How to evaluate integrals like $ int_C frac{z}{e^z-1} dz$ without Residue Theorem?












1














I'm trying to evaluate two integrals:



$$ I_1 = int_{C(0, 1)} frac{z}{e^z-1} dz,quad I_2 = int_{C(1, 3)} frac{sin(iz)}{e^{iz}-1} dz $$



$ C(a, r) $ is a circular contour with radius $r$ centered at $a$, oriented counterclockwise.



In both cases there's a singularity point $ z=0 $. How can I handle it? Can I use some parametrization to get rid of exponential function?



In general, what can I do to evaluate integrals of type:



$$ I_3 = int_{gamma} frac{f(z)}{e^{g(z)}-1} dz$$



Where $f, g$ holomorphic on open, simply connected subset U, $gamma(t) in U$?






complex-analysis exponential-function cauchy-integral-formula







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  • 2




    What do you mean by $C(0,1)$ or $C(1,3)$? Why don't you want to use residues? Note that the singularities at $0$ are removable.
    – Robert Israel
    Nov 28 '18 at 18:09






  • 1




    By C I meant circular contour, I updated the question. About residues: my course did not reach series expansions yet, and I'm wondering if it's doable otherwise.
    – Krzysztof Antoniak
    Nov 28 '18 at 18:16










  • Were removable singularities mentioned? Cauchy integral theorem?
    – Robert Israel
    Nov 28 '18 at 19:58


















1














I'm trying to evaluate two integrals:



$$ I_1 = int_{C(0, 1)} frac{z}{e^z-1} dz,quad I_2 = int_{C(1, 3)} frac{sin(iz)}{e^{iz}-1} dz $$



$ C(a, r) $ is a circular contour with radius $r$ centered at $a$, oriented counterclockwise.



In both cases there's a singularity point $ z=0 $. How can I handle it? Can I use some parametrization to get rid of exponential function?



In general, what can I do to evaluate integrals of type:



$$ I_3 = int_{gamma} frac{f(z)}{e^{g(z)}-1} dz$$



Where $f, g$ holomorphic on open, simply connected subset U, $gamma(t) in U$?






complex-analysis exponential-function cauchy-integral-formula







share|cite|improve this question




















  • 2




    What do you mean by $C(0,1)$ or $C(1,3)$? Why don't you want to use residues? Note that the singularities at $0$ are removable.
    – Robert Israel
    Nov 28 '18 at 18:09






  • 1




    By C I meant circular contour, I updated the question. About residues: my course did not reach series expansions yet, and I'm wondering if it's doable otherwise.
    – Krzysztof Antoniak
    Nov 28 '18 at 18:16










  • Were removable singularities mentioned? Cauchy integral theorem?
    – Robert Israel
    Nov 28 '18 at 19:58
















1












1








1







I'm trying to evaluate two integrals:



$$ I_1 = int_{C(0, 1)} frac{z}{e^z-1} dz,quad I_2 = int_{C(1, 3)} frac{sin(iz)}{e^{iz}-1} dz $$



$ C(a, r) $ is a circular contour with radius $r$ centered at $a$, oriented counterclockwise.



In both cases there's a singularity point $ z=0 $. How can I handle it? Can I use some parametrization to get rid of exponential function?



In general, what can I do to evaluate integrals of type:



$$ I_3 = int_{gamma} frac{f(z)}{e^{g(z)}-1} dz$$



Where $f, g$ holomorphic on open, simply connected subset U, $gamma(t) in U$?






complex-analysis exponential-function cauchy-integral-formula







share|cite|improve this question















I'm trying to evaluate two integrals:



$$ I_1 = int_{C(0, 1)} frac{z}{e^z-1} dz,quad I_2 = int_{C(1, 3)} frac{sin(iz)}{e^{iz}-1} dz $$



$ C(a, r) $ is a circular contour with radius $r$ centered at $a$, oriented counterclockwise.



In both cases there's a singularity point $ z=0 $. How can I handle it? Can I use some parametrization to get rid of exponential function?



In general, what can I do to evaluate integrals of type:



$$ I_3 = int_{gamma} frac{f(z)}{e^{g(z)}-1} dz$$



Where $f, g$ holomorphic on open, simply connected subset U, $gamma(t) in U$?






complex-analysis exponential-function cauchy-integral-formula




complex-analysis exponential-function cauchy-integral-formula






share|cite|improve this question















share|cite|improve this question













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edited Nov 28 '18 at 19:33









Bernard

118k639112




118k639112










asked Nov 28 '18 at 18:04









Krzysztof AntoniakKrzysztof Antoniak

837




837








  • 2




    What do you mean by $C(0,1)$ or $C(1,3)$? Why don't you want to use residues? Note that the singularities at $0$ are removable.
    – Robert Israel
    Nov 28 '18 at 18:09






  • 1




    By C I meant circular contour, I updated the question. About residues: my course did not reach series expansions yet, and I'm wondering if it's doable otherwise.
    – Krzysztof Antoniak
    Nov 28 '18 at 18:16










