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












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












    $begingroup$


    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











          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3034826%2fhow-to-integrate-frac-1e-i-k-xx2%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          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


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3034826%2fhow-to-integrate-frac-1e-i-k-xx2%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          Le Mesnil-Réaume

          Image
          Le Mesnil-Réaume Region Normandie Département Seine-Maritime Arrondissement Dieppe Kanton Eu Gemeindeverband Villes Sœurs Koordinaten 49° 58′  N , 1° 27′  O 49.9725 1.4463888888889 Koordinaten: 49° 58′  N , 1° 27′  O Höhe 92–156 m Fläche 5,55 km 2 Einwohner 793 (1. Januar 2016) Bevölkerungsdichte 143 Einw./km 2 Postleitzahl 76260 INSEE-Code 76435 Website http://www.mesnil-reaume.net/ Rathaus (Mairie) und Schule von Le Mesnil-Réaume Le Mesnil-Réaume ist eine französische Gemeinde mit 793 Einwohnern (Stand: 1. Januar 2016) im Département Seine-Maritime in der Region Normandie (vor 2016 Haute-Normandie). Sie gehört zum Arrondissement Dieppe und zum Kanton Eu. Die Einwohner werden Mesnillais genannt. Inhaltsverzeichnis 1 Geographie 2 Bevölkerungsentwicklung 3 Sehenswürdigkeiten 4 Weblinks Geographie | Le Mesnil-Réaume liegt etwa 43 Kilomet
          Read more

          Ida-Boy-Ed-Garten

          Image
          Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
          Read more

          web3.py web3.isConnected() returns false always

          Image
          1 I am trying an tutorial on web3.py so i created a profile in infura and used its project id.But it shows false always when i try web3.isConnected() PS C:UsersHP> python Python 3.6.3 (v3.6.3:2c5fed8, Oct 3 2017, 18:11:49) [MSC v.1900 64 bit (AMD64)] on win32 Type "help", "copyright", "credits" or "license" for more information. >>> from web3 import Web3 >>> infura_url = "mainnet.infura.io/v3/******************************" >>> web3 = Web3(Web3.HTTPProvider(infura_url)) >>> web3.isConnected() False web3.py share | improve this question
          Read more