Why do we consider an infinite semicircular contour for the integral $int_-infty^inftyf(x) dx$
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I have noticed that $beginalign*int_-infty^inftyf(x) dx endalign*$ can be solved using residue theorem.
My question is that while the actual integral doesn't have a closed path, it's a line integral that goes from $-infty$ to $infty$, so while using residue theorem and after replacing $x$ by $z$ why do we consider a semi-circular contour over positive imaginary plane.
How does an open line integral change into a closed loop integral? It would be better if someone can explain it intuitively.
complex-analysis intuition residue-calculus
edited Aug 25 at 10:55
Henning Makholm
229k16295526
asked Aug 25 at 9:47
paulplusx
437114
add a comment |Â
up vote
0
down vote
favorite
I have noticed that $beginalign*int_-infty^inftyf(x) dx endalign*$ can be solved using residue theorem.
My question is that while the actual integral doesn't have a closed path, it's a line integral that goes from $-infty$ to $infty$, so while using residue theorem and after replacing $x$ by $z$ why do we consider a semi-circular contour over positive imaginary plane.
How does an open line integral change into a closed loop integral? It would be better if someone can explain it intuitively.
complex-analysis intuition residue-calculus
edited Aug 25 at 10:55
Henning Makholm
229k16295526
asked Aug 25 at 9:47
paulplusx
437114
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have noticed that $beginalign*int_-infty^inftyf(x) dx endalign*$ can be solved using residue theorem.
My question is that while the actual integral doesn't have a closed path, it's a line integral that goes from $-infty$ to $infty$, so while using residue theorem and after replacing $x$ by $z$ why do we consider a semi-circular contour over positive imaginary plane.
How does an open line integral change into a closed loop integral? It would be better if someone can explain it intuitively.
complex-analysis intuition residue-calculus
edited Aug 25 at 10:55
Henning Makholm
229k16295526
asked Aug 25 at 9:47
paulplusx
437114
I have noticed that $beginalign*int_-infty^inftyf(x) dx endalign*$ can be solved using residue theorem.
My question is that while the actual integral doesn't have a closed path, it's a line integral that goes from $-infty$ to $infty$, so while using residue theorem and after replacing $x$ by $z$ why do we consider a semi-circular contour over positive imaginary plane.
How does an open line integral change into a closed loop integral? It would be better if someone can explain it intuitively.
complex-analysis intuition residue-calculus
edited Aug 25 at 10:55
Henning Makholm
229k16295526
asked Aug 25 at 9:47
paulplusx
437114
edited Aug 25 at 10:55
Henning Makholm
229k16295526
edited Aug 25 at 10:55
Henning Makholm
229k16295526
edited Aug 25 at 10:55
Henning Makholm
229k16295526
229k16295526
asked Aug 25 at 9:47
paulplusx
437114
asked Aug 25 at 9:47
paulplusx
437114
asked Aug 25 at 9:47
paulplusx
437114
437114
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Assuming the integral exists,
$$int_-infty^infty
= lim_r to infty int_-r^r
= lim_r to infty left( oint_textboundary of semicircle - int_textsemicircular arc right) $$
A typical proof method using this fact is:
- Compute $oint_textboundary of semicircle$ by the residue theorem
- Show that $int_textsemicircular arc$ converges to zero. This is often done by showing the magnitude of the integrand is asymptotically smaller than the arclength
answered Aug 25 at 10:54
Hurkyl
109k9113254
+1, Nice answer. Unfortunately, I still don't understand how the integration around the semi-circular arc converges to $0$, could you please elaborate on that?
– paulplusx
Aug 25 at 11:05
@paulplusx: It doesn't, in general. When applying this technique to solve a particular problem, proving it converges to zero is one of the tasks you need to do.
– Hurkyl
Aug 25 at 11:39
Ohh, understood. So it essentially means that whenever such questions are solved with residue method, the semi-circular arc integral converges to $0$. That's the intuition I required, thank you.
– paulplusx
Aug 25 at 11:42
I searched a bit on this and found something called Jordan's Lemma. I think it shows the proof for the above arc integral to vanish.
– paulplusx
Aug 25 at 13:30
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Assuming the integral exists,
$$int_-infty^infty
= lim_r to infty int_-r^r
= lim_r to infty left( oint_textboundary of semicircle - int_textsemicircular arc right) $$
A typical proof method using this fact is:
- Compute $oint_textboundary of semicircle$ by the residue theorem
- Show that $int_textsemicircular arc$ converges to zero. This is often done by showing the magnitude of the integrand is asymptotically smaller than the arclength
answered Aug 25 at 10:54
Hurkyl
109k9113254
+1, Nice answer. Unfortunately, I still don't understand how the integration around the semi-circular arc converges to $0$, could you please elaborate on that?
– paulplusx
Aug 25 at 11:05
@paulplusx: It doesn't, in general. When applying this technique to solve a particular problem, proving it converges to zero is one of the tasks you need to do.
– Hurkyl
Aug 25 at 11:39
Ohh, understood. So it essentially means that whenever such questions are solved with residue method, the semi-circular arc integral converges to $0$. That's the intuition I required, thank you.
– paulplusx
Aug 25 at 11:42
I searched a bit on this and found something called Jordan's Lemma. I think it shows the proof for the above arc integral to vanish.
