Complex integral with conjugate as an exponent












7












$begingroup$



What is the integral of
$$int_{Gamma}pi e^{pibar{z}}dz$$ where $Gamma$ is the square with vertices at $0,1,1+i,i$ oriented anticlockwise?




I am badly stuck at this problem. I thought of using the Residue theorem by using $bar{z}=frac{|z|^2}{z}$, but we get an essential singularity. Using Laurent series about zero, I get $a_{-1}=pi|z|^2$. Is this correct? how do we proceed? Any hints. Thanks beforehand.






integration complex-analysis contour-integration residue-calculus







share|cite|improve this question











$endgroup$

















    7












    $begingroup$



    What is the integral of
    $$int_{Gamma}pi e^{pibar{z}}dz$$ where $Gamma$ is the square with vertices at $0,1,1+i,i$ oriented anticlockwise?




    I am badly stuck at this problem. I thought of using the Residue theorem by using $bar{z}=frac{|z|^2}{z}$, but we get an essential singularity. Using Laurent series about zero, I get $a_{-1}=pi|z|^2$. Is this correct? how do we proceed? Any hints. Thanks beforehand.






    integration complex-analysis contour-integration residue-calculus







    share|cite|improve this question











    $endgroup$















      7












      7








      7


      5



      $begingroup$



      What is the integral of
      $$int_{Gamma}pi e^{pibar{z}}dz$$ where $Gamma$ is the square with vertices at $0,1,1+i,i$ oriented anticlockwise?




      I am badly stuck at this problem. I thought of using the Residue theorem by using $bar{z}=frac{|z|^2}{z}$, but we get an essential singularity. Using Laurent series about zero, I get $a_{-1}=pi|z|^2$. Is this correct? how do we proceed? Any hints. Thanks beforehand.






      integration complex-analysis contour-integration residue-calculus







      share|cite|improve this question











      $endgroup$





      What is the integral of
      $$int_{Gamma}pi e^{pibar{z}}dz$$ where $Gamma$ is the square with vertices at $0,1,1+i,i$ oriented anticlockwise?




      I am badly stuck at this problem. I thought of using the Residue theorem by using $bar{z}=frac{|z|^2}{z}$, but we get an essential singularity. Using Laurent series about zero, I get $a_{-1}=pi|z|^2$. Is this correct? how do we proceed? Any hints. Thanks beforehand.






      integration complex-analysis contour-integration residue-calculus




      integration complex-analysis contour-integration residue-calculus






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 2 at 15:06









      José Carlos Santos

      171k23132240




      171k23132240










      asked Sep 20 '17 at 8:14









      vidyarthividyarthi

      3,0731833




      3,0731833






















          3 Answers
          3






          active

          oldest

          votes


















          12












          $begingroup$

          In addition to José Carlos Santos' answer, we can also utilize Green's theorem:



          begin{align*}
          int_{gamma} pi e^{pi bar{z}} , dz
          &= int_{gamma} left( pi e^{pi bar{z}} , dx + ipi e^{pi bar{z}} , dy right) \
          &= int_{[0,1]^2} left( frac{partial}{partial x} ipi e^{pi bar{z}} - frac{partial}{partial y} pi e^{pi bar{z}}right) , dxdy \
          &= 2pi^2 i int_{[0,1]^2} e^{pi x}e^{-ipi y} , dxdy \
          &= 4(e^{pi} - 1).
          end{align*}






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Beautiful answer.
            $endgroup$
            – Oria Gruber
            Sep 20 '17 at 9:03



















          7












          $begingroup$

          You can't use the residue theorem in a non-analytic function!



          Your integral can be naturally broken on four pieces. One of them is$$int_gammapi e^{pioverline z},mathrm dz,tag{1}$$with $gammacolon[0,1]longrightarrowmathbb C$ defined by $gamma(t)=t$. But then $(1)$ is equal to$$int_0^1pi e^{pioverline{gamma(t)}}gamma'(t),mathrm dt=int_0^1pi e^{pi t},mathrm dt=left[e^{pi t}right]_{t=0}^{t=1}=e^{pi}-e^0=e^{pi}-1$$Can you compute the other three integrals?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I thought the function is meromorphic with an essential singularity at $0$, isnt it? Anyways, what are four parts you are breaking the integral to: the integral on the four sides of the square?
            $endgroup$
            – vidyarthi
            Sep 20 '17 at 8:44






