How to calculate this simple integral?












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 quantum-mechanics greens-function







share|cite











$endgroup$












  • $begingroup$
    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
    2 hours ago






  • 1




    $begingroup$
    Hint: This is all you need to solve: $$ int_{-infty}^infty e^{-x^2}:dx = sqrt{pi}$$
    $endgroup$
    – DavidG
    2 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 quantum-mechanics greens-function







share|cite











$endgroup$












  • $begingroup$
    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
    2 hours ago






  • 1




    $begingroup$
    Hint: This is all you need to solve: $$ int_{-infty}^infty e^{-x^2}:dx = sqrt{pi}$$
    $endgroup$
    – DavidG
    2 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 quantum-mechanics greens-function







share|cite











$endgroup$




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 quantum-mechanics greens-function




integration quantum-mechanics greens-function






share|cite















share|cite













share|cite




share|cite








edited 27 mins ago







Collin

















asked 3 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
    2 hours ago






  • 1




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






  • 1




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




1




1




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






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





















3












$begingroup$

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%2fhow-to-calculate-this-simple-integral%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
        42 mins 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
        18 mins 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
        42 mins 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
        18 mins 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 1 hour 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
        42 mins 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
        18 mins 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
        42 mins 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
        18 mins 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
      42 mins 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
      42 mins 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
      18 mins 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
      18 mins ago













      3












      $begingroup$

      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









      $endgroup$


















        3












        $begingroup$

        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









        $endgroup$
















          3












          3








          3





          $begingroup$

          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 1 hour 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 1 hour 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%2fhow-to-calculate-this-simple-integral%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

                      SQL Server 17 - Attemping to backup to remote NAS but Access is denied

                      Image
                      .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0; } 0 I was attemping to use Ola Hallengrens scripts to backup to a remote NAS, but it was failing and not giving any direct errors, so I tried doing it manually using Server Management Studio and got the following error: TITLE: Microsoft SQL Server Management Studio Failed to retrieve data for this request. (Microsoft.SqlServer.Management.Sdk.Sfc) For help, click: http://go.microsoft.com/fwlink?ProdName=Microsoft%20SQL%20Server&LinkId=20476 ADDITIONAL INFORMATION: An exception occurred while executing a Transact-SQL statement or batch. (Microsoft.SqlServer.ConnectionInfo) The operating system returned the error '0x80070005(Access is denied.)&#
                      Read more

                      Always On Availability groups resolving state after failover - Remote harden of transaction...

                      Image
                      0 My explanation will be opinion based as it is based on my findings. If more information is needed, I am always happy to add it. The setup We have 4 SQL Server 2016 Instances, and all 4 underlying servers are in a WSFC (In my eyes a strange setup since Always On AG's are 2 by 2. But that is a different question.) 4 nodes + Fileshare witness Version: 13.0.5026.0 (all nodes) The instances that are having the failover issues are instances 1 & 2. They have 3 AG's between them. AG1 is primary on instance 1 and is having the issue. AG2 and AG3 are primary on instance 2. Instances 3 and 4 have their own separate AG's (2). AG1's settings The week before Last week, after checking for long running queries (none) and the synchronization state (green accross the board) we initiated a
                      Read more

                      Restoring from pg_dump with foreign key constraints

                      Image
                      5 2 In restoring a database from a pg_dump , a number of errors are being generated and the whole table is subsequently being ignored. An example: ERROR: insert or update on table "channelproducts" violates foreign key constraint "fk_rails_dfaae373a5" DETAIL: Key (channel_id)=(1) is not present in table "channels". Interestingly enough, I've noted that all these instances are popping up because of the sequence of loading. channels is after channelproducts both alphabetically and in the file and thus I can understand why postgres complains about having to create a child without a parent. caveat: the foreign key is being generated automatically by rails 4.2: I could remove the problem at the source but that still does not really solve the problem... version: PostgreSQL
                      Read more