Calculate complex integral $int_{-infty}^infty e^{-ix^2}d x=?$












3












$begingroup$


How to calculate this elementary complex integral? This is what we would encounter if we are studying the Green's function for Schroedinger's equation.
$$int_{-infty}^infty e^{-ix^2}d x=?$$
However, I think there should be someone that posted similar question on Math SE, though I don't know how to search by equation.



Thank you very much if you can help me out! And I would be grateful if you can give more than one approach




P.S.: The equation $int _{-infty}^{infty}e^{-kt^2}d sqrt{k}t=sqrt{pi}$ surely comes to my mind, but I don't know why it holds for $kinmathbb{C}$, because for me, the above integral is over real line, however, the question here is like integral on $y=e^{i pi/4}x$ ( So I think it's the problem with my complex integral knowledge.)



I tried to rotate this integral path by $pi/4$, but the two arcs at $Rrightarrow infty$ seem not easy to handle either.






integration complex-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hi Collin - Welcome to MSE - in order for the community to be able to assist you, you need to provide all working you have done so far. If you are looking for a starting point, please ask - but this site is not a 'homework for free' site.
    $endgroup$
    – DavidG
    4 hours ago






  • 1




    $begingroup$
    Hint: This is all you need to solve: $$ int_{-infty}^infty e^{-x^2}:dx = sqrt{pi}$$
    $endgroup$
    – DavidG
    4 hours ago












  • $begingroup$
    Assuming the jump into the complex domain is valid. I always do.
    $endgroup$
    – marty cohen
    2 hours ago


















3












$begingroup$


How to calculate this elementary complex integral? This is what we would encounter if we are studying the Green's function for Schroedinger's equation.
$$int_{-infty}^infty e^{-ix^2}d x=?$$
However, I think there should be someone that posted similar question on Math SE, though I don't know how to search by equation.



Thank you very much if you can help me out! And I would be grateful if you can give more than one approach




P.S.: The equation $int _{-infty}^{infty}e^{-kt^2}d sqrt{k}t=sqrt{pi}$ surely comes to my mind, but I don't know why it holds for $kinmathbb{C}$, because for me, the above integral is over real line, however, the question here is like integral on $y=e^{i pi/4}x$ ( So I think it's the problem with my complex integral knowledge.)



I tried to rotate this integral path by $pi/4$, but the two arcs at $Rrightarrow infty$ seem not easy to handle either.






integration complex-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hi Collin - Welcome to MSE - in order for the community to be able to assist you, you need to provide all working you have done so far. If you are looking for a starting point, please ask - but this site is not a 'homework for free' site.
    $endgroup$
    – DavidG
    4 hours ago






  • 1




    $begingroup$
    Hint: This is all you need to solve: $$ int_{-infty}^infty e^{-x^2}:dx = sqrt{pi}$$
    $endgroup$
    – DavidG
    4 hours ago












  • $begingroup$
    Assuming the jump into the complex domain is valid. I always do.
    $endgroup$
    – marty cohen
    2 hours ago
















3












3








3





$begingroup$


How to calculate this elementary complex integral? This is what we would encounter if we are studying the Green's function for Schroedinger's equation.
$$int_{-infty}^infty e^{-ix^2}d x=?$$
However, I think there should be someone that posted similar question on Math SE, though I don't know how to search by equation.



Thank you very much if you can help me out! And I would be grateful if you can give more than one approach




P.S.: The equation $int _{-infty}^{infty}e^{-kt^2}d sqrt{k}t=sqrt{pi}$ surely comes to my mind, but I don't know why it holds for $kinmathbb{C}$, because for me, the above integral is over real line, however, the question here is like integral on $y=e^{i pi/4}x$ ( So I think it's the problem with my complex integral knowledge.)



I tried to rotate this integral path by $pi/4$, but the two arcs at $Rrightarrow infty$ seem not easy to handle either.






integration complex-analysis







share|cite|improve this question











$endgroup$




How to calculate this elementary complex integral? This is what we would encounter if we are studying the Green's function for Schroedinger's equation.
$$int_{-infty}^infty e^{-ix^2}d x=?$$
However, I think there should be someone that posted similar question on Math SE, though I don't know how to search by equation.



Thank you very much if you can help me out! And I would be grateful if you can give more than one approach




P.S.: The equation $int _{-infty}^{infty}e^{-kt^2}d sqrt{k}t=sqrt{pi}$ surely comes to my mind, but I don't know why it holds for $kinmathbb{C}$, because for me, the above integral is over real line, however, the question here is like integral on $y=e^{i pi/4}x$ ( So I think it's the problem with my complex integral knowledge.)



