About the proof of $sumlimits_n=-infty^infty f(n)=-pisumlimits_k=1^mtextres [f(z)cot(pi z)]_z=a_k$?

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Let $f(z)$ be a meromorphic function with a finite number of poles $a_1,dots,a_m$, where $a_inotinmathbb Zcup0$.
Prove that if there exists a sequence of contours $C_n$ that goes to the point infinity and is such that $$displaystylelim_ntoinftyint_C_nf(z)cot (pi z)dz=0,$$ then $$displaystylesum_n=-infty^infty f(n)=-pisum_k=1^mtextres [f(z)cot(pi z)]_z=a_k$$
From Marsden book:

What does it mean the part that says
Taking limits on both sides...
If you take the limits,
$displaystylelim_Ntoinftysum_n=-N^N f(n)=lim_Ntoinftysumtextres [f(z)picot(pi z)]_z=ndots (*)$
and how do you pass from $(*)$ to $displaystyle-sum_k=1^mtextres [pi f(z)cot(pi z)]_z=a_k$ ?
Could anyone explain please?
Thank you
complex-analysis analysis proof-explanation meromorphic-functions
add a comment |Â
up vote
1
down vote
favorite
Let $f(z)$ be a meromorphic function with a finite number of poles $a_1,dots,a_m$, where $a_inotinmathbb Zcup0$.
Prove that if there exists a sequence of contours $C_n$ that goes to the point infinity and is such that $$displaystylelim_ntoinftyint_C_nf(z)cot (pi z)dz=0,$$ then $$displaystylesum_n=-infty^infty f(n)=-pisum_k=1^mtextres [f(z)cot(pi z)]_z=a_k$$
From Marsden book:

What does it mean the part that says
Taking limits on both sides...
If you take the limits,
$displaystylelim_Ntoinftysum_n=-N^N f(n)=lim_Ntoinftysumtextres [f(z)picot(pi z)]_z=ndots (*)$
and how do you pass from $(*)$ to $displaystyle-sum_k=1^mtextres [pi f(z)cot(pi z)]_z=a_k$ ?
Could anyone explain please?
Thank you
complex-analysis analysis proof-explanation meromorphic-functions
There is an example computation at this MSE link.
â Marko Riedel
Sep 8 at 11:59
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $f(z)$ be a meromorphic function with a finite number of poles $a_1,dots,a_m$, where $a_inotinmathbb Zcup0$.
Prove that if there exists a sequence of contours $C_n$ that goes to the point infinity and is such that $$displaystylelim_ntoinftyint_C_nf(z)cot (pi z)dz=0,$$ then $$displaystylesum_n=-infty^infty f(n)=-pisum_k=1^mtextres [f(z)cot(pi z)]_z=a_k$$
From Marsden book:

What does it mean the part that says
Taking limits on both sides...
If you take the limits,
$displaystylelim_Ntoinftysum_n=-N^N f(n)=lim_Ntoinftysumtextres [f(z)picot(pi z)]_z=ndots (*)$
and how do you pass from $(*)$ to $displaystyle-sum_k=1^mtextres [pi f(z)cot(pi z)]_z=a_k$ ?
Could anyone explain please?
Thank you
complex-analysis analysis proof-explanation meromorphic-functions
Let $f(z)$ be a meromorphic function with a finite number of poles $a_1,dots,a_m$, where $a_inotinmathbb Zcup0$.
Prove that if there exists a sequence of contours $C_n$ that goes to the point infinity and is such that $$displaystylelim_ntoinftyint_C_nf(z)cot (pi z)dz=0,$$ then $$displaystylesum_n=-infty^infty f(n)=-pisum_k=1^mtextres [f(z)cot(pi z)]_z=a_k$$
From Marsden book:

What does it mean the part that says
Taking limits on both sides...
If you take the limits,
$displaystylelim_Ntoinftysum_n=-N^N f(n)=lim_Ntoinftysumtextres [f(z)picot(pi z)]_z=ndots (*)$
and how do you pass from $(*)$ to $displaystyle-sum_k=1^mtextres [pi f(z)cot(pi z)]_z=a_k$ ?
