Problem in showing that contours $\gamma_2$ is equivalent to $ \gamma $

Question

Let $\gamma_2(t)= e^{-it^2}, t\in[0,\sqrt{2\pi}]$ and $\gamma(t)=e^{2\pi it}, t\in[0,1]$ Show that $\gamma_2 \sim \gamma $.

I think that for the latter to be true $\gamma_2$ should be $\gamma_2(t)= e^{it^2}, t\in[0,\sqrt{2\pi}]$

I have tried to prove $\gamma_2 \sim \gamma $ for the original $\gamma_2$ by showing that $ \int_{\gamma_2}f(z)dz=\int_{\gamma}f(z)dz $ for an arbitrary $f$ ,

but that $-$ in $e^{-it^2}$ messes up the integral limits producing

$ \int_{\gamma_2}f(z)dz=\int_0^{-1}f(\gamma(t))\gamma(t)'dt $ instead of

$ \int_{\gamma_2}f(z)dz=\int_0^1f(\gamma(t))\gamma(t)'dt=\int_{\gamma}f(z)dz $ .

What am I doing wrong?

Answer 1

Can you show the following are equivalent? What would the reparametrizations be, explicitly?

1. $a:[0,\sqrt{2\pi}]\to \Bbb C:t\mapsto e^{-it^2}$
2. $b:[0,2\pi]\to \Bbb C:t\mapsto e^{-it}$
3. $c:[0,2\pi]\to \Bbb C:t\mapsto e^{it}$
4. $d:[0,1]\to\Bbb C:t\mapsto e^{2\pi it}$

It might help to review what exactly a reparametrization is.

By the way, your integrals should have been

$$\int_0^{\sqrt{2\pi}}f(\gamma_2(t))\gamma_2'(t)dt \quad{\rm vs}\quad \int_0^1f(\gamma(t))\gamma'(t)dt,$$

where $\gamma_2(t)=e^{-it^2}$ and $\gamma(t)=e^{2\pi it}$, exactly as you defined them. Can you simplify them now?