Mixing
Showing that a path in the complex plane converges so we can use the residue theorem
$begingroup$
I wanted to calculate the Fourier transform of $$f(t):=frac{1}{t^2+6t+13}
tag{1}$$
so basically calculating
$$F[f](t)=hat{f}(omega)=int_{-infty}^infty f(t) e^{-iomega t}dt$$
and doing so by using the residue theorem. Now let's consider the path $gamma_1=gamma_{1,R}+gamma_{1,I}$ along the upper half-circle with radius $R$.
Note that this question is only about the convergence of the non-real path, everything else is more or less clear.
$$gamma_{1,R}: [0,1] to mathbb R, quad tmapsto 2Rt-R, qquad dot{gamma}_{1,R}(t)=2R tag{2}$$
$$gamma_{1,I}: [0,1] to mathbb R, quad tmapsto Re^{pi i t}, qquad dot{gamma}_{1,I}(t)=pi i Re^{pi i t}$$
Now we have
$$int_{gamma_1}f(t)e^{-iomega t}dt=int_{gamma_{1,R}}f(t)e^{-iomega t}dt+int_{gamma_{1,I}}f(t)e^{-iomega t}dt=2pi i Res(f;t_1)tag{3}$$
with $t_1$ being the singularity enclosed by $gamma_1$.
Now we want to check the convergence of $int_{gamma_{1,I}}f(t)e^{-iomega t}dt$:
$$int_{gamma_{1,I}}f(z)e^{-iomega t}dzsimint_{gamma_{1,I}}frac{e^{-iomega t}}{t^2}dt=int_0^1frac{e^{-iomega Re^{pi i t}}}{Re^{2pi i t}}pi i Re^{pi i t}dt=pi iint_0^1frac{e^{-iomega Re^{pi i t}}}{Re^{pi i t}}dt tag{4}$$
Now, I I'd like to somehow show that the last term in (4) converges to 0 for $Rtoinfty$ but I can't really see how. I know there are probably other way to approach this, some lemmas and what not (you are welcome to mention it), but I'd like to find an estimate for it which shows me the convergence. :)
How would I proceed to show the convergence?
fourier-analysis
asked Jan 31 at 20:09
xotixxotix
291411
$endgroup$
add a comment |
$begingroup$
I wanted to calculate the Fourier transform of $$f(t):=frac{1}{t^2+6t+13}
tag{1}$$
so basically calculating
$$F[f](t)=hat{f}(omega)=int_{-infty}^infty f(t) e^{-iomega t}dt$$
and doing so by using the residue theorem. Now let's consider the path $gamma_1=gamma_{1,R}+gamma_{1,I}$ along the upper half-circle with radius $R$.
Note that this question is only about the convergence of the non-real path, everything else is more or less clear.
$$gamma_{1,R}: [0,1] to mathbb R, quad tmapsto 2Rt-R, qquad dot{gamma}_{1,R}(t)=2R tag{2}$$
$$gamma_{1,I}: [0,1] to mathbb R, quad tmapsto Re^{pi i t}, qquad dot{gamma}_{1,I}(t)=pi i Re^{pi i t}$$
Now we have
$$int_{gamma_1}f(t)e^{-iomega t}dt=int_{gamma_{1,R}}f(t)e^{-iomega t}dt+int_{gamma_{1,I}}f(t)e^{-iomega t}dt=2pi i Res(f;t_1)tag{3}$$
with $t_1$ being the singularity enclosed by $gamma_1$.
Now we want to check the convergence of $int_{gamma_{1,I}}f(t)e^{-iomega t}dt$:
$$int_{gamma_{1,I}}f(z)e^{-iomega t}dzsimint_{gamma_{1,I}}frac{e^{-iomega t}}{t^2}dt=int_0^1frac{e^{-iomega Re^{pi i t}}}{Re^{2pi i t}}pi i Re^{pi i t}dt=pi iint_0^1frac{e^{-iomega Re^{pi i t}}}{Re^{pi i t}}dt tag{4}$$
Now, I I'd like to somehow show that the last term in (4) converges to 0 for $Rtoinfty$ but I can't really see how. I know there are probably other way to approach this, some lemmas and what not (you are welcome to mention it), but I'd like to find an estimate for it which shows me the convergence. :)
How would I proceed to show the convergence?
fourier-analysis
asked Jan 31 at 20:09
xotixxotix
291411
$endgroup$
$begingroup$
Yeah, that's the question I asked, how to properly show that.
