how to integrate $frac{-1+e^{-i k x}}{x^2}$












1












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How do I integrate the following?$$int_{-infty }^{infty } frac{-1+e^{-i k x}}{x^2} , dx$$



I am not very familiar with complex analysis, but I did try to use contour integral to do this but I couldn't get any success with that.






contour-integration calculator







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    How do I integrate the following?$$int_{-infty }^{infty } frac{-1+e^{-i k x}}{x^2} , dx$$



    I am not very familiar with complex analysis, but I did try to use contour integral to do this but I couldn't get any success with that.






    contour-integration calculator







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      How do I integrate the following?$$int_{-infty }^{infty } frac{-1+e^{-i k x}}{x^2} , dx$$



      I am not very familiar with complex analysis, but I did try to use contour integral to do this but I couldn't get any success with that.






      contour-integration calculator







      share|cite|improve this question











      $endgroup$




      How do I integrate the following?$$int_{-infty }^{infty } frac{-1+e^{-i k x}}{x^2} , dx$$



      I am not very familiar with complex analysis, but I did try to use contour integral to do this but I couldn't get any success with that.






      contour-integration calculator




      contour-integration calculator






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 11 '18 at 3:21









      Lau

      517315




      517315










      asked Dec 11 '18 at 3:03









      xi daixi dai

      63




      63






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          If we interpret the integral as A Cauchy Principal Value, then we have



          $$int_{-infty}^inftyfrac{-1+e^{-ikx}}{x^2} ,dx=int_{-infty}^infty frac{-1+cos(kx)}{x^2},dx$$



          Using the trigonometric identity $sin^2(kx/2)=frac12(1-cos(kx))$ and enforcing the substitution $kx/2mapsto x$, we see that



          $$ int_{-infty}^infty frac{-1+e^{ikx}}{x^2},dx=-kint_{-infty}^infty frac{sin^2(x)}{x^2},dx$$



          See THIS. Can you finish now?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Yes. Thanks a lot!
            $endgroup$
            – xi dai
            Dec 11 '18 at 4:45










          • $begingroup$
            Sorry why do we need Cauchy principal value for the first line to hold?
            $endgroup$
            – xi dai
            Dec 11 '18 at 4:55










          • $begingroup$
            You're welcome. My pleasure.
            $endgroup$
            – Mark Viola
            Dec 11 '18 at 14:20










          • $begingroup$
            We need a CPV for the first part since $int_{-infty}^infty frac{sin(x)}{x^2},dx$ fails to converges due to the $frac1x$ type singularity around $x=0$.
            $endgroup$
            – Mark Viola
            Dec 11 '18 at 14:22



















          1












          $begingroup$

          Another way to do it if you want to use contour integration is to use an indented semi circle in the upper half plane going around the singularity at $z=0$. For this contour, the function $f(z)=(e^{-ikz}-1)/z^2$ will integrate to 0 by Cauchy's theorem since it is holomorphic in the region enclosed by the indented semi circle. Now you simply split $int_gamma=int_{C_R}+int_{C_epsilon}+int_{-R}^{-epsilon}+int_{epsilon}^{R}$ and you're left with justifying the exchange of limit and integrals when you want to let $epsilon$ (radius of small semi circle) go to $0$ and R go to $infty$ (radius of larger semi-circle).
          Here are some examples : https://web.williams.edu/Mathematics/sjmiller/public_html/302/coursenotes/Trapper_MethodsContourIntegrals.pdf



          There are also nice people explaining it in full detail on youtube, have fun !






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This pdf is exactly the one I have been following. Thanks!
            $endgroup$
            – xi dai
            Dec 11 '18 at 4:47











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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          If we interpret the integral as A Cauchy Principal Value, then we have



          $$int_{-infty}^inftyfrac{-1+e^{-ikx}}{x^2} ,dx=int_{-infty}^infty frac{-1+cos(kx)}{x^2},dx$$



          Using the trigonometric identity $sin^2(kx/2)=frac12(1-cos(kx))$ and enforcing the substitution $kx/2mapsto x$, we see that



          $$ int_{-infty}^infty frac{-1+e^{ikx}}{x^2},dx=-kint_{-infty}^infty frac{sin^2(x)}{x^2},dx$$



          See THIS. Can you finish now?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Yes. Thanks a lot!
            $endgroup$
            – xi dai
            Dec 11 '18 at 4:45










          • $begingroup$
            Sorry why do we need Cauchy principal value for the first line to hold?
            $endgroup$
            – xi dai
            Dec 11 '18 at 4:55










          • $begingroup$
            You're welcome. My pleasure.
            $endgroup$
            – Mark Viola
            Dec 11 '18 at 14:20










          • $begingroup$
            We need a CPV for the first part since $int_{-infty}^infty frac{sin(x)}{x^2},dx$ fails to converges due to the $frac1x$ type singularity around $x=0$.
            $endgroup$
            – Mark Viola
            Dec 11 '18 at 14:22
