I tried to rotate this integral path by $pi/4$, but the two arcs at $Rrightarrow infty$ seem not easy to handle either.






integration complex-analysis




integration complex-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 57 mins ago









Andrews

5931318




5931318










asked 4 hours ago









CollinCollin

1427




1427












  • $begingroup$
    Hi Collin - Welcome to MSE - in order for the community to be able to assist you, you need to provide all working you have done so far. If you are looking for a starting point, please ask - but this site is not a 'homework for free' site.
    $endgroup$
    – DavidG
    4 hours ago






  • 1




    $begingroup$
    Hint: This is all you need to solve: $$ int_{-infty}^infty e^{-x^2}:dx = sqrt{pi}$$
    $endgroup$
    – DavidG
    4 hours ago












  • $begingroup$
    Assuming the jump into the complex domain is valid. I always do.
    $endgroup$
    – marty cohen
    2 hours ago




















  • $begingroup$
    Hi Collin - Welcome to MSE - in order for the community to be able to assist you, you need to provide all working you have done so far. If you are looking for a starting point, please ask - but this site is not a 'homework for free' site.
    $endgroup$
    – DavidG
    4 hours ago






  • 1




    $begingroup$
    Hint: This is all you need to solve: $$ int_{-infty}^infty e^{-x^2}:dx = sqrt{pi}$$
    $endgroup$
    – DavidG
    4 hours ago












  • $begingroup$
    Assuming the jump into the complex domain is valid. I always do.
    $endgroup$
    – marty cohen
    2 hours ago


















$begingroup$
Hi Collin - Welcome to MSE - in order for the community to be able to assist you, you need to provide all working you have done so far. If you are looking for a starting point, please ask - but this site is not a 'homework for free' site.
$endgroup$
– DavidG
4 hours ago




$begingroup$
Hi Collin - Welcome to MSE - in order for the community to be able to assist you, you need to provide all working you have done so far. If you are looking for a starting point, please ask - but this site is not a 'homework for free' site.
$endgroup$
– DavidG
4 hours ago




1




1




$begingroup$
Hint: This is all you need to solve: $$ int_{-infty}^infty e^{-x^2}:dx = sqrt{pi}$$
$endgroup$
– DavidG
4 hours ago






$begingroup$
Hint: This is all you need to solve: $$ int_{-infty}^infty e^{-x^2}:dx = sqrt{pi}$$
$endgroup$
– DavidG
4 hours ago














$begingroup$
Assuming the jump into the complex domain is valid. I always do.
$endgroup$
– marty cohen
2 hours ago






$begingroup$
Assuming the jump into the complex domain is valid. I always do.
$endgroup$
– marty cohen
2 hours ago












3 Answers
3






active

oldest

votes


















5












$begingroup$

Hint
$$int e^{-k x^2},dx=frac{sqrt{pi } }{2 sqrt{k}},text{erf}left(sqrt{k} xright)$$
$$f(k)=int_{-infty}^infty e^{-k x^2},dx=frac{sqrt{pi }}{sqrt{k}}$$
$$f(i)=frac{sqrt{pi }}{sqrt{i}}=(1-i) sqrt{frac{pi }{2}}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Hi! I think it's because I'm not familiar with complex integral; so how can you guys play so freely on the $mathbb{C}$ plane, without worrying much? The very reason I ask this is because I don't know why your second equation holds for k being a complex number. I tried to rotate this integral path by $pi/4$ on the complex plane, but the two arcs at $|x|rightarrow infty$ seem not trying to vanish.
    $endgroup$
    – Collin
    1 hour ago












  • $begingroup$
    @Collin, The reasoning behind this 'extrapolation' is that $f$ defines a holomorphic function on the right half-plane $H={k:operatorname{Re}(k) > 0}$ and a continuous function on $overline{H}setminus{0} = {k : operatorname{Re}(k) geq 0, k neq 0 }$. Now, in view of identity theorem, if you can identify another holomorphic function $g$ on $H$ such that $f = g$ on $(0,infty)$, then $f = g$ on all of $H$ as well. We can check that this argument works with $g(k)=sqrt{pi/k}$, and so, $f = g$ on $H$ by identity theorem and on all of $overline{H}setminus{0}$ by continuity.
    $endgroup$
    – Sangchul Lee
    1 hour ago





















4












$begingroup$

Trying to avoid complex funniness.