          • 3




            $begingroup$
            @vidyarthi Your function ($e^{pioverline z}$) is differentiable nowhere! Forget about meromorphic. Yes, I was talking about the four sides of the square.
            $endgroup$
            – José Carlos Santos
            Sep 20 '17 at 8:45












          • $begingroup$
            why do you use the path $gamma(t)=it$? and how do you determine the limits?
            $endgroup$
            – vidyarthi
            Sep 20 '17 at 8:52










          • $begingroup$
            @vidyarthi Its limits are $gamma(0)=0$ and $gamma(1)=i$. Actually, I made a mistake. Since you want to go around the square counterclockwise, I should have defined $gamma(t)=t$, in order to go from $0$ to $1$. Then you should go from $1$ to $1+i$, then from $1+i$ to $i$ and finally go back to $0$, along the four sides of a square.
            $endgroup$
            – José Carlos Santos
            Sep 20 '17 at 8:56






          • 1




            $begingroup$
            @vidyarthi, The contour integral $int_{gamma} f(z) , dz$ depends only on the value of the function $f$ along the curve $gamma$. So as long as you can identify one analytic function that matches $f$ on $gamma$, you can initiate the residue theorem. For your function $f(z) = pi e^{pi bar{z}}$ it is simply that the circular contour allows one such choice, and the square contour doesn't.
            $endgroup$
            – Sangchul Lee
            Sep 20 '17 at 9:35





















          3












          $begingroup$

          Hint. Note that if $gamma$ is a segment from $P$ to $Q$ then
          $gamma(t)=P+(Q-P)t$ for $tin [0,1]$ and
          $$int_{gamma}pi e^{pioverline{z}}dz=int_{t=0}^1pi e^{pi(overline{P}+(overline{Q}-overline{P})t)}(Q-P)dt=frac{Q-P}{overline{Q}-overline{P}}[e^{pi(overline{P}+(overline{Q}-bar{P})t)}]_0^1=frac{(Q-P)(e^{pi overline{Q}}-e^{pi overline{P}})}{overline{Q-P}}.$$






          share|cite|improve this answer









          $endgroup$














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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            12












            $begingroup$

            In addition to José Carlos Santos' answer, we can also utilize Green's theorem:



            begin{align*}
            int_{gamma} pi e^{pi bar{z}} , dz
            &= int_{gamma} left( pi e^{pi bar{z}} , dx + ipi e^{pi bar{z}} , dy right) \
            &= int_{[0,1]^2} left( frac{partial}{partial x} ipi e^{pi bar{z}} - frac{partial}{partial y} pi e^{pi bar{z}}right) , dxdy \
            &= 2pi^2 i int_{[0,1]^2} e^{pi x}e^{-ipi y} , dxdy \
            &= 4(e^{pi} - 1).
            end{align*}






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Beautiful answer.
              $endgroup$
              – Oria Gruber
              Sep 20 '17 at 9:03
















            12












            $begingroup$

            In addition to José Carlos Santos' answer, we can also utilize Green's theorem:



            begin{align*}
            int_{gamma} pi e^{pi bar{z}} , dz
            &= int_{gamma} left( pi e^{pi bar{z}} , dx + ipi e^{pi bar{z}} , dy right) \
            &= int_{[0,1]^2} left( frac{partial}{partial x} ipi e^{pi bar{z}} - frac{partial}{partial y} pi e^{pi bar{z}}right) , dxdy \
            &= 2pi^2 i int_{[0,1]^2} e^{pi x}e^{-ipi y} , dxdy \
            &= 4(e^{pi} - 1).
            end{align*}






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Beautiful answer.
              $endgroup$
              – Oria Gruber
              Sep 20 '17 at 9:03