– paulplusx
Aug 25 at 13:30
add a comment |Â
up vote
2
down vote
accepted
Assuming the integral exists,
$$int_-infty^infty
= lim_r to infty int_-r^r
= lim_r to infty left( oint_textboundary of semicircle - int_textsemicircular arc right) $$
A typical proof method using this fact is:
- Compute $oint_textboundary of semicircle$ by the residue theorem
- Show that $int_textsemicircular arc$ converges to zero. This is often done by showing the magnitude of the integrand is asymptotically smaller than the arclength
answered Aug 25 at 10:54
Hurkyl
109k9113254
+1, Nice answer. Unfortunately, I still don't understand how the integration around the semi-circular arc converges to $0$, could you please elaborate on that?
– paulplusx
Aug 25 at 11:05
@paulplusx: It doesn't, in general. When applying this technique to solve a particular problem, proving it converges to zero is one of the tasks you need to do.
– Hurkyl
Aug 25 at 11:39
Ohh, understood. So it essentially means that whenever such questions are solved with residue method, the semi-circular arc integral converges to $0$. That's the intuition I required, thank you.
– paulplusx
Aug 25 at 11:42
I searched a bit on this and found something called Jordan's Lemma. I think it shows the proof for the above arc integral to vanish.
– paulplusx
Aug 25 at 13:30
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Assuming the integral exists,
$$int_-infty^infty
= lim_r to infty int_-r^r
= lim_r to infty left( oint_textboundary of semicircle - int_textsemicircular arc right) $$
A typical proof method using this fact is:
- Compute $oint_textboundary of semicircle$ by the residue theorem
- Show that $int_textsemicircular arc$ converges to zero. This is often done by showing the magnitude of the integrand is asymptotically smaller than the arclength
answered Aug 25 at 10:54
Hurkyl
109k9113254
Assuming the integral exists,
$$int_-infty^infty
= lim_r to infty int_-r^r
= lim_r to infty left( oint_textboundary of semicircle - int_textsemicircular arc right) $$
A typical proof method using this fact is:
- Compute $oint_textboundary of semicircle$ by the residue theorem
- Show that $int_textsemicircular arc$ converges to zero. This is often done by showing the magnitude of the integrand is asymptotically smaller than the arclength
answered Aug 25 at 10:54
Hurkyl
109k9113254
answered Aug 25 at 10:54
Hurkyl
109k9113254
answered Aug 25 at 10:54
Hurkyl
109k9113254
answered Aug 25 at 10:54
Hurkyl
109k9113254
109k9113254
+1, Nice answer. Unfortunately, I still don't understand how the integration around the semi-circular arc converges to $0$, could you please elaborate on that?
– paulplusx
Aug 25 at 11:05
@paulplusx: It doesn't, in general. When applying this technique to solve a particular problem, proving it converges to zero is one of the tasks you need to do.
– Hurkyl
Aug 25 at 11:39
Ohh, understood. So it essentially means that whenever such questions are solved with residue method, the semi-circular arc integral converges to $0$. That's the intuition I required, thank you.
– paulplusx
Aug 25 at 11:42
I searched a bit on this and found something called Jordan's Lemma. I think it shows the proof for the above arc integral to vanish.
– paulplusx
Aug 25 at 13:30
add a comment |Â
+1, Nice answer. Unfortunately, I still don't understand how the integration around the semi-circular arc converges to $0$, could you please elaborate on that?
– paulplusx
Aug 25 at 11:05
@paulplusx: It doesn't, in general. When applying this technique to solve a particular problem, proving it converges to zero is one of the tasks you need to do.
– Hurkyl
Aug 25 at 11:39
Ohh, understood. So it essentially means that whenever such questions are solved with residue method, the semi-circular arc integral converges to $0$. That's the intuition I required, thank you.
– paulplusx
Aug 25 at 11:42
I searched a bit on this and found something called Jordan's Lemma. I think it shows the proof for the above arc integral to vanish.
– paulplusx
Aug 25 at 13:30
+1, Nice answer. Unfortunately, I still don't understand how the integration around the semi-circular arc converges to $0$, could you please elaborate on that?
– paulplusx
Aug 25 at 11:05
+1, Nice answer. Unfortunately, I still don't understand how the integration around the semi-circular arc converges to $0$, could you please elaborate on that?
– paulplusx
Aug 25 at 11:05
@paulplusx: It doesn't, in general. When applying this technique to solve a particular problem, proving it converges to zero is one of the tasks you need to do.
– Hurkyl
Aug 25 at 11:39
@paulplusx: It doesn't, in general. When applying this technique to solve a particular problem, proving it converges to zero is one of the tasks you need to do.
– Hurkyl
Aug 25 at 11:39
Ohh, understood. So it essentially means that whenever such questions are solved with residue method, the semi-circular arc integral converges to $0$. That's the intuition I required, thank you.
– paulplusx
Aug 25 at 11:42
Ohh, understood. So it essentially means that whenever such questions are solved with residue method, the semi-circular arc integral converges to $0$. That's the intuition I required, thank you.
– paulplusx
Aug 25 at 11:42
I searched a bit on this and found something called Jordan's Lemma. I think it shows the proof for the above arc integral to vanish.
– paulplusx
Aug 25 at 13:30
I searched a bit on this and found something called Jordan's Lemma. I think it shows the proof for the above arc integral to vanish.
– paulplusx
Aug 25 at 13:30
add a comment |Â
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