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
    51 mins 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
    51 mins 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 15 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
    51 mins 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
    51 mins 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
51 mins ago






$begingroup$
Assuming the jump into the complex domain is valid. I always do.
$endgroup$
– marty cohen
51 mins 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
    30 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
    6 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





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      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
        30 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
        6 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
        30 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
        6 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
        30 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
        6 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
        30 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
        6 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
      30 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
      30 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
      6 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
      6 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 52 mins 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



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                  answered 55 mins ago









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