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Thread: Evaluate the path integral in C

  1. #1
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    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...
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  2. #2
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    Re: Evaluate the path integral in C

    we can assume that $r(t)= \sqrt{2}(t+1)$
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  3. #3
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    Re: Evaluate the path integral in C

    Quote Originally Posted by Idea View Post
    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 $\log(2)+i4\pi$.
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  4. #4
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    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
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    Re: Evaluate the path integral in C

    Quote Originally Posted by bkbowser View Post
    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
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  7. #7
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    Re: Evaluate the path integral in C

    OK.

    Thanks then.
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