Evaluating an integral using residue theorem











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I'm evaluating $frac{1}{2pi i}oint_C frac{e^{zt}}{z(z^2+1)}dz$, $t > 0$ where $C$ is defined by the vertices $2+2i$, $-2+2i$, $-2-2i$, and $2-2i$. I'm pretty sure this is done by finding the poles in $C$ and the residues at the poles and then adding them together, but the answer is supposed to be $1-cos t$. Here is what I've done:



$$frac{1}{2pi i}oint_Cf(z)dz=a_{-1}$$
$$a_{-1}=lim_{z to a}frac{1}{(m-1)!}frac{d^{m-1}}{dz^{m-1}}{(z-a)^mf(z)}$$



The poles are found where $z(z^2+1)$ would be zero, which are $z=0,i,-i$. I need to take the above limit as $z$ goes to each pole.
$$lim_{z to 0}frac{1}{0!}frac{d^0}{dz^0}{(z-0)^1frac{e^{zt}}{z(z^2+1)}}=lim_{z to 0}frac{e^{zt}}{z^2+1}=1$$



$$lim_{z to i}frac{1}{0!}frac{d^0}{dz^0}{(z-i)^1frac{e^z}{z(z^2+1)}}=lim_{z to i}frac{e^{zt}+(z-i)e^{zt}t}{2z}=frac{e^{it}}{2i}$$



$$lim_{z to -i}frac{1}{0!}frac{d^0}{dz^0}{(z+i)^1frac{e^z}{z(z^2+1)}}=lim_{z to -i}frac{e^{zt}+(z+i)e^{zt}t}{2z}=frac{e^{-it}}{-2i}$$



If I were correct, the answer would be $1+frac{e^{it}}{2i}-frac{e^{-it}}{2i}$.






complex-analysis proof-verification residue-calculus







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    I'm evaluating $frac{1}{2pi i}oint_C frac{e^{zt}}{z(z^2+1)}dz$, $t > 0$ where $C$ is defined by the vertices $2+2i$, $-2+2i$, $-2-2i$, and $2-2i$. I'm pretty sure this is done by finding the poles in $C$ and the residues at the poles and then adding them together, but the answer is supposed to be $1-cos t$. Here is what I've done:



    $$frac{1}{2pi i}oint_Cf(z)dz=a_{-1}$$
    $$a_{-1}=lim_{z to a}frac{1}{(m-1)!}frac{d^{m-1}}{dz^{m-1}}{(z-a)^mf(z)}$$



    The poles are found where $z(z^2+1)$ would be zero, which are $z=0,i,-i$. I need to take the above limit as $z$ goes to each pole.
    $$lim_{z to 0}frac{1}{0!}frac{d^0}{dz^0}{(z-0)^1frac{e^{zt}}{z(z^2+1)}}=lim_{z to 0}frac{e^{zt}}{z^2+1}=1$$



    $$lim_{z to i}frac{1}{0!}frac{d^0}{dz^0}{(z-i)^1frac{e^z}{z(z^2+1)}}=lim_{z to i}frac{e^{zt}+(z-i)e^{zt}t}{2z}=frac{e^{it}}{2i}$$



    $$lim_{z to -i}frac{1}{0!}frac{d^0}{dz^0}{(z+i)^1frac{e^z}{z(z^2+1)}}=lim_{z to -i}frac{e^{zt}+(z+i)e^{zt}t}{2z}=frac{e^{-it}}{-2i}$$



    If I were correct, the answer would be $1+frac{e^{it}}{2i}-frac{e^{-it}}{2i}$.






    complex-analysis proof-verification residue-calculus







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      I'm evaluating $frac{1}{2pi i}oint_C frac{e^{zt}}{z(z^2+1)}dz$, $t > 0$ where $C$ is defined by the vertices $2+2i$, $-2+2i$, $-2-2i$, and $2-2i$. I'm pretty sure this is done by finding the poles in $C$ and the residues at the poles and then adding them together, but the answer is supposed to be $1-cos t$. Here is what I've done:



      $$frac{1}{2pi i}oint_Cf(z)dz=a_{-1}$$
      $$a_{-1}=lim_{z to a}frac{1}{(m-1)!}frac{d^{m-1}}{dz^{m-1}}{(z-a)^mf(z)}$$