          1












          $begingroup$

          If we interpret the integral as A Cauchy Principal Value, then we have



          $$int_{-infty}^inftyfrac{-1+e^{-ikx}}{x^2} ,dx=int_{-infty}^infty frac{-1+cos(kx)}{x^2},dx$$



          Using the trigonometric identity $sin^2(kx/2)=frac12(1-cos(kx))$ and enforcing the substitution $kx/2mapsto x$, we see that



          $$ int_{-infty}^infty frac{-1+e^{ikx}}{x^2},dx=-kint_{-infty}^infty frac{sin^2(x)}{x^2},dx$$



          See THIS. Can you finish now?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Yes. Thanks a lot!
            $endgroup$
            – xi dai
            Dec 11 '18 at 4:45










          • $begingroup$
            Sorry why do we need Cauchy principal value for the first line to hold?
            $endgroup$
            – xi dai
            Dec 11 '18 at 4:55










          • $begingroup$
            You're welcome. My pleasure.
            $endgroup$
            – Mark Viola
            Dec 11 '18 at 14:20










          • $begingroup$
            We need a CPV for the first part since $int_{-infty}^infty frac{sin(x)}{x^2},dx$ fails to converges due to the $frac1x$ type singularity around $x=0$.
            $endgroup$
            – Mark Viola
            Dec 11 '18 at 14:22














          1












          1








          1





          $begingroup$

          If we interpret the integral as A Cauchy Principal Value, then we have



          $$int_{-infty}^inftyfrac{-1+e^{-ikx}}{x^2} ,dx=int_{-infty}^infty frac{-1+cos(kx)}{x^2},dx$$



          Using the trigonometric identity $sin^2(kx/2)=frac12(1-cos(kx))$ and enforcing the substitution $kx/2mapsto x$, we see that



          $$ int_{-infty}^infty frac{-1+e^{ikx}}{x^2},dx=-kint_{-infty}^infty frac{sin^2(x)}{x^2},dx$$



          See THIS. Can you finish now?






          share|cite|improve this answer











          $endgroup$



          If we interpret the integral as A Cauchy Principal Value, then we have



          $$int_{-infty}^inftyfrac{-1+e^{-ikx}}{x^2} ,dx=int_{-infty}^infty frac{-1+cos(kx)}{x^2},dx$$



          Using the trigonometric identity $sin^2(kx/2)=frac12(1-cos(kx))$ and enforcing the substitution $kx/2mapsto x$, we see that



          $$ int_{-infty}^infty frac{-1+e^{ikx}}{x^2},dx=-kint_{-infty}^infty frac{sin^2(x)}{x^2},dx$$



          See THIS. Can you finish now?







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 11 '18 at 3:28

























          answered Dec 11 '18 at 3:22









          Mark ViolaMark Viola

          132k1276174




          132k1276174












          • $begingroup$
            Yes. Thanks a lot!
            $endgroup$
            – xi dai
            Dec 11 '18 at 4:45










          • $begingroup$
            Sorry why do we need Cauchy principal value for the first line to hold?
            $endgroup$
            – xi dai
            Dec 11 '18 at 4:55










          • $begingroup$
            You're welcome. My pleasure.
            $endgroup$
            – Mark Viola
            Dec 11 '18 at 14:20










          • $begingroup$
            We need a CPV for the first part since $int_{-infty}^infty frac{sin(x)}{x^2},dx$ fails to converges due to the $frac1x$ type singularity around $x=0$.
            $endgroup$
            – Mark Viola
            Dec 11 '18 at 14:22


















          • $begingroup$
            Yes. Thanks a lot!
            $endgroup$
            – xi dai
            Dec 11 '18 at 4:45










          • $begingroup$
            Sorry why do we need Cauchy principal value for the first line to hold?
            $endgroup$
            – xi dai
            Dec 11 '18 at 4:55










          • $begingroup$
            You're welcome. My pleasure.
            $endgroup$
            – Mark Viola
            Dec 11 '18 at 14:20










          • $begingroup$
            We need a CPV for the first part since $int_{-infty}^infty frac{sin(x)}{x^2},dx$ fails to converges due to the $frac1x$ type singularity around $x=0$.
            $endgroup$
            – Mark Viola
            Dec 11 '18 at 14:22
















          $begingroup$
          Yes. Thanks a lot!
          $endgroup$
          – xi dai
          Dec 11 '18 at 4:45




          $begingroup$
          Yes. Thanks a lot!
          $endgroup$
          – xi dai
          Dec 11 '18 at 4:45












          $begingroup$
          Sorry why do we need Cauchy principal value for the first line to hold?
          $endgroup$
          – xi dai
          Dec 11 '18 at 4:55




          $begingroup$
          Sorry why do we need Cauchy principal value for the first line to hold?
          $endgroup$
          – xi dai
          Dec 11 '18 at 4:55












          $begingroup$
          You're welcome. My pleasure.
          $endgroup$
          – Mark Viola
          Dec 11 '18 at 14:20




          $begingroup$
          You're welcome. My pleasure.
          $endgroup$
          – Mark Viola
          Dec 11 '18 at 14:20