Could anyone explain please?
Thank you
complex-analysis analysis proof-explanation meromorphic-functions
complex-analysis analysis proof-explanation meromorphic-functions
edited Sep 13 at 13:13
Martin Sleziak
43.7k6113261
43.7k6113261
asked Sep 8 at 6:04
Isabella
67317
67317
There is an example computation at this MSE link.
â Marko Riedel
Sep 8 at 11:59
add a comment |Â
There is an example computation at this MSE link.
â Marko Riedel
Sep 8 at 11:59
There is an example computation at this MSE link.
â Marko Riedel
Sep 8 at 11:59
There is an example computation at this MSE link.
â Marko Riedel
Sep 8 at 11:59
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Taking limits on both sides of the preceding displayed equation for the integral $oint_C_N(picotpi z)f(z)dz$
This means taking the limit $Ntoinfty$ on both sides of
$$
beginalign
oint_C_Nf(z)dz
&=2pi isumtextRes $(picotpi z)f(z)$ at $-N,cdots,+N$ \
&~~~~+2pi isumtextRes $(picotpi z)f(z)$ at singularities of $f$ \
endalign
$$
which is
$$
beginalign
lim_Ntoinftyoint_C_Nf(z)dz
&=lim_Ntoinfty2pi isumtextRes $(picotpi z)f(z)$ at $-N,cdots,+N$ \
&~~~~+lim_Ntoinfty2pi isumtextRes $(picotpi z)f(z)$ at singularities of $f$ \
endalign
$$
and using the fact that $oint_C_N(picotpi z)f(z)dzto0$ as $Ntoinfty$,
Therefore,
$$
beginalign
0
&=lim_Ntoinfty2pi isumtextRes $(picotpi z)f(z)$ at $-N,cdots,+N$ \
&~~~~+lim_Ntoinfty2pi isumtextRes $(picotpi z)f(z)$ at singularities of $f$ qquad(*)\
endalign
$$
Since for an integer $k$ $$textRes $(picotpi z)f(z)$ at $k$=f(k)$$
we can rewrite $(*)$ as
$$
beginalign
0
&=lim_Ntoinfty2pi isum_k=-N^N f(k) \
&~~~~+lim_Ntoinfty2pi isumtextRes $(picotpi z)f(z)$ at singularities of $f$ \
endalign
$$
Furthermore, recognizing that the second sum does not depend on $N$ for large enough $N$, we have
$$beginalign
0
&=lim_Ntoinfty2pi isum_k=-N^N f(k)
+2pi isumtextRes $(picotpi z)f(z)$ at singularities of $f$ \
0
&=lim_Ntoinftysum_k=-N^N f(k)
+sumtextRes $(picotpi z)f(z)$ at singularities of $f$ \
endalign
$$
$$
colorred
lim_Ntoinftysum_k=-N^N f(k)
=-sumtextRes $(picotpi z)f(z)$ at singularities of $f$
$$
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Taking limits on both sides of the preceding displayed equation for the integral $oint_C_N(picotpi z)f(z)dz$
This means taking the limit $Ntoinfty$ on both sides of
$$
beginalign
oint_C_Nf(z)dz
&=2pi isumtextRes $(picotpi z)f(z)$ at $-N,cdots,+N$ \
&~~~~+2pi isumtextRes $(picotpi z)f(z)$ at singularities of $f$ \
endalign
$$
which is
$$
beginalign
lim_Ntoinftyoint_C_Nf(z)dz
&=lim_Ntoinfty2pi isumtextRes $(picotpi z)f(z)$ at $-N,cdots,+N$ \
&~~~~+lim_Ntoinfty2pi isumtextRes $(picotpi z)f(z)$ at singularities of $f$ \
endalign
$$
and using the fact that $oint_C_N(picotpi z)f(z)dzto0$ as $Ntoinfty$,