$begin{array}\
int_{-infty}^infty e^{-ix^2}dx
&=int_{-infty}^infty (cos(x^2)-isin(x^2))dx\
&=2int_{0}^infty (cos(x^2)-isin(x^2))dx\
&=2int_{0}^infty cos(x^2)dx-2iint_{0}^inftysin(x^2))dx\
end{array}
$



and these are the
Fresnel integrals
$C(x)$ and $S(x)$
both of which approach
$dfrac{sqrt{pi}}{8}
$
as $x to infty$.



Therefore the result is
$(1-i)sqrt{frac{pi}{2}}
$
as Claude Leibovici
got.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Hint:$$int_{-infty}^infty e^{-kx^2}dx=int_{-infty}^infty e^{-left(xsqrt kright)^2}dx$$
    Use the $u$-substution $u=xsqrt k$ and this transforms the integral into the form given in DavidG's suggestion. Can you take it from here?






    share|cite|improve this answer









    $endgroup$













      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%2f3125571%2fcalculate-complex-integral-int-infty-infty-e-ix2d-x%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      Hint
      $$int e^{-k x^2},dx=frac{sqrt{pi } }{2 sqrt{k}},text{erf}left(sqrt{k} xright)$$
      $$f(k)=int_{-infty}^infty e^{-k x^2},dx=frac{sqrt{pi }}{sqrt{k}}$$
      $$f(i)=frac{sqrt{pi }}{sqrt{i}}=(1-i) sqrt{frac{pi }{2}}$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Hi! I think it's because I'm not familiar with complex integral; so how can you guys play so freely on the $mathbb{C}$ plane, without worrying much? The very reason I ask this is because I don't know why your second equation holds for k being a complex number. I tried to rotate this integral path by $pi/4$ on the complex plane, but the two arcs at $|x|rightarrow infty$ seem not trying to vanish.
        $endgroup$
        – Collin
        1 hour ago












      • $begingroup$
        @Collin, The reasoning behind this 'extrapolation' is that $f$ defines a holomorphic function on the right half-plane $H={k:operatorname{Re}(k) > 0}$ and a continuous function on $overline{H}setminus{0} = {k : operatorname{Re}(k) geq 0, k neq 0 }$. Now, in view of identity theorem, if you can identify another holomorphic function $g$ on $H$ such that $f = g$ on $(0,infty)$, then $f = g$ on all of $H$ as well. We can check that this argument works with $g(k)=sqrt{pi/k}$, and so, $f = g$ on $H$ by identity theorem and on all of $overline{H}setminus{0}$ by continuity.
        $endgroup$
        – Sangchul Lee
        1 hour ago


















      5












      $begingroup$

      Hint
      $$int e^{-k x^2},dx=frac{sqrt{pi } }{2 sqrt{k}},text{erf}left(sqrt{k} xright)$$
      $$f(k)=int_{-infty}^infty e^{-k x^2},dx=frac{sqrt{pi }}{sqrt{k}}$$
      $$f(i)=frac{sqrt{pi }}{sqrt{i}}=(1-i) sqrt{frac{pi }{2}}$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Hi! I think it's because I'm not familiar with complex integral; so how can you guys play so freely on the $mathbb{C}$ plane, without worrying much? The very reason I ask this is because I don't know why your second equation holds for k being a complex number. I tried to rotate this integral path by $pi/4$ on the complex plane, but the two arcs at $|x|rightarrow infty$ seem not trying to vanish.
        $endgroup$
        – Collin
        1 hour ago












      • $begingroup$
        @Collin, The reasoning behind this 'extrapolation' is that $f$ defines a holomorphic function on the right half-plane $H={k:operatorname{Re}(k) > 0}$ and a continuous function on $overline{H}setminus{0} = {k : operatorname{Re}(k) geq 0, k neq 0 }$. Now, in view of identity theorem, if you can identify another holomorphic function $g$ on $H$ such that $f = g$ on $(0,infty)$, then $f = g$ on all of $H$ as well. We can check that this argument works with $g(k)=sqrt{pi/k}$, and so, $f = g$ on $H$ by identity theorem and on all of $overline{H}setminus{0}$ by continuity.
        $endgroup$
        – Sangchul Lee
        1 hour ago
