          $begingroup$
          We need a CPV for the first part since $int_{-infty}^infty frac{sin(x)}{x^2},dx$ fails to converges due to the $frac1x$ type singularity around $x=0$.
          $endgroup$
          – Mark Viola
          Dec 11 '18 at 14:22




          $begingroup$
          We need a CPV for the first part since $int_{-infty}^infty frac{sin(x)}{x^2},dx$ fails to converges due to the $frac1x$ type singularity around $x=0$.
          $endgroup$
          – Mark Viola
          Dec 11 '18 at 14:22











          1












          $begingroup$

          Another way to do it if you want to use contour integration is to use an indented semi circle in the upper half plane going around the singularity at $z=0$. For this contour, the function $f(z)=(e^{-ikz}-1)/z^2$ will integrate to 0 by Cauchy's theorem since it is holomorphic in the region enclosed by the indented semi circle. Now you simply split $int_gamma=int_{C_R}+int_{C_epsilon}+int_{-R}^{-epsilon}+int_{epsilon}^{R}$ and you're left with justifying the exchange of limit and integrals when you want to let $epsilon$ (radius of small semi circle) go to $0$ and R go to $infty$ (radius of larger semi-circle).
          Here are some examples : https://web.williams.edu/Mathematics/sjmiller/public_html/302/coursenotes/Trapper_MethodsContourIntegrals.pdf



          There are also nice people explaining it in full detail on youtube, have fun !






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This pdf is exactly the one I have been following. Thanks!
            $endgroup$
            – xi dai
            Dec 11 '18 at 4:47
















          1












          $begingroup$

          Another way to do it if you want to use contour integration is to use an indented semi circle in the upper half plane going around the singularity at $z=0$. For this contour, the function $f(z)=(e^{-ikz}-1)/z^2$ will integrate to 0 by Cauchy's theorem since it is holomorphic in the region enclosed by the indented semi circle. Now you simply split $int_gamma=int_{C_R}+int_{C_epsilon}+int_{-R}^{-epsilon}+int_{epsilon}^{R}$ and you're left with justifying the exchange of limit and integrals when you want to let $epsilon$ (radius of small semi circle) go to $0$ and R go to $infty$ (radius of larger semi-circle).
          Here are some examples : https://web.williams.edu/Mathematics/sjmiller/public_html/302/coursenotes/Trapper_MethodsContourIntegrals.pdf



          There are also nice people explaining it in full detail on youtube, have fun !






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This pdf is exactly the one I have been following. Thanks!
            $endgroup$
            – xi dai
            Dec 11 '18 at 4:47














          1












          1








          1





          $begingroup$

          Another way to do it if you want to use contour integration is to use an indented semi circle in the upper half plane going around the singularity at $z=0$. For this contour, the function $f(z)=(e^{-ikz}-1)/z^2$ will integrate to 0 by Cauchy's theorem since it is holomorphic in the region enclosed by the indented semi circle. Now you simply split $int_gamma=int_{C_R}+int_{C_epsilon}+int_{-R}^{-epsilon}+int_{epsilon}^{R}$ and you're left with justifying the exchange of limit and integrals when you want to let $epsilon$ (radius of small semi circle) go to $0$ and R go to $infty$ (radius of larger semi-circle).
          Here are some examples : https://web.williams.edu/Mathematics/sjmiller/public_html/302/coursenotes/Trapper_MethodsContourIntegrals.pdf



          There are also nice people explaining it in full detail on youtube, have fun !






          share|cite|improve this answer









          $endgroup$



          Another way to do it if you want to use contour integration is to use an indented semi circle in the upper half plane going around the singularity at $z=0$. For this contour, the function $f(z)=(e^{-ikz}-1)/z^2$ will integrate to 0 by Cauchy's theorem since it is holomorphic in the region enclosed by the indented semi circle. Now you simply split $int_gamma=int_{C_R}+int_{C_epsilon}+int_{-R}^{-epsilon}+int_{epsilon}^{R}$ and you're left with justifying the exchange of limit and integrals when you want to let $epsilon$ (radius of small semi circle) go to $0$ and R go to $infty$ (radius of larger semi-circle).
          Here are some examples : https://web.williams.edu/Mathematics/sjmiller/public_html/302/coursenotes/Trapper_MethodsContourIntegrals.pdf



          There are also nice people explaining it in full detail on youtube, have fun !







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 11 '18 at 3:35









          MalikMalik

          1018




          1018












          • $begingroup$
            This pdf is exactly the one I have been following. Thanks!
            $endgroup$
            – xi dai
            Dec 11 '18 at 4:47


















          • $begingroup$
            This pdf is exactly the one I have been following. Thanks!
            $endgroup$
            – xi dai
            Dec 11 '18 at 4:47
















          $begingroup$
          This pdf is exactly the one I have been following. Thanks!
          $endgroup$
          – xi dai
          Dec 11 '18 at 4:47




          $begingroup$
          This pdf is exactly the one I have been following. Thanks!
          $endgroup$
          – xi dai
          Dec 11 '18 at 4:47


















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