Therefore,
$$
beginalign
0
&=lim_Ntoinfty2pi isumtextRes $(picotpi z)f(z)$ at $-N,cdots,+N$ \
&~~~~+lim_Ntoinfty2pi isumtextRes $(picotpi z)f(z)$ at singularities of $f$ qquad(*)\
endalign
$$
Since for an integer $k$ $$textRes $(picotpi z)f(z)$ at $k$=f(k)$$
we can rewrite $(*)$ as
$$
beginalign
0
&=lim_Ntoinfty2pi isum_k=-N^N f(k) \
&~~~~+lim_Ntoinfty2pi isumtextRes $(picotpi z)f(z)$ at singularities of $f$ \
endalign
$$
Furthermore, recognizing that the second sum does not depend on $N$ for large enough $N$, we have
$$beginalign
0
&=lim_Ntoinfty2pi isum_k=-N^N f(k)
+2pi isumtextRes $(picotpi z)f(z)$ at singularities of $f$ \
0
&=lim_Ntoinftysum_k=-N^N f(k)
+sumtextRes $(picotpi z)f(z)$ at singularities of $f$ \
endalign
$$
$$
colorred
lim_Ntoinftysum_k=-N^N f(k)
=-sumtextRes $(picotpi z)f(z)$ at singularities of $f$
$$
add a comment |Â
up vote
1
down vote
accepted
Taking limits on both sides of the preceding displayed equation for the integral $oint_C_N(picotpi z)f(z)dz$
This means taking the limit $Ntoinfty$ on both sides of
$$
beginalign
oint_C_Nf(z)dz
&=2pi isumtextRes $(picotpi z)f(z)$ at $-N,cdots,+N$ \
&~~~~+2pi isumtextRes $(picotpi z)f(z)$ at singularities of $f$ \
endalign
$$
which is
$$
beginalign
lim_Ntoinftyoint_C_Nf(z)dz
&=lim_Ntoinfty2pi isumtextRes $(picotpi z)f(z)$ at $-N,cdots,+N$ \
&~~~~+lim_Ntoinfty2pi isumtextRes $(picotpi z)f(z)$ at singularities of $f$ \
endalign
$$
and using the fact that $oint_C_N(picotpi z)f(z)dzto0$ as $Ntoinfty$,
Therefore,
$$
beginalign
0
&=lim_Ntoinfty2pi isumtextRes $(picotpi z)f(z)$ at $-N,cdots,+N$ \
&~~~~+lim_Ntoinfty2pi isumtextRes $(picotpi z)f(z)$ at singularities of $f$ qquad(*)\
endalign
$$
Since for an integer $k$ $$textRes $(picotpi z)f(z)$ at $k$=f(k)$$
we can rewrite $(*)$ as
$$
beginalign
0
&=lim_Ntoinfty2pi isum_k=-N^N f(k) \
&~~~~+lim_Ntoinfty2pi isumtextRes $(picotpi z)f(z)$ at singularities of $f$ \
endalign
$$
Furthermore, recognizing that the second sum does not depend on $N$ for large enough $N$, we have
$$beginalign
0
&=lim_Ntoinfty2pi isum_k=-N^N f(k)
+2pi isumtextRes $(picotpi z)f(z)$ at singularities of $f$ \
0
&=lim_Ntoinftysum_k=-N^N f(k)
+sumtextRes $(picotpi z)f(z)$ at singularities of $f$ \
endalign
$$
$$
colorred
lim_Ntoinftysum_k=-N^N f(k)
=-sumtextRes $(picotpi z)f(z)$ at singularities of $f$
$$
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Taking limits on both sides of the preceding displayed equation for the integral $oint_C_N(picotpi z)f(z)dz$
This means taking the limit $Ntoinfty$ on both sides of
$$
beginalign
oint_C_Nf(z)dz
&=2pi isumtextRes $(picotpi z)f(z)$ at $-N,cdots,+N$ \
&~~~~+2pi isumtextRes $(picotpi z)f(z)$ at singularities of $f$ \
endalign
$$
which is
$$
beginalign