            12












            12








            12





            $begingroup$

            In addition to José Carlos Santos' answer, we can also utilize Green's theorem:



            begin{align*}
            int_{gamma} pi e^{pi bar{z}} , dz
            &= int_{gamma} left( pi e^{pi bar{z}} , dx + ipi e^{pi bar{z}} , dy right) \
            &= int_{[0,1]^2} left( frac{partial}{partial x} ipi e^{pi bar{z}} - frac{partial}{partial y} pi e^{pi bar{z}}right) , dxdy \
            &= 2pi^2 i int_{[0,1]^2} e^{pi x}e^{-ipi y} , dxdy \
            &= 4(e^{pi} - 1).
            end{align*}






            share|cite|improve this answer









            $endgroup$



            In addition to José Carlos Santos' answer, we can also utilize Green's theorem:



            begin{align*}
            int_{gamma} pi e^{pi bar{z}} , dz
            &= int_{gamma} left( pi e^{pi bar{z}} , dx + ipi e^{pi bar{z}} , dy right) \
            &= int_{[0,1]^2} left( frac{partial}{partial x} ipi e^{pi bar{z}} - frac{partial}{partial y} pi e^{pi bar{z}}right) , dxdy \
            &= 2pi^2 i int_{[0,1]^2} e^{pi x}e^{-ipi y} , dxdy \
            &= 4(e^{pi} - 1).
            end{align*}







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Sep 20 '17 at 9:01









            Sangchul LeeSangchul Lee

            96.3k12171282




            96.3k12171282












            • $begingroup$
              Beautiful answer.
              $endgroup$
              – Oria Gruber
              Sep 20 '17 at 9:03


















            • $begingroup$
              Beautiful answer.
              $endgroup$
              – Oria Gruber
              Sep 20 '17 at 9:03
















            $begingroup$
            Beautiful answer.
            $endgroup$
            – Oria Gruber
            Sep 20 '17 at 9:03




            $begingroup$
            Beautiful answer.
            $endgroup$
            – Oria Gruber
            Sep 20 '17 at 9:03











            7












            $begingroup$

            You can't use the residue theorem in a non-analytic function!



            Your integral can be naturally broken on four pieces. One of them is$$int_gammapi e^{pioverline z},mathrm dz,tag{1}$$with $gammacolon[0,1]longrightarrowmathbb C$ defined by $gamma(t)=t$. But then $(1)$ is equal to$$int_0^1pi e^{pioverline{gamma(t)}}gamma'(t),mathrm dt=int_0^1pi e^{pi t},mathrm dt=left[e^{pi t}right]_{t=0}^{t=1}=e^{pi}-e^0=e^{pi}-1$$Can you compute the other three integrals?






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I thought the function is meromorphic with an essential singularity at $0$, isnt it? Anyways, what are four parts you are breaking the integral to: the integral on the four sides of the square?
              $endgroup$
              – vidyarthi
              Sep 20 '17 at 8:44






            • 3




              $begingroup$
              @vidyarthi Your function ($e^{pioverline z}$) is differentiable nowhere! Forget about meromorphic. Yes, I was talking about the four sides of the square.
              $endgroup$
              – José Carlos Santos
              Sep 20 '17 at 8:45












            • $begingroup$
              why do you use the path $gamma(t)=it$? and how do you determine the limits?
              $endgroup$
              – vidyarthi
              Sep 20 '17 at 8:52










            • $begingroup$
              @vidyarthi Its limits are $gamma(0)=0$ and $gamma(1)=i$. Actually, I made a mistake. Since you want to go around the square counterclockwise, I should have defined $gamma(t)=t$, in order to go from $0$ to $1$. Then you should go from $1$ to $1+i$, then from $1+i$ to $i$ and finally go back to $0$, along the four sides of a square.
              $endgroup$
              – José Carlos Santos
              Sep 20 '17 at 8:56






            • 1




              $begingroup$
              @vidyarthi, The contour integral $int_{gamma} f(z) , dz$ depends only on the value of the function $f$ along the curve $gamma$. So as long as you can identify one analytic function that matches $f$ on $gamma$, you can initiate the residue theorem. For your function $f(z) = pi e^{pi bar{z}}$ it is simply that the circular contour allows one such choice, and the square contour doesn't.
              $endgroup$
              – Sangchul Lee
              Sep 20 '17 at 9:35


















            7












            $begingroup$

            You can't use the residue theorem in a non-analytic function!



            Your integral can be naturally broken on four pieces. One of them is$$int_gammapi e^{pioverline z},mathrm dz,tag{1}$$with $gammacolon[0,1]longrightarrowmathbb C$ defined by $gamma(t)=t$. But then $(1)$ is equal to$$int_0^1pi e^{pioverline{gamma(t)}}gamma'(t),mathrm dt=int_0^1pi e^{pi t},mathrm dt=left[e^{pi t}right]_{t=0}^{t=1}=e^{pi}-e^0=e^{pi}-1$$Can you compute the other three integrals?