$endgroup$
– xotix
Jan 31 at 21:18
add a comment |
$begingroup$
I wanted to calculate the Fourier transform of $$f(t):=frac{1}{t^2+6t+13}
tag{1}$$
so basically calculating
$$F[f](t)=hat{f}(omega)=int_{-infty}^infty f(t) e^{-iomega t}dt$$
and doing so by using the residue theorem. Now let's consider the path $gamma_1=gamma_{1,R}+gamma_{1,I}$ along the upper half-circle with radius $R$.
Note that this question is only about the convergence of the non-real path, everything else is more or less clear.
$$gamma_{1,R}: [0,1] to mathbb R, quad tmapsto 2Rt-R, qquad dot{gamma}_{1,R}(t)=2R tag{2}$$
$$gamma_{1,I}: [0,1] to mathbb R, quad tmapsto Re^{pi i t}, qquad dot{gamma}_{1,I}(t)=pi i Re^{pi i t}$$
Now we have
$$int_{gamma_1}f(t)e^{-iomega t}dt=int_{gamma_{1,R}}f(t)e^{-iomega t}dt+int_{gamma_{1,I}}f(t)e^{-iomega t}dt=2pi i Res(f;t_1)tag{3}$$
with $t_1$ being the singularity enclosed by $gamma_1$.
Now we want to check the convergence of $int_{gamma_{1,I}}f(t)e^{-iomega t}dt$:
$$int_{gamma_{1,I}}f(z)e^{-iomega t}dzsimint_{gamma_{1,I}}frac{e^{-iomega t}}{t^2}dt=int_0^1frac{e^{-iomega Re^{pi i t}}}{Re^{2pi i t}}pi i Re^{pi i t}dt=pi iint_0^1frac{e^{-iomega Re^{pi i t}}}{Re^{pi i t}}dt tag{4}$$
Now, I I'd like to somehow show that the last term in (4) converges to 0 for $Rtoinfty$ but I can't really see how. I know there are probably other way to approach this, some lemmas and what not (you are welcome to mention it), but I'd like to find an estimate for it which shows me the convergence. :)
How would I proceed to show the convergence?
fourier-analysis
asked Jan 31 at 20:09
xotixxotix
291411
$endgroup$
I wanted to calculate the Fourier transform of $$f(t):=frac{1}{t^2+6t+13}
tag{1}$$
so basically calculating
$$F[f](t)=hat{f}(omega)=int_{-infty}^infty f(t) e^{-iomega t}dt$$
and doing so by using the residue theorem. Now let's consider the path $gamma_1=gamma_{1,R}+gamma_{1,I}$ along the upper half-circle with radius $R$.
Note that this question is only about the convergence of the non-real path, everything else is more or less clear.
$$gamma_{1,R}: [0,1] to mathbb R, quad tmapsto 2Rt-R, qquad dot{gamma}_{1,R}(t)=2R tag{2}$$
$$gamma_{1,I}: [0,1] to mathbb R, quad tmapsto Re^{pi i t}, qquad dot{gamma}_{1,I}(t)=pi i Re^{pi i t}$$
Now we have
$$int_{gamma_1}f(t)e^{-iomega t}dt=int_{gamma_{1,R}}f(t)e^{-iomega t}dt+int_{gamma_{1,I}}f(t)e^{-iomega t}dt=2pi i Res(f;t_1)tag{3}$$
with $t_1$ being the singularity enclosed by $gamma_1$.