  • Were removable singularities mentioned? Cauchy integral theorem?
    – Robert Israel
    Nov 28 '18 at 19:58
















  • 2




    What do you mean by $C(0,1)$ or $C(1,3)$? Why don't you want to use residues? Note that the singularities at $0$ are removable.
    – Robert Israel
    Nov 28 '18 at 18:09






  • 1




    By C I meant circular contour, I updated the question. About residues: my course did not reach series expansions yet, and I'm wondering if it's doable otherwise.
    – Krzysztof Antoniak
    Nov 28 '18 at 18:16










  • Were removable singularities mentioned? Cauchy integral theorem?
    – Robert Israel
    Nov 28 '18 at 19:58










2




2




What do you mean by $C(0,1)$ or $C(1,3)$? Why don't you want to use residues? Note that the singularities at $0$ are removable.
– Robert Israel
Nov 28 '18 at 18:09




What do you mean by $C(0,1)$ or $C(1,3)$? Why don't you want to use residues? Note that the singularities at $0$ are removable.
– Robert Israel
Nov 28 '18 at 18:09




1




1




By C I meant circular contour, I updated the question. About residues: my course did not reach series expansions yet, and I'm wondering if it's doable otherwise.
– Krzysztof Antoniak
Nov 28 '18 at 18:16




By C I meant circular contour, I updated the question. About residues: my course did not reach series expansions yet, and I'm wondering if it's doable otherwise.
– Krzysztof Antoniak
Nov 28 '18 at 18:16












Were removable singularities mentioned? Cauchy integral theorem?
– Robert Israel
Nov 28 '18 at 19:58






Were removable singularities mentioned? Cauchy integral theorem?
– Robert Israel
Nov 28 '18 at 19:58












1 Answer
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This is really using the fact that the singularity is removable without explicitly mentioning it.



Somewhat more generally, consider an integral of the form



$$J = oint_Gamma frac{f(z)}{g(z)}; dz$$



where $f$ and $g$ are analytic in a simply connected region $U$ containing the piecewise smooth simple closed contour $Gamma$, $f(p) = g(p) = 0$ for some point $p$ on $Gamma$, and $g$ has no other zeros in or on $Gamma$. So except for the singularity at $p$ we would have $J=0$ by the Cauchy integral theorem.



We might compare this to another integral



$$ widetilde{J} = oint_widetilde{Gamma} frac{f(z)}{g(z)}; dz = 0$$



where $widetilde{Gamma}$ is obtained by modifying $Gamma$ near $p$ so that $p$ is outside $widetilde{Gamma}$.



The difference between them is the integral over a small contour $widetilde{widetilde Gamma}$ near $p$.



Now suppose we know that $f(z)/g(z)$ is bounded for $z$ near $p$, say $|f(z)/g(z)| < B$ for $|z - p| < epsilon$. If $widetilde{widetilde Gamma}$ is all in the region $|z-p|<epsilon$ and has length at most $k epsilon$ for some constant $k$, then we can
estimate



$$ |J |= |J - widetilde{J}| = left|oint_{widetilde{widetilde Gamma}} frac{f(z)}{g(z)}; dz right| le B cdot text{length}(widetilde{widetilde Gamma}) le B k epsilon $$



and taking the limit as $epsilon to 0$ we conclude that $J = 0$.






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    active

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    1














    This is really using the fact that the singularity is removable without explicitly mentioning it.



    Somewhat more generally, consider an integral of the form



    $$J = oint_Gamma frac{f(z)}{g(z)}; dz$$



    where $f$ and $g$ are analytic in a simply connected region $U$ containing the piecewise smooth simple closed contour $Gamma$, $f(p) = g(p) = 0$ for some point $p$ on $Gamma$, and $g$ has no other zeros in or on $Gamma$. So except for the singularity at $p$ we would have $J=0$ by the Cauchy integral theorem.



    We might compare this to another integral



    $$ widetilde{J} = oint_widetilde{Gamma} frac{f(z)}{g(z)}; dz = 0$$



    where $widetilde{Gamma}$ is obtained by modifying $Gamma$ near $p$ so that $p$ is outside $widetilde{Gamma}$.



    The difference between them is the integral over a small contour $widetilde{widetilde Gamma}$ near $p$.



    Now suppose we know that $f(z)/g(z)$ is bounded for $z$ near $p$, say $|f(z)/g(z)| < B$ for $|z - p| < epsilon$. If $widetilde{widetilde Gamma}$ is all in the region $|z-p|<epsilon$ and has length at most $k epsilon$ for some constant $k$, then we can
    estimate



    $$ |J |= |J - widetilde{J}| = left|oint_{widetilde{widetilde Gamma}} frac{f(z)}{g(z)}; dz right| le B cdot text{length}(widetilde{widetilde Gamma}) le B k epsilon $$



    and taking the limit as $epsilon to 0$ we conclude that $J = 0$.






    share|cite|improve this answer


























      1














      This is really using the fact that the singularity is removable without explicitly mentioning it.