      5












      5








      5





      $begingroup$

      Hint
      $$int e^{-k x^2},dx=frac{sqrt{pi } }{2 sqrt{k}},text{erf}left(sqrt{k} xright)$$
      $$f(k)=int_{-infty}^infty e^{-k x^2},dx=frac{sqrt{pi }}{sqrt{k}}$$
      $$f(i)=frac{sqrt{pi }}{sqrt{i}}=(1-i) sqrt{frac{pi }{2}}$$






      share|cite|improve this answer









      $endgroup$



      Hint
      $$int e^{-k x^2},dx=frac{sqrt{pi } }{2 sqrt{k}},text{erf}left(sqrt{k} xright)$$
      $$f(k)=int_{-infty}^infty e^{-k x^2},dx=frac{sqrt{pi }}{sqrt{k}}$$
      $$f(i)=frac{sqrt{pi }}{sqrt{i}}=(1-i) sqrt{frac{pi }{2}}$$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered 3 hours ago









      Claude LeiboviciClaude Leibovici

      122k1157134




      122k1157134












      • $begingroup$
        Hi! I think it's because I'm not familiar with complex integral; so how can you guys play so freely on the $mathbb{C}$ plane, without worrying much? The very reason I ask this is because I don't know why your second equation holds for k being a complex number. I tried to rotate this integral path by $pi/4$ on the complex plane, but the two arcs at $|x|rightarrow infty$ seem not trying to vanish.
        $endgroup$
        – Collin
        1 hour ago












      • $begingroup$
        @Collin, The reasoning behind this 'extrapolation' is that $f$ defines a holomorphic function on the right half-plane $H={k:operatorname{Re}(k) > 0}$ and a continuous function on $overline{H}setminus{0} = {k : operatorname{Re}(k) geq 0, k neq 0 }$. Now, in view of identity theorem, if you can identify another holomorphic function $g$ on $H$ such that $f = g$ on $(0,infty)$, then $f = g$ on all of $H$ as well. We can check that this argument works with $g(k)=sqrt{pi/k}$, and so, $f = g$ on $H$ by identity theorem and on all of $overline{H}setminus{0}$ by continuity.
        $endgroup$
        – Sangchul Lee
        1 hour ago




















      • $begingroup$
        Hi! I think it's because I'm not familiar with complex integral; so how can you guys play so freely on the $mathbb{C}$ plane, without worrying much? The very reason I ask this is because I don't know why your second equation holds for k being a complex number. I tried to rotate this integral path by $pi/4$ on the complex plane, but the two arcs at $|x|rightarrow infty$ seem not trying to vanish.
        $endgroup$
        – Collin
        1 hour ago












      • $begingroup$
        @Collin, The reasoning behind this 'extrapolation' is that $f$ defines a holomorphic function on the right half-plane $H={k:operatorname{Re}(k) > 0}$ and a continuous function on $overline{H}setminus{0} = {k : operatorname{Re}(k) geq 0, k neq 0 }$. Now, in view of identity theorem, if you can identify another holomorphic function $g$ on $H$ such that $f = g$ on $(0,infty)$, then $f = g$ on all of $H$ as well. We can check that this argument works with $g(k)=sqrt{pi/k}$, and so, $f = g$ on $H$ by identity theorem and on all of $overline{H}setminus{0}$ by continuity.
        $endgroup$
        – Sangchul Lee
        1 hour ago


















      $begingroup$
      Hi! I think it's because I'm not familiar with complex integral; so how can you guys play so freely on the $mathbb{C}$ plane, without worrying much? The very reason I ask this is because I don't know why your second equation holds for k being a complex number. I tried to rotate this integral path by $pi/4$ on the complex plane, but the two arcs at $|x|rightarrow infty$ seem not trying to vanish.
      $endgroup$
      – Collin
      1 hour ago






      $begingroup$
      Hi! I think it's because I'm not familiar with complex integral; so how can you guys play so freely on the $mathbb{C}$ plane, without worrying much? The very reason I ask this is because I don't know why your second equation holds for k being a complex number. I tried to rotate this integral path by $pi/4$ on the complex plane, but the two arcs at $|x|rightarrow infty$ seem not trying to vanish.
      $endgroup$
      – Collin
      1 hour ago