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I thought the function is meromorphic with an essential singularity at $0$, isnt it? Anyways, what are four parts you are breaking the integral to: the integral on the four sides of the square?
              $endgroup$
              – vidyarthi
              Sep 20 '17 at 8:44






            • 3




              $begingroup$
              @vidyarthi Your function ($e^{pioverline z}$) is differentiable nowhere! Forget about meromorphic. Yes, I was talking about the four sides of the square.
              $endgroup$
              – José Carlos Santos
              Sep 20 '17 at 8:45












            • $begingroup$
              why do you use the path $gamma(t)=it$? and how do you determine the limits?
              $endgroup$
              – vidyarthi
              Sep 20 '17 at 8:52










            • $begingroup$
              @vidyarthi Its limits are $gamma(0)=0$ and $gamma(1)=i$. Actually, I made a mistake. Since you want to go around the square counterclockwise, I should have defined $gamma(t)=t$, in order to go from $0$ to $1$. Then you should go from $1$ to $1+i$, then from $1+i$ to $i$ and finally go back to $0$, along the four sides of a square.
              $endgroup$
              – José Carlos Santos
              Sep 20 '17 at 8:56






            • 1




              $begingroup$
              @vidyarthi, The contour integral $int_{gamma} f(z) , dz$ depends only on the value of the function $f$ along the curve $gamma$. So as long as you can identify one analytic function that matches $f$ on $gamma$, you can initiate the residue theorem. For your function $f(z) = pi e^{pi bar{z}}$ it is simply that the circular contour allows one such choice, and the square contour doesn't.
              $endgroup$
              – Sangchul Lee
              Sep 20 '17 at 9:35
















            7












            7








            7





            $begingroup$

            You can't use the residue theorem in a non-analytic function!



            Your integral can be naturally broken on four pieces. One of them is$$int_gammapi e^{pioverline z},mathrm dz,tag{1}$$with $gammacolon[0,1]longrightarrowmathbb C$ defined by $gamma(t)=t$. But then $(1)$ is equal to$$int_0^1pi e^{pioverline{gamma(t)}}gamma'(t),mathrm dt=int_0^1pi e^{pi t},mathrm dt=left[e^{pi t}right]_{t=0}^{t=1}=e^{pi}-e^0=e^{pi}-1$$Can you compute the other three integrals?






            share|cite|improve this answer











            $endgroup$



            You can't use the residue theorem in a non-analytic function!



            Your integral can be naturally broken on four pieces. One of them is$$int_gammapi e^{pioverline z},mathrm dz,tag{1}$$with $gammacolon[0,1]longrightarrowmathbb C$ defined by $gamma(t)=t$. But then $(1)$ is equal to$$int_0^1pi e^{pioverline{gamma(t)}}gamma'(t),mathrm dt=int_0^1pi e^{pi t},mathrm dt=left[e^{pi t}right]_{t=0}^{t=1}=e^{pi}-e^0=e^{pi}-1$$Can you compute the other three integrals?







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Sep 20 '17 at 9:04









            vidyarthi

            3,0731833




            3,0731833










            answered Sep 20 '17 at 8:37









            José Carlos SantosJosé Carlos Santos

            171k23132240




            171k23132240












            • $begingroup$
              I thought the function is meromorphic with an essential singularity at $0$, isnt it? Anyways, what are four parts you are breaking the integral to: the integral on the four sides of the square?
              $endgroup$
              – vidyarthi
              Sep 20 '17 at 8:44






            • 3




              $begingroup$
              @vidyarthi Your function ($e^{pioverline z}$) is differentiable nowhere! Forget about meromorphic. Yes, I was talking about the four sides of the square.
              $endgroup$
              – José Carlos Santos
              Sep 20 '17 at 8:45












            • $begingroup$
              why do you use the path $gamma(t)=it$? and how do you determine the limits?
              $endgroup$
              – vidyarthi
              Sep 20 '17 at 8:52