Now we want to check the convergence of $int_{gamma_{1,I}}f(t)e^{-iomega t}dt$:
$$int_{gamma_{1,I}}f(z)e^{-iomega t}dzsimint_{gamma_{1,I}}frac{e^{-iomega t}}{t^2}dt=int_0^1frac{e^{-iomega Re^{pi i t}}}{Re^{2pi i t}}pi i Re^{pi i t}dt=pi iint_0^1frac{e^{-iomega Re^{pi i t}}}{Re^{pi i t}}dt tag{4}$$
Now, I I'd like to somehow show that the last term in (4) converges to 0 for $Rtoinfty$ but I can't really see how. I know there are probably other way to approach this, some lemmas and what not (you are welcome to mention it), but I'd like to find an estimate for it which shows me the convergence. :)
How would I proceed to show the convergence?
fourier-analysis
fourier-analysis
asked Jan 31 at 20:09
xotixxotix
291411
asked Jan 31 at 20:09
xotixxotix
291411
asked Jan 31 at 20:09
xotixxotix
291411
asked Jan 31 at 20:09
xotixxotix
291411
asked Jan 31 at 20:09
xotixxotix
291411
291411
$begingroup$
Yeah, that's the question I asked, how to properly show that.
$endgroup$
– xotix
Jan 31 at 21:18
add a comment |
$begingroup$
Yeah, that's the question I asked, how to properly show that.
$endgroup$
– xotix
Jan 31 at 21:18
$begingroup$
Yeah, that's the question I asked, how to properly show that.
$endgroup$
– xotix
Jan 31 at 21:18
$begingroup$
Yeah, that's the question I asked, how to properly show that.
$endgroup$
– xotix
Jan 31 at 21:18
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Suppose that $f$ is meromorphic and doesnt have poles in the real line, also suppose that $lim_{|z|toinfty} zf(z)=0$ and $omegale 0$. Then choosing the arc above the real line we have by the residue theorem that
$$int_{-R}^R f(x)e^{-iomega x}, dx+iint_0^pi f(R e^{it})e^{-iomega Re^{it}}Re^{i t}, dt=2pi isum_{zin V}operatorname{Res}(f e^{-iomegacdot},z)$$
where $V:=RBbb Dcap{zinBbb C:Im(z)> 0}$. But now note that
$$left|int_0^pi f(R e^{it})e^{-iomega Re^{it}}Re^{i t}, dtright|lepimax_{tin[0,pi]} |f(Re^{it})Re^{omega Rsin t}|lepimax_{|z|=R}|zf(z)|$$
so the integral in the arc goes to zero as $R$ goes to infinity. For the case $omega >0$ we choose the arc below the real line, with a similar result.
answered Jan 31 at 21:46
MasacrosoMasacroso
13.1k41747
$endgroup$
$begingroup$
How do you get from $e^{-iomega R e^{it}}$ to $e^{omega Rsin(t)}$? I mean, I see that you somehow used Euler's Formula but I can't figure out how to get there.
$endgroup$
– xotix
Jan 31 at 21:52
$begingroup$
took absolute values, note that $|e^z|= e^{Re(z)}$, that is, if $z=a+ib$ for some $a,binBbb R$ then $|e^z|=|e^a|cdot|e^{ib}|=e^a$.
$endgroup$
– Masacroso
Jan 31 at 22:55
$begingroup$
ah, of course. That's how you end up with the sin term. Thanks!
$endgroup$
– xotix
Feb 1 at 9:38
add a comment |
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1 Answer
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$begingroup$
Suppose that $f$ is meromorphic and doesnt have poles in the real line, also suppose that $lim_{|z|toinfty} zf(z)=0$ and $omegale 0$. Then choosing the arc above the real line we have by the residue theorem that
$$int_{-R}^R f(x)e^{-iomega x}, dx+iint_0^pi f(R e^{it})e^{-iomega Re^{it}}Re^{i t}, dt=2pi isum_{zin V}operatorname{Res}(f e^{-iomegacdot},z)$$
where $V:=RBbb Dcap{zinBbb C:Im(z)> 0}$. But now note that
$$left|int_0^pi f(R e^{it})e^{-iomega Re^{it}}Re^{i t}, dtright|lepimax_{tin[0,pi]} |f(Re^{it})Re^{omega Rsin t}|lepimax_{|z|=R}|zf(z)|$$
so the integral in the arc goes to zero as $R$ goes to infinity. For the case $omega >0$ we choose the arc below the real line, with a similar result.
answered Jan 31 at 21:46
MasacrosoMasacroso
13.1k41747
$endgroup$
$begingroup$
How do you get from $e^{-iomega R e^{it}}$ to $e^{omega Rsin(t)}$? I mean, I see that you somehow used Euler's Formula but I can't figure out how to get there.