      $begingroup$
      @Collin, The reasoning behind this 'extrapolation' is that $f$ defines a holomorphic function on the right half-plane $H={k:operatorname{Re}(k) > 0}$ and a continuous function on $overline{H}setminus{0} = {k : operatorname{Re}(k) geq 0, k neq 0 }$. Now, in view of identity theorem, if you can identify another holomorphic function $g$ on $H$ such that $f = g$ on $(0,infty)$, then $f = g$ on all of $H$ as well. We can check that this argument works with $g(k)=sqrt{pi/k}$, and so, $f = g$ on $H$ by identity theorem and on all of $overline{H}setminus{0}$ by continuity.
      $endgroup$
      – Sangchul Lee
      1 hour ago






      $begingroup$
      @Collin, The reasoning behind this 'extrapolation' is that $f$ defines a holomorphic function on the right half-plane $H={k:operatorname{Re}(k) > 0}$ and a continuous function on $overline{H}setminus{0} = {k : operatorname{Re}(k) geq 0, k neq 0 }$. Now, in view of identity theorem, if you can identify another holomorphic function $g$ on $H$ such that $f = g$ on $(0,infty)$, then $f = g$ on all of $H$ as well. We can check that this argument works with $g(k)=sqrt{pi/k}$, and so, $f = g$ on $H$ by identity theorem and on all of $overline{H}setminus{0}$ by continuity.
      $endgroup$
      – Sangchul Lee
      1 hour ago













      4












      $begingroup$

      Trying to avoid complex funniness.



      $begin{array}\
      int_{-infty}^infty e^{-ix^2}dx
      &=int_{-infty}^infty (cos(x^2)-isin(x^2))dx\
      &=2int_{0}^infty (cos(x^2)-isin(x^2))dx\
      &=2int_{0}^infty cos(x^2)dx-2iint_{0}^inftysin(x^2))dx\
      end{array}
      $



      and these are the
      Fresnel integrals
      $C(x)$ and $S(x)$
      both of which approach
      $dfrac{sqrt{pi}}{8}
      $
      as $x to infty$.



      Therefore the result is
      $(1-i)sqrt{frac{pi}{2}}
      $
      as Claude Leibovici
      got.






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        Trying to avoid complex funniness.



        $begin{array}\
        int_{-infty}^infty e^{-ix^2}dx
        &=int_{-infty}^infty (cos(x^2)-isin(x^2))dx\
        &=2int_{0}^infty (cos(x^2)-isin(x^2))dx\
        &=2int_{0}^infty cos(x^2)dx-2iint_{0}^inftysin(x^2))dx\
        end{array}
        $



        and these are the
        Fresnel integrals
        $C(x)$ and $S(x)$
        both of which approach
        $dfrac{sqrt{pi}}{8}
        $
        as $x to infty$.



        Therefore the result is
        $(1-i)sqrt{frac{pi}{2}}
        $
        as Claude Leibovici
        got.






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          Trying to avoid complex funniness.



          $begin{array}\
          int_{-infty}^infty e^{-ix^2}dx
          &=int_{-infty}^infty (cos(x^2)-isin(x^2))dx\
          &=2int_{0}^infty (cos(x^2)-isin(x^2))dx\
          &=2int_{0}^infty cos(x^2)dx-2iint_{0}^inftysin(x^2))dx\
          end{array}
          $



          and these are the
          Fresnel integrals
          $C(x)$ and $S(x)$
          both of which approach
          $dfrac{sqrt{pi}}{8}
          $
          as $x to infty$.



          Therefore the result is
          $(1-i)sqrt{frac{pi}{2}}
          $
          as Claude Leibovici
          got.






          share|cite|improve this answer









          $endgroup$



          Trying to avoid complex funniness.



          $begin{array}\
          int_{-infty}^infty e^{-ix^2}dx
          &=int_{-infty}^infty (cos(x^2)-isin(x^2))dx\
          &=2int_{0}^infty (cos(x^2)-isin(x^2))dx\
          &=2int_{0}^infty cos(x^2)dx-2iint_{0}^inftysin(x^2))dx\
          end{array}
          $



          and these are the
          Fresnel integrals
          $C(x)$ and $S(x)$
          both of which approach
          $dfrac{sqrt{pi}}{8}
          $
          as $x to infty$.



          Therefore the result is
          $(1-i)sqrt{frac{pi}{2}}
          $
          as Claude Leibovici
          got.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 hours ago









          marty cohenmarty cohen

          73.9k549128




          73.9k549128























              2












              $begingroup$

              Hint:$$int_{-infty}^infty e^{-kx^2}dx=int_{-infty}^infty e^{-left(xsqrt kright)^2}dx$$
              Use the $u$-substution $u=xsqrt k$ and this transforms the integral into the form given in DavidG's suggestion. Can you take it from here?