            • $begingroup$
              @vidyarthi Its limits are $gamma(0)=0$ and $gamma(1)=i$. Actually, I made a mistake. Since you want to go around the square counterclockwise, I should have defined $gamma(t)=t$, in order to go from $0$ to $1$. Then you should go from $1$ to $1+i$, then from $1+i$ to $i$ and finally go back to $0$, along the four sides of a square.
              $endgroup$
              – José Carlos Santos
              Sep 20 '17 at 8:56






            • 1




              $begingroup$
              @vidyarthi, The contour integral $int_{gamma} f(z) , dz$ depends only on the value of the function $f$ along the curve $gamma$. So as long as you can identify one analytic function that matches $f$ on $gamma$, you can initiate the residue theorem. For your function $f(z) = pi e^{pi bar{z}}$ it is simply that the circular contour allows one such choice, and the square contour doesn't.
              $endgroup$
              – Sangchul Lee
              Sep 20 '17 at 9:35




















            • $begingroup$
              I thought the function is meromorphic with an essential singularity at $0$, isnt it? Anyways, what are four parts you are breaking the integral to: the integral on the four sides of the square?
              $endgroup$
              – vidyarthi
              Sep 20 '17 at 8:44






            • 3




              $begingroup$
              @vidyarthi Your function ($e^{pioverline z}$) is differentiable nowhere! Forget about meromorphic. Yes, I was talking about the four sides of the square.
              $endgroup$
              – José Carlos Santos
              Sep 20 '17 at 8:45












            • $begingroup$
              why do you use the path $gamma(t)=it$? and how do you determine the limits?
              $endgroup$
              – vidyarthi
              Sep 20 '17 at 8:52










            • $begingroup$
              @vidyarthi Its limits are $gamma(0)=0$ and $gamma(1)=i$. Actually, I made a mistake. Since you want to go around the square counterclockwise, I should have defined $gamma(t)=t$, in order to go from $0$ to $1$. Then you should go from $1$ to $1+i$, then from $1+i$ to $i$ and finally go back to $0$, along the four sides of a square.
              $endgroup$
              – José Carlos Santos
              Sep 20 '17 at 8:56






            • 1




              $begingroup$
              @vidyarthi, The contour integral $int_{gamma} f(z) , dz$ depends only on the value of the function $f$ along the curve $gamma$. So as long as you can identify one analytic function that matches $f$ on $gamma$, you can initiate the residue theorem. For your function $f(z) = pi e^{pi bar{z}}$ it is simply that the circular contour allows one such choice, and the square contour doesn't.
              $endgroup$
              – Sangchul Lee
              Sep 20 '17 at 9:35


















            $begingroup$
            I thought the function is meromorphic with an essential singularity at $0$, isnt it? Anyways, what are four parts you are breaking the integral to: the integral on the four sides of the square?
            $endgroup$
            – vidyarthi
            Sep 20 '17 at 8:44




            $begingroup$
            I thought the function is meromorphic with an essential singularity at $0$, isnt it? Anyways, what are four parts you are breaking the integral to: the integral on the four sides of the square?
            $endgroup$
            – vidyarthi
            Sep 20 '17 at 8:44




            3




            3




            $begingroup$
            @vidyarthi Your function ($e^{pioverline z}$) is differentiable nowhere! Forget about meromorphic. Yes, I was talking about the four sides of the square.
            $endgroup$
            – José Carlos Santos
            Sep 20 '17 at 8:45






            $begingroup$
            @vidyarthi Your function ($e^{pioverline z}$) is differentiable nowhere! Forget about meromorphic. Yes, I was talking about the four sides of the square.
            $endgroup$
            – José Carlos Santos
            Sep 20 '17 at 8:45














            $begingroup$
            why do you use the path $gamma(t)=it$? and how do you determine the limits?
            $endgroup$
            – vidyarthi
            Sep 20 '17 at 8:52




            $begingroup$
            why do you use the path $gamma(t)=it$? and how do you determine the limits?
            $endgroup$
            – vidyarthi
            Sep 20 '17 at 8:52












            $begingroup$
            @vidyarthi Its limits are $gamma(0)=0$ and $gamma(1)=i$. Actually, I made a mistake. Since you want to go around the square counterclockwise, I should have defined $gamma(t)=t$, in order to go from $0$ to $1$. Then you should go from $1$ to $1+i$, then from $1+i$ to $i$ and finally go back to $0$, along the four sides of a square.
            $endgroup$
            – José Carlos Santos
            Sep 20 '17 at 8:56