      The poles are found where $z(z^2+1)$ would be zero, which are $z=0,i,-i$. I need to take the above limit as $z$ goes to each pole.
      $$lim_{z to 0}frac{1}{0!}frac{d^0}{dz^0}{(z-0)^1frac{e^{zt}}{z(z^2+1)}}=lim_{z to 0}frac{e^{zt}}{z^2+1}=1$$



      $$lim_{z to i}frac{1}{0!}frac{d^0}{dz^0}{(z-i)^1frac{e^z}{z(z^2+1)}}=lim_{z to i}frac{e^{zt}+(z-i)e^{zt}t}{2z}=frac{e^{it}}{2i}$$



      $$lim_{z to -i}frac{1}{0!}frac{d^0}{dz^0}{(z+i)^1frac{e^z}{z(z^2+1)}}=lim_{z to -i}frac{e^{zt}+(z+i)e^{zt}t}{2z}=frac{e^{-it}}{-2i}$$



      If I were correct, the answer would be $1+frac{e^{it}}{2i}-frac{e^{-it}}{2i}$.






      complex-analysis proof-verification residue-calculus







      share|cite|improve this question















      I'm evaluating $frac{1}{2pi i}oint_C frac{e^{zt}}{z(z^2+1)}dz$, $t > 0$ where $C$ is defined by the vertices $2+2i$, $-2+2i$, $-2-2i$, and $2-2i$. I'm pretty sure this is done by finding the poles in $C$ and the residues at the poles and then adding them together, but the answer is supposed to be $1-cos t$. Here is what I've done:



      $$frac{1}{2pi i}oint_Cf(z)dz=a_{-1}$$
      $$a_{-1}=lim_{z to a}frac{1}{(m-1)!}frac{d^{m-1}}{dz^{m-1}}{(z-a)^mf(z)}$$



      The poles are found where $z(z^2+1)$ would be zero, which are $z=0,i,-i$. I need to take the above limit as $z$ goes to each pole.
      $$lim_{z to 0}frac{1}{0!}frac{d^0}{dz^0}{(z-0)^1frac{e^{zt}}{z(z^2+1)}}=lim_{z to 0}frac{e^{zt}}{z^2+1}=1$$



      $$lim_{z to i}frac{1}{0!}frac{d^0}{dz^0}{(z-i)^1frac{e^z}{z(z^2+1)}}=lim_{z to i}frac{e^{zt}+(z-i)e^{zt}t}{2z}=frac{e^{it}}{2i}$$



      $$lim_{z to -i}frac{1}{0!}frac{d^0}{dz^0}{(z+i)^1frac{e^z}{z(z^2+1)}}=lim_{z to -i}frac{e^{zt}+(z+i)e^{zt}t}{2z}=frac{e^{-it}}{-2i}$$



      If I were correct, the answer would be $1+frac{e^{it}}{2i}-frac{e^{-it}}{2i}$.






      complex-analysis proof-verification residue-calculus




      complex-analysis proof-verification residue-calculus






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      edited yesterday

























      asked Nov 18 at 1:33









      AdamK

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          I think that you made a simple mistake in evaluating the second limit, which can be rewritten as:



          $lim_{zto i}frac{(z-i)e^{zt}}{z(z+i)(z-i)} = -frac{1}{2}e^{it}$ so you missed a minus sign






          share|cite|improve this answer





















          • Yes, thank you. I was just giving it another try, and realized there were a few mistakes and typos, including taking a derivative incorrectly when applying L'Hopital's rule. So the sum of residues should be $1-frac{e^{it}}{2}-frac{e^{-it}}{2}$ according to what I have done. I suspect (hope) this can be worked around to the correct answer, but am currently struggling with this.
            – AdamK
            yesterday






          • 1




            You may have forgotten: $$frac{e^{it}+ e^{-it}}{2}=cos t.$$
            – zhw.
            yesterday













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          up vote
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          I think that you made a simple mistake in evaluating the second limit, which can be rewritten as:



          $lim_{zto i}frac{(z-i)e^{zt}}{z(z+i)(z-i)} = -frac{1}{2}e^{it}$ so you missed a minus sign






          share|cite|improve this answer





















          • Yes, thank you. I was just giving it another try, and realized there were a few mistakes and typos, including taking a derivative incorrectly when applying L'Hopital's rule. So the sum of residues should be $1-frac{e^{it}}{2}-frac{e^{-it}}{2}$ according to what I have done. I suspect (hope) this can be worked around to the correct answer, but am currently struggling with this.
            – AdamK
            yesterday






          • 1




            You may have forgotten: $$frac{e^{it}+ e^{-it}}{2}=cos t.$$
            – zhw.
            yesterday

















          up vote
          1
          down vote













          I think that you made a simple mistake in evaluating the second limit, which can be rewritten as:



          $lim_{zto i}frac{(z-i)e^{zt}}{z(z+i)(z-i)} = -frac{1}{2}e^{it}$ so you missed a minus sign






          share|cite|improve this answer





















          • Yes, thank you. I was just giving it another try, and realized there were a few mistakes and typos, including taking a derivative incorrectly when applying L'Hopital's rule. So the sum of residues should be $1-frac{e^{it}}{2}-frac{e^{-it}}{2}$ according to what I have done. I suspect (hope) this can be worked around to the correct answer, but am currently struggling with this.
            – AdamK
            yesterday






          • 1




            You may have forgotten: $$frac{e^{it}+ e^{-it}}{2}=cos t.$$
            – zhw.
            yesterday















          up vote
          1
          down vote










          up vote
          1
          down vote









          I think that you made a simple mistake in evaluating the second limit, which can be rewritten as:



          $lim_{zto i}frac{(z-i)e^{zt}}{z(z+i)(z-i)} = -frac{1}{2}e^{it}$ so you missed a minus sign






          share|cite|improve this answer












          I think that you made a simple mistake in evaluating the second limit, which can be rewritten as:



          $lim_{zto i}frac{(z-i)e^{zt}}{z(z+i)(z-i)} = -frac{1}{2}e^{it}$ so you missed a minus sign







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered yesterday









          Sorin Tirc

          5039




          5039












          • Yes, thank you. I was just giving it another try, and realized there were a few mistakes and typos, including taking a derivative incorrectly when applying L'Hopital's rule. So the sum of residues should be $1-frac{e^{it}}{2}-frac{e^{-it}}{2}$ according to what I have done. I suspect (hope) this can be worked around to the correct answer, but am currently struggling with this.
            – AdamK
            yesterday






          • 1




            You may have forgotten: $$frac{e^{it}+ e^{-it}}{2}=cos t.$$
            – zhw.
            yesterday




















          • Yes, thank you. I was just giving it another try, and realized there were a few mistakes and typos, including taking a derivative incorrectly when applying L'Hopital's rule. So the sum of residues should be $1-frac{e^{it}}{2}-frac{e^{-it}}{2}$ according to what I have done. I suspect (hope) this can be worked around to the correct answer, but am currently struggling with this.
            – AdamK
            yesterday






          • 1




            You may have forgotten: $$frac{e^{it}+ e^{-it}}{2}=cos t.$$
            – zhw.
            yesterday


















          Yes, thank you. I was just giving it another try, and realized there were a few mistakes and typos, including taking a derivative incorrectly when applying L'Hopital's rule. So the sum of residues should be $1-frac{e^{it}}{2}-frac{e^{-it}}{2}$ according to what I have done. I suspect (hope) this can be worked around to the correct answer, but am currently struggling with this.
          – AdamK
          yesterday




          Yes, thank you. I was just giving it another try, and realized there were a few mistakes and typos, including taking a derivative incorrectly when applying L'Hopital's rule. So the sum of residues should be $1-frac{e^{it}}{2}-frac{e^{-it}}{2}$ according to what I have done. I suspect (hope) this can be worked around to the correct answer, but am currently struggling with this.
          – AdamK
          yesterday




          1




          1




          You may have forgotten: $$frac{e^{it}+ e^{-it}}{2}=cos t.$$
          – zhw.
          yesterday






          You may have forgotten: $$frac{e^{it}+ e^{-it}}{2}=cos t.$$
          – zhw.
          yesterday




















           

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