      Somewhat more generally, consider an integral of the form



      $$J = oint_Gamma frac{f(z)}{g(z)}; dz$$



      where $f$ and $g$ are analytic in a simply connected region $U$ containing the piecewise smooth simple closed contour $Gamma$, $f(p) = g(p) = 0$ for some point $p$ on $Gamma$, and $g$ has no other zeros in or on $Gamma$. So except for the singularity at $p$ we would have $J=0$ by the Cauchy integral theorem.



      We might compare this to another integral



      $$ widetilde{J} = oint_widetilde{Gamma} frac{f(z)}{g(z)}; dz = 0$$



      where $widetilde{Gamma}$ is obtained by modifying $Gamma$ near $p$ so that $p$ is outside $widetilde{Gamma}$.



      The difference between them is the integral over a small contour $widetilde{widetilde Gamma}$ near $p$.



      Now suppose we know that $f(z)/g(z)$ is bounded for $z$ near $p$, say $|f(z)/g(z)| < B$ for $|z - p| < epsilon$. If $widetilde{widetilde Gamma}$ is all in the region $|z-p|<epsilon$ and has length at most $k epsilon$ for some constant $k$, then we can
      estimate



      $$ |J |= |J - widetilde{J}| = left|oint_{widetilde{widetilde Gamma}} frac{f(z)}{g(z)}; dz right| le B cdot text{length}(widetilde{widetilde Gamma}) le B k epsilon $$



      and taking the limit as $epsilon to 0$ we conclude that $J = 0$.






      share|cite|improve this answer
























        1












        1








        1






        This is really using the fact that the singularity is removable without explicitly mentioning it.



        Somewhat more generally, consider an integral of the form



        $$J = oint_Gamma frac{f(z)}{g(z)}; dz$$



        where $f$ and $g$ are analytic in a simply connected region $U$ containing the piecewise smooth simple closed contour $Gamma$, $f(p) = g(p) = 0$ for some point $p$ on $Gamma$, and $g$ has no other zeros in or on $Gamma$. So except for the singularity at $p$ we would have $J=0$ by the Cauchy integral theorem.



        We might compare this to another integral



        $$ widetilde{J} = oint_widetilde{Gamma} frac{f(z)}{g(z)}; dz = 0$$



        where $widetilde{Gamma}$ is obtained by modifying $Gamma$ near $p$ so that $p$ is outside $widetilde{Gamma}$.



        The difference between them is the integral over a small contour $widetilde{widetilde Gamma}$ near $p$.



        Now suppose we know that $f(z)/g(z)$ is bounded for $z$ near $p$, say $|f(z)/g(z)| < B$ for $|z - p| < epsilon$. If $widetilde{widetilde Gamma}$ is all in the region $|z-p|<epsilon$ and has length at most $k epsilon$ for some constant $k$, then we can
        estimate



        $$ |J |= |J - widetilde{J}| = left|oint_{widetilde{widetilde Gamma}} frac{f(z)}{g(z)}; dz right| le B cdot text{length}(widetilde{widetilde Gamma}) le B k epsilon $$



        and taking the limit as $epsilon to 0$ we conclude that $J = 0$.






        share|cite|improve this answer












        This is really using the fact that the singularity is removable without explicitly mentioning it.



        Somewhat more generally, consider an integral of the form



        $$J = oint_Gamma frac{f(z)}{g(z)}; dz$$



        where $f$ and $g$ are analytic in a simply connected region $U$ containing the piecewise smooth simple closed contour $Gamma$, $f(p) = g(p) = 0$ for some point $p$ on $Gamma$, and $g$ has no other zeros in or on $Gamma$. So except for the singularity at $p$ we would have $J=0$ by the Cauchy integral theorem.



        We might compare this to another integral



        $$ widetilde{J} = oint_widetilde{Gamma} frac{f(z)}{g(z)}; dz = 0$$



        where $widetilde{Gamma}$ is obtained by modifying $Gamma$ near $p$ so that $p$ is outside $widetilde{Gamma}$.



        The difference between them is the integral over a small contour $widetilde{widetilde Gamma}$ near $p$.



        Now suppose we know that $f(z)/g(z)$ is bounded for $z$ near $p$, say $|f(z)/g(z)| < B$ for $|z - p| < epsilon$. If $widetilde{widetilde Gamma}$ is all in the region $|z-p|<epsilon$ and has length at most $k epsilon$ for some constant $k$, then we can
        estimate



        $$ |J |= |J - widetilde{J}| = left|oint_{widetilde{widetilde Gamma}} frac{f(z)}{g(z)}; dz right| le B cdot text{length}(widetilde{widetilde Gamma}) le B k epsilon $$



        and taking the limit as $epsilon to 0$ we conclude that $J = 0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 28 '18 at 20:24









        Robert IsraelRobert Israel

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