- #1
latentcorpse
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two questions here:
(i) my notes say that \frac{1}{e^{\frac{1}{z}}-1} has an isolated singularity at z=\frac{1}{2 \pi i n}, n \in \mathbb{Z} \backslash \{0\}
i can't see this though...
(ii) let b \in \mathbb{R}. show
\int_{-\infty}^{\infty} e^{-x^2} \cos{(2bx)} dx = e^{-b^2} \sqrt{\pi}
we take a rectangular contour \Gamma with vertices (R,0) , (R,b) , (-R,b) , (-R,0)
this integrand is holomorphic on the entire complex plane so Cauchy's Theorem tells us that
\int_{\Gamma}e^{-z^2} \cos{(2bz)} dz=0[/latex<br /> <br /> now we define <br /> \gamma_1 between (-R,0) and (R,0) <br /> \gamma_2 between (R,0) and (R,b) <br /> \gamma_3 between (R,b) and (-R,b) <br /> \gamma_4 between (-R,b) and (-R,0)<br /> so that \Gamma= \gamma_1 \cup \gamma_2 \cup \gamma_3 \cup \gamma_4<br /> <br /> now on \gamma_2 we use Jordan's lemma<br /> <br /> the length of the curve is b<br /> <br /> and we have |e^{-z^2} \cos{(2bz)}| \leq |e^{-z^2}| as |\cos{(2bz)}| \leq 1<br /> <br /> we parameterise \gamma_2 by z=R+it, t \in [0,b]<br /> <br /> so |e^{-z^2}| = |e^{-(R+it)^2}| \leq e^{t^2-R^2} \leq e^{b^2-R^2}<br /> <br /> so \int_{\gamma_2} e^{-z^2} \cos{(2bz)} dz \leq be^{b^2-R^2} \rightarrow 0 as R \rightarrow \infty<br /> <br /> and \gamma_4 vanishes simlarly<br /> <br /> my problem is how to treat \gamma_3. is it meant to be rearranged so that we get the integral over \gamma_3 is proportional to the integral over \gamma_1. i was going to say it ISN'T because we want it to be non-zero so that the integral over \gamma_1 doesn't end up as 0.<br /> <br /> this is what I've been trying:<br /> <br /> parameterise \gamma_3 by z=z+ib , x \in [-R,R]<br /> <br /> \int_{\gamma_3} f(z) dz = \int_R^{-R} e^{-(x+ib)^2} \cos{(2b(x+ib))} dx<br /> <br /> do i continue down this road or am i supposed to be using Jordan's lemma here:<br /> <br /> |e^{-z^2} \cos{(2bz)}| \leq |e^{-z^2}| = |e^{-x^2} e^{-2ixb} e^{b^2}| = |e^{b^2-x^2}| \leq e^{b^2} as x \in [-R,R] and the length of this contour is b so<br /> <br /> we'd end up with our desired integral being equal to -be^{b^2} which is a bit off<br /> <br /> any advice...
(i) my notes say that \frac{1}{e^{\frac{1}{z}}-1} has an isolated singularity at z=\frac{1}{2 \pi i n}, n \in \mathbb{Z} \backslash \{0\}
i can't see this though...
(ii) let b \in \mathbb{R}. show
\int_{-\infty}^{\infty} e^{-x^2} \cos{(2bx)} dx = e^{-b^2} \sqrt{\pi}
we take a rectangular contour \Gamma with vertices (R,0) , (R,b) , (-R,b) , (-R,0)
this integrand is holomorphic on the entire complex plane so Cauchy's Theorem tells us that
\int_{\Gamma}e^{-z^2} \cos{(2bz)} dz=0[/latex<br /> <br /> now we define <br /> \gamma_1 between (-R,0) and (R,0) <br /> \gamma_2 between (R,0) and (R,b) <br /> \gamma_3 between (R,b) and (-R,b) <br /> \gamma_4 between (-R,b) and (-R,0)<br /> so that \Gamma= \gamma_1 \cup \gamma_2 \cup \gamma_3 \cup \gamma_4<br /> <br /> now on \gamma_2 we use Jordan's lemma<br /> <br /> the length of the curve is b<br /> <br /> and we have |e^{-z^2} \cos{(2bz)}| \leq |e^{-z^2}| as |\cos{(2bz)}| \leq 1<br /> <br /> we parameterise \gamma_2 by z=R+it, t \in [0,b]<br /> <br /> so |e^{-z^2}| = |e^{-(R+it)^2}| \leq e^{t^2-R^2} \leq e^{b^2-R^2}<br /> <br /> so \int_{\gamma_2} e^{-z^2} \cos{(2bz)} dz \leq be^{b^2-R^2} \rightarrow 0 as R \rightarrow \infty<br /> <br /> and \gamma_4 vanishes simlarly<br /> <br /> my problem is how to treat \gamma_3. is it meant to be rearranged so that we get the integral over \gamma_3 is proportional to the integral over \gamma_1. i was going to say it ISN'T because we want it to be non-zero so that the integral over \gamma_1 doesn't end up as 0.<br /> <br /> this is what I've been trying:<br /> <br /> parameterise \gamma_3 by z=z+ib , x \in [-R,R]<br /> <br /> \int_{\gamma_3} f(z) dz = \int_R^{-R} e^{-(x+ib)^2} \cos{(2b(x+ib))} dx<br /> <br /> do i continue down this road or am i supposed to be using Jordan's lemma here:<br /> <br /> |e^{-z^2} \cos{(2bz)}| \leq |e^{-z^2}| = |e^{-x^2} e^{-2ixb} e^{b^2}| = |e^{b^2-x^2}| \leq e^{b^2} as x \in [-R,R] and the length of this contour is b so<br /> <br /> we'd end up with our desired integral being equal to -be^{b^2} which is a bit off<br /> <br /> any advice...