$endgroup$
– xotix
Jan 31 at 21:52
$begingroup$
took absolute values, note that $|e^z|= e^{Re(z)}$, that is, if $z=a+ib$ for some $a,binBbb R$ then $|e^z|=|e^a|cdot|e^{ib}|=e^a$.
$endgroup$
– Masacroso
Jan 31 at 22:55
$begingroup$
ah, of course. That's how you end up with the sin term. Thanks!
$endgroup$
– xotix
Feb 1 at 9:38
add a comment |
$begingroup$
Suppose that $f$ is meromorphic and doesnt have poles in the real line, also suppose that $lim_{|z|toinfty} zf(z)=0$ and $omegale 0$. Then choosing the arc above the real line we have by the residue theorem that
$$int_{-R}^R f(x)e^{-iomega x}, dx+iint_0^pi f(R e^{it})e^{-iomega Re^{it}}Re^{i t}, dt=2pi isum_{zin V}operatorname{Res}(f e^{-iomegacdot},z)$$
where $V:=RBbb Dcap{zinBbb C:Im(z)> 0}$. But now note that
$$left|int_0^pi f(R e^{it})e^{-iomega Re^{it}}Re^{i t}, dtright|lepimax_{tin[0,pi]} |f(Re^{it})Re^{omega Rsin t}|lepimax_{|z|=R}|zf(z)|$$
so the integral in the arc goes to zero as $R$ goes to infinity. For the case $omega >0$ we choose the arc below the real line, with a similar result.
answered Jan 31 at 21:46
MasacrosoMasacroso
13.1k41747
$endgroup$
$begingroup$
How do you get from $e^{-iomega R e^{it}}$ to $e^{omega Rsin(t)}$? I mean, I see that you somehow used Euler's Formula but I can't figure out how to get there.
$endgroup$
– xotix
Jan 31 at 21:52
$begingroup$
took absolute values, note that $|e^z|= e^{Re(z)}$, that is, if $z=a+ib$ for some $a,binBbb R$ then $|e^z|=|e^a|cdot|e^{ib}|=e^a$.
$endgroup$
– Masacroso
Jan 31 at 22:55
$begingroup$
ah, of course. That's how you end up with the sin term. Thanks!
$endgroup$
– xotix
Feb 1 at 9:38
add a comment |
$begingroup$
Suppose that $f$ is meromorphic and doesnt have poles in the real line, also suppose that $lim_{|z|toinfty} zf(z)=0$ and $omegale 0$. Then choosing the arc above the real line we have by the residue theorem that
$$int_{-R}^R f(x)e^{-iomega x}, dx+iint_0^pi f(R e^{it})e^{-iomega Re^{it}}Re^{i t}, dt=2pi isum_{zin V}operatorname{Res}(f e^{-iomegacdot},z)$$
where $V:=RBbb Dcap{zinBbb C:Im(z)> 0}$. But now note that
$$left|int_0^pi f(R e^{it})e^{-iomega Re^{it}}Re^{i t}, dtright|lepimax_{tin[0,pi]} |f(Re^{it})Re^{omega Rsin t}|lepimax_{|z|=R}|zf(z)|$$
so the integral in the arc goes to zero as $R$ goes to infinity. For the case $omega >0$ we choose the arc below the real line, with a similar result.
