How to change the limits of integration












4












$begingroup$


I am attempting to solve the integral of the following...



$$int_{0}^{2 pi}int_{0}^{infty}e^{-r^2}rdrTheta $$



So I do the following step...



$$=2 piint_{0}^{infty}e^{-r^2}rdr$$



but then the next step is to substitute $s = -r^2$ which results in...



$$=2 piint_{- infty}^{0}frac{1}{2}e^{s}ds$$



The limits of integration are reversed now and the $r$ somehow results in $1/2$.



Can someone explain why this works? Why did substituting cause the limits change and result in the integration above?






calculus integration limits







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    $$s = -r^{2} implies ds = -2r dr implies -ds/2 = rdr$$ and as $r to infty$, $s to - infty$ and as $r to 0$, $s to 0$.
    $endgroup$
    – Mattos
    3 hours ago


















4












$begingroup$


I am attempting to solve the integral of the following...



$$int_{0}^{2 pi}int_{0}^{infty}e^{-r^2}rdrTheta $$



So I do the following step...



$$=2 piint_{0}^{infty}e^{-r^2}rdr$$



but then the next step is to substitute $s = -r^2$ which results in...



$$=2 piint_{- infty}^{0}frac{1}{2}e^{s}ds$$



The limits of integration are reversed now and the $r$ somehow results in $1/2$.



Can someone explain why this works? Why did substituting cause the limits change and result in the integration above?






calculus integration limits







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    $$s = -r^{2} implies ds = -2r dr implies -ds/2 = rdr$$ and as $r to infty$, $s to - infty$ and as $r to 0$, $s to 0$.
    $endgroup$
    – Mattos
    3 hours ago
















4












4








4





$begingroup$


I am attempting to solve the integral of the following...



$$int_{0}^{2 pi}int_{0}^{infty}e^{-r^2}rdrTheta $$



So I do the following step...



$$=2 piint_{0}^{infty}e^{-r^2}rdr$$



but then the next step is to substitute $s = -r^2$ which results in...



$$=2 piint_{- infty}^{0}frac{1}{2}e^{s}ds$$



The limits of integration are reversed now and the $r$ somehow results in $1/2$.



Can someone explain why this works? Why did substituting cause the limits change and result in the integration above?






calculus integration limits







share|cite|improve this question









$endgroup$




I am attempting to solve the integral of the following...



$$int_{0}^{2 pi}int_{0}^{infty}e^{-r^2}rdrTheta $$



So I do the following step...



$$=2 piint_{0}^{infty}e^{-r^2}rdr$$



but then the next step is to substitute $s = -r^2$ which results in...



$$=2 piint_{- infty}^{0}frac{1}{2}e^{s}ds$$



The limits of integration are reversed now and the $r$ somehow results in $1/2$.



Can someone explain why this works? Why did substituting cause the limits change and result in the integration above?






calculus integration limits




calculus integration limits






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 3 hours ago









BolboaBolboa

398516




398516








  • 1




    $begingroup$
    $$s = -r^{2} implies ds = -2r dr implies -ds/2 = rdr$$ and as $r to infty$, $s to - infty$ and as $r to 0$, $s to 0$.
    $endgroup$
    – Mattos
    3 hours ago
















  • 1




    $begingroup$
    $$s = -r^{2} implies ds = -2r dr implies -ds/2 = rdr$$ and as $r to infty$, $s to - infty$ and as $r to 0$, $s to 0$.
    $endgroup$
    – Mattos
    3 hours ago










1




1




$begingroup$
$$s = -r^{2} implies ds = -2r dr implies -ds/2 = rdr$$ and as $r to infty$, $s to - infty$ and as $r to 0$, $s to 0$.
$endgroup$
– Mattos
3 hours ago






$begingroup$
$$s = -r^{2} implies ds = -2r dr implies -ds/2 = rdr$$ and as $r to infty$, $s to - infty$ and as $r to 0$, $s to 0$.
$endgroup$
– Mattos
3 hours ago












2 Answers
2






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6












$begingroup$

$s=-r^{2}$ gives $ds=-2rdr$ so $dr =-frac 1 {2r} ds$. Also, as $r$ increases from $0$ to $infty$, $s$ decreases from $0$ to $-infty$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    It should be noted that the minus sign from the substitution is then used to reverse the order of the limits.
    $endgroup$
    – John Doe
    3 hours ago