lim_Ntoinftyoint_C_Nf(z)dz
&=lim_Ntoinfty2pi isumtextRes $(picotpi z)f(z)$ at $-N,cdots,+N$ \
&~~~~+lim_Ntoinfty2pi isumtextRes $(picotpi z)f(z)$ at singularities of $f$ \
endalign
$$
and using the fact that $oint_C_N(picotpi z)f(z)dzto0$ as $Ntoinfty$,
Therefore,
$$
beginalign
0
&=lim_Ntoinfty2pi isumtextRes $(picotpi z)f(z)$ at $-N,cdots,+N$ \
&~~~~+lim_Ntoinfty2pi isumtextRes $(picotpi z)f(z)$ at singularities of $f$ qquad(*)\
endalign
$$
Since for an integer $k$ $$textRes $(picotpi z)f(z)$ at $k$=f(k)$$
we can rewrite $(*)$ as
$$
beginalign
0
&=lim_Ntoinfty2pi isum_k=-N^N f(k) \
&~~~~+lim_Ntoinfty2pi isumtextRes $(picotpi z)f(z)$ at singularities of $f$ \
endalign
$$
Furthermore, recognizing that the second sum does not depend on $N$ for large enough $N$, we have
$$beginalign
0
&=lim_Ntoinfty2pi isum_k=-N^N f(k)
+2pi isumtextRes $(picotpi z)f(z)$ at singularities of $f$ \
0
&=lim_Ntoinftysum_k=-N^N f(k)
+sumtextRes $(picotpi z)f(z)$ at singularities of $f$ \
endalign
$$
$$
colorred
lim_Ntoinftysum_k=-N^N f(k)
=-sumtextRes $(picotpi z)f(z)$ at singularities of $f$
$$
Taking limits on both sides of the preceding displayed equation for the integral $oint_C_N(picotpi z)f(z)dz$
This means taking the limit $Ntoinfty$ on both sides of
$$
beginalign
oint_C_Nf(z)dz
&=2pi isumtextRes $(picotpi z)f(z)$ at $-N,cdots,+N$ \
&~~~~+2pi isumtextRes $(picotpi z)f(z)$ at singularities of $f$ \
endalign
$$
which is
$$
beginalign
lim_Ntoinftyoint_C_Nf(z)dz
&=lim_Ntoinfty2pi isumtextRes $(picotpi z)f(z)$ at $-N,cdots,+N$ \
&~~~~+lim_Ntoinfty2pi isumtextRes $(picotpi z)f(z)$ at singularities of $f$ \
endalign
$$
and using the fact that $oint_C_N(picotpi z)f(z)dzto0$ as $Ntoinfty$,
Therefore,
$$
beginalign
0
&=lim_Ntoinfty2pi isumtextRes $(picotpi z)f(z)$ at $-N,cdots,+N$ \
&~~~~+lim_Ntoinfty2pi isumtextRes $(picotpi z)f(z)$ at singularities of $f$ qquad(*)\
endalign
$$
Since for an integer $k$ $$textRes $(picotpi z)f(z)$ at $k$=f(k)$$
we can rewrite $(*)$ as
$$
beginalign
0
&=lim_Ntoinfty2pi isum_k=-N^N f(k) \
&~~~~+lim_Ntoinfty2pi isumtextRes $(picotpi z)f(z)$ at singularities of $f$ \
endalign
$$
Furthermore, recognizing that the second sum does not depend on $N$ for large enough $N$, we have
$$beginalign
0
&=lim_Ntoinfty2pi isum_k=-N^N f(k)
+2pi isumtextRes $(picotpi z)f(z)$ at singularities of $f$ \
0
&=lim_Ntoinftysum_k=-N^N f(k)
+sumtextRes $(picotpi z)f(z)$ at singularities of $f$ \
endalign
$$
$$
colorred
lim_Ntoinftysum_k=-N^N f(k)
=-sumtextRes $(picotpi z)f(z)$ at singularities of $f$
$$
answered Sep 16 at 4:22
Szeto
5,0541623
5,0541623
add a comment |Â
add a comment |Â
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There is an example computation at this MSE link.
â Marko Riedel
Sep 8 at 11:59