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












2












$begingroup$


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
    3 hours ago






  • 1




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












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


















2












$begingroup$


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
    3 hours ago






  • 1




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












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
















2












2








2





$begingroup$


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 36 mins ago









Andrews

5931318




5931318










asked 4 hours ago









CollinCollin

1377




1377












  • $begingroup$
    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
    3 hours ago






  • 1




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












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






  • 1




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












  • $begingroup$
    Assuming the jump into the complex domain is valid. I always do.
    $endgroup$
    – marty cohen
    1 hour 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
3 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
3 hours ago




1




1




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






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














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






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

                      Loup dans la culture

                      Image
                      Loups sur une enluminure du Bestiaire d'Aberdeen Le loup est l'animal le plus emblématique de l'histoire de l'Eurasie, il était à l'honneur durant l'Antiquité chez la totalité des anciens peuples européens [ 1 ] . Sommaire 1 Le loup dans la mythologie 2 Les loups dans le folklore 2.1 L'évolution des mentalités 2.2 Le loup dans la psychanalyse 3 Le loup dans les œuvres culturelles 3.1 Le loup dans la littérature 3.1.1 Fables 3.1.2 Poésie 3.1.3 Romans 3.1.4 Littérature jeunesse 3.1.4.1 Quelques titres d'albums contemporains 3.1.5 Bande dessinée 3.2 Le loup à l'écran 3.2.1 Le cinéma et la télévision 3.2.2 Documentaires 3.2.3 Dessins animés 3.2.4 Jeux vidéo 3.3 Le loup en peinture et en sculpture 3.4 Le loup dans les œuvres musicales 4 Le loup et les noms propres 5 Bibliographie 6 Notes et références 7 Voir aussi 7.1 Articles connexes ...
                      Read more

                      How to solve the problem of ntp “Unable to contact time server” from KDE?

                      Image
                      1 0 I want to set date and time automatically but this error appears when I try to .. I'm using openSUSE, I tried different servers but it didn't work either, help please! It doesn't seem to be a network problem: $ ping pool.ntp.org 64 bytes from 41.78.128.17: icmp_seq=1 ttl=41 time=260 ms To the question of whether ntpd is installed and running: $ ps -C ntpd PID TTY TIME CMD i.e. there's no ntpd process running. $ rpm -qa | grep ntp yast2-ntp-client-3.1.12-1.7.noarch ntp-4.2.6p5-25.2.1.i586 Further information from the comments: $ sudo ntpdate pool.ntp.org 25 Sep 19:41:51 ntpdate[3162]: no server suitable for synchronization found networking kde opensuse ntp ntpd sh...
                      Read more

                      ASUS Zenbook UX433/UX333 — Configure Touchpad-embedded numpad on Linux

                      Image
                      0 1 I have an ASUS Zenbook UX433,on which ASUS decided to innovate by embedding a Numpad inside the touchpad (see picture 1). Toggling between classic touchpad mode and numpad (also turns the touchpad LEDs on) can be done through the golden button in the top-right corner of the touchpad. However, on my Manjaro arch Linux config, this behavior does not work by default and I can only use the touchpad, no numpad. Has anyone been able to configure the numpad to work on Linux, or have any idea how this could be achieved? EDIT1: I have had a look at the official Windows drivers for the UX433FA TouchPad, but was not able to make anything of it. I tried installing them on Linux using tools like NDISWrapper, but they do not detect the TouchPad/Numpad hardware, and my understanding of drivers is too limited. ...
                      Read more