  • 1




    $begingroup$
    @JohnDoe Right. I didn't mention it explicitly but that is what I meant.
    $endgroup$
    – Kavi Rama Murthy
    3 hours ago



















0












$begingroup$

Do you really need substitution. We already know the antiderivative of $re^{-r^2}$ and it is $-e^{-r^2}over 2$






share|cite|improve this answer











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    2 Answers
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    2 Answers
    2






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    6












    $begingroup$

    $s=-r^{2}$ gives $ds=-2rdr$ so $dr =-frac 1 {2r} ds$. Also, as $r$ increases from $0$ to $infty$, $s$ decreases from $0$ to $-infty$.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      It should be noted that the minus sign from the substitution is then used to reverse the order of the limits.
      $endgroup$
      – John Doe
      3 hours ago






    • 1




      $begingroup$
      @JohnDoe Right. I didn't mention it explicitly but that is what I meant.
      $endgroup$
      – Kavi Rama Murthy
      3 hours ago
















    6












    $begingroup$

    $s=-r^{2}$ gives $ds=-2rdr$ so $dr =-frac 1 {2r} ds$. Also, as $r$ increases from $0$ to $infty$, $s$ decreases from $0$ to $-infty$.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      It should be noted that the minus sign from the substitution is then used to reverse the order of the limits.
      $endgroup$
      – John Doe
      3 hours ago






    • 1




      $begingroup$
      @JohnDoe Right. I didn't mention it explicitly but that is what I meant.
      $endgroup$
      – Kavi Rama Murthy
      3 hours ago














    6












    6








    6





    $begingroup$

    $s=-r^{2}$ gives $ds=-2rdr$ so $dr =-frac 1 {2r} ds$. Also, as $r$ increases from $0$ to $infty$, $s$ decreases from $0$ to $-infty$.






    share|cite|improve this answer









    $endgroup$



    $s=-r^{2}$ gives $ds=-2rdr$ so $dr =-frac 1 {2r} ds$. Also, as $r$ increases from $0$ to $infty$, $s$ decreases from $0$ to $-infty$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 3 hours ago









    Kavi Rama MurthyKavi Rama Murthy

    73.6k53170




    73.6k53170








    • 1




      $begingroup$
      It should be noted that the minus sign from the substitution is then used to reverse the order of the limits.
      $endgroup$
      – John Doe
      3 hours ago






    • 1




      $begingroup$
      @JohnDoe Right. I didn't mention it explicitly but that is what I meant.
      $endgroup$
      – Kavi Rama Murthy
      3 hours ago














    • 1




      $begingroup$
      It should be noted that the minus sign from the substitution is then used to reverse the order of the limits.
      $endgroup$
      – John Doe
      3 hours ago






    • 1




      $begingroup$
      @JohnDoe Right. I didn't mention it explicitly but that is what I meant.
      $endgroup$
      – Kavi Rama Murthy
      3 hours ago








    1




    1




    $begingroup$
    It should be noted that the minus sign from the substitution is then used to reverse the order of the limits.
    $endgroup$
    – John Doe
    3 hours ago




    $begingroup$
    It should be noted that the minus sign from the substitution is then used to reverse the order of the limits.
    $endgroup$
    – John Doe
    3 hours ago




    1




    1




    $begingroup$
    @JohnDoe Right. I didn't mention it explicitly but that is what I meant.
    $endgroup$
    – Kavi Rama Murthy
    3 hours ago




    $begingroup$
    @JohnDoe Right. I didn't mention it explicitly but that is what I meant.
    $endgroup$
    – Kavi Rama Murthy
    3 hours ago











    0












    $begingroup$

    Do you really need substitution. We already know the antiderivative of $re^{-r^2}$ and it is $-e^{-r^2}over 2$






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Do you really need substitution. We already know the antiderivative of $re^{-r^2}$ and it is $-e^{-r^2}over 2$






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Do you really need substitution. We already know the antiderivative of $re^{-r^2}$ and it is $-e^{-r^2}over 2$






        share|cite|improve this answer











        $endgroup$



        Do you really need substitution. We already know the antiderivative of $re^{-r^2}$ and it is $-e^{-r^2}over 2$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 3 hours ago

























        answered 3 hours ago









        HAMIDINE SOUMAREHAMIDINE SOUMARE

        1,836212




        1,836212






























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