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Hint:$$int_{-infty}^infty e^{-kx^2}dx=int_{-infty}^infty e^{-left(xsqrt kright)^2}dx$$
                Use the $u$-substution $u=xsqrt k$ and this transforms the integral into the form given in DavidG's suggestion. Can you take it from here?






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Hint:$$int_{-infty}^infty e^{-kx^2}dx=int_{-infty}^infty e^{-left(xsqrt kright)^2}dx$$
                  Use the $u$-substution $u=xsqrt k$ and this transforms the integral into the form given in DavidG's suggestion. Can you take it from here?






                  share|cite|improve this answer









                  $endgroup$



                  Hint:$$int_{-infty}^infty e^{-kx^2}dx=int_{-infty}^infty e^{-left(xsqrt kright)^2}dx$$
                  Use the $u$-substution $u=xsqrt k$ and this transforms the integral into the form given in DavidG's suggestion. Can you take it from here?







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 hours ago









                  csch2csch2

                  2571311




                  2571311






























                      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%2f3125571%2fcalculate-complex-integral-int-infty-infty-e-ix2d-x%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

                      How to make a Squid Proxy server?

                      Image
                      1 I want to set up squid proxy server for monitoring user activity in a network and Internet Web URL filtering for user. I also want to configure the user login, user log, user access group, and block unwanted web sites in my network. Please help me and guide me the full process to do this. server proxy squid share | improve this question edited Jul 31 '12 at 14:45 Jorge Castro 37.1k 107 422 617 asked Jul 31 '12 at 5:16 Goivind Tiwari Goivind Tiwari ...
                      Read more

                      Is this a new Fibonacci Identity?

                      Image
                      6 1 $begingroup$ I have found the following Fibonacci Identity (and proved it). If $F_n$ denotes the nth Fibonacci Number, we have the following identity begin{equation} F_{n-r+h}F_{n+k+g+1} - F_{n-r+g}F_{n+k+h+1} = (-1)^{n+r+h+1} F_{g-h}F_{k+r+1} end{equation} where $F_1 = F_2 = 1$ , $r leq n$ , $h leq g$ , and $n, g, k in mathbb{N}$ . It is not too hard to show that this identity subsumes Cassini's Identity, Catalan's Identity, Vajda's Idenity, and d'Ocagne's identity to name a few. I have done a pretty thorough literature review, and I have not found anything like this, but I am still wondering if anyone has seen this identity before? I found this by accident after noticing some patterns in some analysis work I was doing, so if this is already known I would be curious to see w...
                      Read more

                      19世紀

                      Image
                      この記事は検証可能な参考文献や出典が全く示されていないか、不十分です。 出典を追加して記事の信頼性向上にご協力ください。 （ 2012年3月 ） 千年紀: 2千年紀 世紀: 18世紀 - 19世紀 - 20世紀 十年紀: 1800年代 1810年代 1820年代 1830年代 1840年代 1850年代 1860年代 1870年代 1880年代 1890年代 19世紀に君臨した大英帝国。 19世紀 （じゅうきゅうせいき）は、西暦1801年から西暦1900年までの100年間を指す世紀。 目次 1 19世紀の歴史 1.1 国民国家の成立 1.2 帝国主義の興隆 1.3 列強の植民地争奪戦 2 できごと 2.1 1800年代 2.2 1810年代 2.3 1820年代 2.4 1830年代 2.5 1840年代 2.6 1850年代 2.7 1860年代 2.8 1870年代 2.9 1880年代 2.10 1890年代 2.11 1900年代 3 文化 3.1 文学 3.2 音楽 3.3 思想 3.4 科学 3.5 技術 4 人物 4.1 ヨーロッパ 4.1.1 政治家・王族 4.1.1.1 フランス 4.1.1.2 オーストリア＝ハンガリー 4.1.1.3 ロシア 4.1.1.4 イギリス 4.1.1.5 ドイツ（プロイセンほかドイツ領邦を含む） 4.1.1.6 北欧 4.1.1.7 イタリア 4.1.1.8 スペイン 4.1.1.9 ベルギー 4.1.1.10 ギリシア 4.1.2 軍人 4.1.3 実業家 4.1.4 科学と技術 4.1.5 思想と哲学・人文諸学 4.1.6 宗教 4.1.7 文学 4.1.8 美術 4.1.9 音楽 4.1.10 社会事業家 4.1.11 探検家・旅行家 4.1.12 料...
                      Read more