answered Jan 31 at 21:46
MasacrosoMasacroso
13.1k41747
$endgroup$
Suppose that $f$ is meromorphic and doesnt have poles in the real line, also suppose that $lim_{|z|toinfty} zf(z)=0$ and $omegale 0$. Then choosing the arc above the real line we have by the residue theorem that
$$int_{-R}^R f(x)e^{-iomega x}, dx+iint_0^pi f(R e^{it})e^{-iomega Re^{it}}Re^{i t}, dt=2pi isum_{zin V}operatorname{Res}(f e^{-iomegacdot},z)$$
where $V:=RBbb Dcap{zinBbb C:Im(z)> 0}$. But now note that
$$left|int_0^pi f(R e^{it})e^{-iomega Re^{it}}Re^{i t}, dtright|lepimax_{tin[0,pi]} |f(Re^{it})Re^{omega Rsin t}|lepimax_{|z|=R}|zf(z)|$$
so the integral in the arc goes to zero as $R$ goes to infinity. For the case $omega >0$ we choose the arc below the real line, with a similar result.
answered Jan 31 at 21:46
MasacrosoMasacroso
13.1k41747
answered Jan 31 at 21:46
MasacrosoMasacroso
13.1k41747
answered Jan 31 at 21:46
MasacrosoMasacroso
13.1k41747
answered Jan 31 at 21:46
MasacrosoMasacroso
13.1k41747
13.1k41747
$begingroup$
How do you get from $e^{-iomega R e^{it}}$ to $e^{omega Rsin(t)}$? I mean, I see that you somehow used Euler's Formula but I can't figure out how to get there.
$endgroup$
– xotix
Jan 31 at 21:52
$begingroup$
took absolute values, note that $|e^z|= e^{Re(z)}$, that is, if $z=a+ib$ for some $a,binBbb R$ then $|e^z|=|e^a|cdot|e^{ib}|=e^a$.
$endgroup$
– Masacroso
Jan 31 at 22:55
$begingroup$
ah, of course. That's how you end up with the sin term. Thanks!
$endgroup$
– xotix
Feb 1 at 9:38
add a comment |
$begingroup$
How do you get from $e^{-iomega R e^{it}}$ to $e^{omega Rsin(t)}$? I mean, I see that you somehow used Euler's Formula but I can't figure out how to get there.
$endgroup$
– xotix
Jan 31 at 21:52
$begingroup$
took absolute values, note that $|e^z|= e^{Re(z)}$, that is, if $z=a+ib$ for some $a,binBbb R$ then $|e^z|=|e^a|cdot|e^{ib}|=e^a$.
$endgroup$
– Masacroso
Jan 31 at 22:55
$begingroup$
ah, of course. That's how you end up with the sin term. Thanks!
$endgroup$
– xotix
Feb 1 at 9:38
$begingroup$
How do you get from $e^{-iomega R e^{it}}$ to $e^{omega Rsin(t)}$? I mean, I see that you somehow used Euler's Formula but I can't figure out how to get there.
$endgroup$
– xotix
Jan 31 at 21:52
$begingroup$
How do you get from $e^{-iomega R e^{it}}$ to $e^{omega Rsin(t)}$? I mean, I see that you somehow used Euler's Formula but I can't figure out how to get there.
$endgroup$
– xotix
Jan 31 at 21:52
$begingroup$
took absolute values, note that $|e^z|= e^{Re(z)}$, that is, if $z=a+ib$ for some $a,binBbb R$ then $|e^z|=|e^a|cdot|e^{ib}|=e^a$.
$endgroup$
– Masacroso
Jan 31 at 22:55
$begingroup$
took absolute values, note that $|e^z|= e^{Re(z)}$, that is, if $z=a+ib$ for some $a,binBbb R$ then $|e^z|=|e^a|cdot|e^{ib}|=e^a$.
$endgroup$
– Masacroso
Jan 31 at 22:55
$begingroup$
ah, of course. That's how you end up with the sin term. Thanks!
$endgroup$
– xotix
Feb 1 at 9:38
$begingroup$
ah, of course. That's how you end up with the sin term. Thanks!
$endgroup$
– xotix
Feb 1 at 9:38
add a comment |
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$begingroup$
Yeah, that's the question I asked, how to properly show that.
$endgroup$
– xotix
Jan 31 at 21:18