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:




enter image description here




enter image description here



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










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  • There is an example computation at this MSE link.
    – Marko Riedel
    Sep 8 at 11:59














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1
down vote

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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:




enter image description here




enter image description here



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










share|cite|improve this question























  • There is an example computation at this MSE link.
    – Marko Riedel
    Sep 8 at 11:59












up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1






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:




enter image description here




enter image description here



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










share|cite|improve this question
















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:




enter image description here




enter image description here



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






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edited Sep 13 at 13:13









Martin Sleziak

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asked Sep 8 at 6:04









Isabella

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  • There is an example computation at this MSE link.
    – Marko Riedel
    Sep 8 at 11:59
















  • There is an example computation at this MSE link.
    – Marko Riedel
    Sep 8 at 11:59















There is an example computation at this MSE link.
– Marko Riedel
Sep 8 at 11:59




There is an example computation at this MSE link.
– Marko Riedel
Sep 8 at 11:59










1 Answer
1






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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



$$beginalign
0
&=lim_Ntoinfty2pi isum_k=-N^N f(k)
+2pi isumtextRes $(picotpi z)f(z)$ at singularities of $f$ \
0
&=lim_Ntoinftysum_k=-N^N f(k)
+sumtextRes $(picotpi z)f(z)$ at singularities of $f$ \
endalign
$$
$$
colorred
lim_Ntoinftysum_k=-N^N f(k)
=-sumtextRes $(picotpi z)f(z)$ at singularities of $f$
$$






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    1 Answer
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    active

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    1 Answer
    1






    active

    oldest

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    active

    oldest

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    active

    oldest

    votes








    up vote
    1
    down vote



    accepted
    +50











    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



    $$beginalign
    0
    &=lim_Ntoinfty2pi isum_k=-N^N f(k)
    +2pi isumtextRes $(picotpi z)f(z)$ at singularities of $f$ \
    0
    &=lim_Ntoinftysum_k=-N^N f(k)
    +sumtextRes $(picotpi z)f(z)$ at singularities of $f$ \
    endalign
    $$
    $$
    colorred
    lim_Ntoinftysum_k=-N^N f(k)
    =-sumtextRes $(picotpi z)f(z)$ at singularities of $f$
    $$






    share|cite|improve this answer
























      up vote
      1
      down vote



      accepted
      +50











      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



      $$beginalign
      0
      &=lim_Ntoinfty2pi isum_k=-N^N f(k)
      +2pi isumtextRes $(picotpi z)f(z)$ at singularities of $f$ \
      0
      &=lim_Ntoinftysum_k=-N^N f(k)
      +sumtextRes $(picotpi z)f(z)$ at singularities of $f$ \
      endalign
      $$
      $$
      colorred
      lim_Ntoinftysum_k=-N^N f(k)
      =-sumtextRes $(picotpi z)f(z)$ at singularities of $f$
      $$






      share|cite|improve this answer






















        up vote
        1
        down vote



        accepted
        +50







        up vote
        1
        down vote



        accepted
        +50




        +50





        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



        $$beginalign
        0
        &=lim_Ntoinfty2pi isum_k=-N^N f(k)
        +2pi isumtextRes $(picotpi z)f(z)$ at singularities of $f$ \
        0
        &=lim_Ntoinftysum_k=-N^N f(k)
        +sumtextRes $(picotpi z)f(z)$ at singularities of $f$ \
        endalign
        $$
        $$
        colorred
        lim_Ntoinftysum_k=-N^N f(k)
        =-sumtextRes $(picotpi z)f(z)$ at singularities of $f$
        $$






        share|cite|improve this answer













        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



        $$beginalign
        0
        &=lim_Ntoinfty2pi isum_k=-N^N f(k)
        +2pi isumtextRes $(picotpi z)f(z)$ at singularities of $f$ \
        0
        &=lim_Ntoinftysum_k=-N^N f(k)
        +sumtextRes $(picotpi z)f(z)$ at singularities of $f$ \
        endalign
        $$
        $$
        colorred
        lim_Ntoinftysum_k=-N^N f(k)
        =-sumtextRes $(picotpi z)f(z)$ at singularities of $f$
        $$







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        answered Sep 16 at 4:22









        Szeto

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