            $begingroup$
            @vidyarthi Its limits are $gamma(0)=0$ and $gamma(1)=i$. Actually, I made a mistake. Since you want to go around the square counterclockwise, I should have defined $gamma(t)=t$, in order to go from $0$ to $1$. Then you should go from $1$ to $1+i$, then from $1+i$ to $i$ and finally go back to $0$, along the four sides of a square.
            $endgroup$
            – José Carlos Santos
            Sep 20 '17 at 8:56




            1




            1




            $begingroup$
            @vidyarthi, The contour integral $int_{gamma} f(z) , dz$ depends only on the value of the function $f$ along the curve $gamma$. So as long as you can identify one analytic function that matches $f$ on $gamma$, you can initiate the residue theorem. For your function $f(z) = pi e^{pi bar{z}}$ it is simply that the circular contour allows one such choice, and the square contour doesn't.
            $endgroup$
            – Sangchul Lee
            Sep 20 '17 at 9:35






            $begingroup$
            @vidyarthi, The contour integral $int_{gamma} f(z) , dz$ depends only on the value of the function $f$ along the curve $gamma$. So as long as you can identify one analytic function that matches $f$ on $gamma$, you can initiate the residue theorem. For your function $f(z) = pi e^{pi bar{z}}$ it is simply that the circular contour allows one such choice, and the square contour doesn't.
            $endgroup$
            – Sangchul Lee
            Sep 20 '17 at 9:35













            3












            $begingroup$

            Hint. Note that if $gamma$ is a segment from $P$ to $Q$ then
            $gamma(t)=P+(Q-P)t$ for $tin [0,1]$ and
            $$int_{gamma}pi e^{pioverline{z}}dz=int_{t=0}^1pi e^{pi(overline{P}+(overline{Q}-overline{P})t)}(Q-P)dt=frac{Q-P}{overline{Q}-overline{P}}[e^{pi(overline{P}+(overline{Q}-bar{P})t)}]_0^1=frac{(Q-P)(e^{pi overline{Q}}-e^{pi overline{P}})}{overline{Q-P}}.$$






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              Hint. Note that if $gamma$ is a segment from $P$ to $Q$ then
              $gamma(t)=P+(Q-P)t$ for $tin [0,1]$ and
              $$int_{gamma}pi e^{pioverline{z}}dz=int_{t=0}^1pi e^{pi(overline{P}+(overline{Q}-overline{P})t)}(Q-P)dt=frac{Q-P}{overline{Q}-overline{P}}[e^{pi(overline{P}+(overline{Q}-bar{P})t)}]_0^1=frac{(Q-P)(e^{pi overline{Q}}-e^{pi overline{P}})}{overline{Q-P}}.$$






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                Hint. Note that if $gamma$ is a segment from $P$ to $Q$ then
                $gamma(t)=P+(Q-P)t$ for $tin [0,1]$ and
                $$int_{gamma}pi e^{pioverline{z}}dz=int_{t=0}^1pi e^{pi(overline{P}+(overline{Q}-overline{P})t)}(Q-P)dt=frac{Q-P}{overline{Q}-overline{P}}[e^{pi(overline{P}+(overline{Q}-bar{P})t)}]_0^1=frac{(Q-P)(e^{pi overline{Q}}-e^{pi overline{P}})}{overline{Q-P}}.$$






                share|cite|improve this answer









                $endgroup$



                Hint. Note that if $gamma$ is a segment from $P$ to $Q$ then
                $gamma(t)=P+(Q-P)t$ for $tin [0,1]$ and
                $$int_{gamma}pi e^{pioverline{z}}dz=int_{t=0}^1pi e^{pi(overline{P}+(overline{Q}-overline{P})t)}(Q-P)dt=frac{Q-P}{overline{Q}-overline{P}}[e^{pi(overline{P}+(overline{Q}-bar{P})t)}]_0^1=frac{(Q-P)(e^{pi overline{Q}}-e^{pi overline{P}})}{overline{Q-P}}.$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Sep 20 '17 at 9:05









                Robert ZRobert Z

                101k1070143




                101k1070143






























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