting the same two points

the path r(t) (>0) is in the upper half-plane so the theorem applies

6. ## Re: Evaluate the path integral in C

Originally Posted by bkbowser
That would give $r(0)=\sqrt{2}$ and $r(1)=2\sqrt{2}$.

How do I justify it though? Something about integrals over paths being independent of parameterization though?

Anyways, $$\int_0^1 \frac{r'(t)}{r(t)}dt=\int_0^1 \frac{\sqrt{2}}{\sqrt{2}(t+1)}dt= \log(t+1)\Big|_{t=0}^{t=1}=\log(2)-\log(1).$$

There may be some problems with using logarithms. I don't know.

But then the integral would evaluate to $\log(2)+i4\pi$.
This is correct. you don't need to use the Cauchy integral theorem.
No problem with the logarithm since r(t) is a real valued function with r(t)>0

OK.

Thanks then.