How to change the limits of integration












5












$begingroup$


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







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  • 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
    Apr 9 at 23:44












  • $begingroup$
    This constellation of consequences is why I recommend to my students to set $s = r^2$. This gives fewer minus signs and since the substituent is monotonically increasing, does not reverse the order of the bounds of integration.
    $endgroup$
    – Eric Towers
    Apr 10 at 4:30
















5












$begingroup$


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
    Apr 9 at 23:44












  • $begingroup$
    This constellation of consequences is why I recommend to my students to set $s = r^2$. This gives fewer minus signs and since the substituent is monotonically increasing, does not reverse the order of the bounds of integration.
    $endgroup$
    – Eric Towers
    Apr 10 at 4:30














5












5








5





$begingroup$


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 Apr 9 at 23:40









BolboaBolboa

408616




408616








  • 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
    Apr 9 at 23:44












  • $begingroup$
    This constellation of consequences is why I recommend to my students to set $s = r^2$. This gives fewer minus signs and since the substituent is monotonically increasing, does not reverse the order of the bounds of integration.
    $endgroup$
    – Eric Towers
    Apr 10 at 4:30














  • 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
    Apr 9 at 23:44












  • $begingroup$
    This constellation of consequences is why I recommend to my students to set $s = r^2$. This gives fewer minus signs and since the substituent is monotonically increasing, does not reverse the order of the bounds of integration.
    $endgroup$
    – Eric Towers
    Apr 10 at 4:30








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
Apr 9 at 23:44






$begingroup$
$$s = -r^{2} implies ds = -2r dr implies -ds/2 = rdr$$ and as $r to infty$, $s to - infty$ and as $r to 0$, $s to 0$.
$endgroup$
– Mattos
Apr 9 at 23:44














$begingroup$
This constellation of consequences is why I recommend to my students to set $s = r^2$. This gives fewer minus signs and since the substituent is monotonically increasing, does not reverse the order of the bounds of integration.
$endgroup$
– Eric Towers
Apr 10 at 4:30




$begingroup$
This constellation of consequences is why I recommend to my students to set $s = r^2$. This gives fewer minus signs and since the substituent is monotonically increasing, does not reverse the order of the bounds of integration.
$endgroup$
– Eric Towers
Apr 10 at 4:30










2 Answers
2






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8












$begingroup$

$s=-r^{2}$ 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
    Apr 9 at 23:45






  • 1




    $begingroup$
    @JohnDoe Right. I didn't mention it explicitly but that is what I meant.
    $endgroup$
    – Kavi Rama Murthy
    Apr 9 at 23:49



















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
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









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    active

    oldest

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    8












    $begingroup$

    $s=-r^{2}$ 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
      Apr 9 at 23:45






    • 1




      $begingroup$
      @JohnDoe Right. I didn't mention it explicitly but that is what I meant.
      $endgroup$
      – Kavi Rama Murthy
      Apr 9 at 23:49
















    8












    $begingroup$

    $s=-r^{2}$ 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
      Apr 9 at 23:45






    • 1




      $begingroup$
      @JohnDoe Right. I didn't mention it explicitly but that is what I meant.
      $endgroup$
      – Kavi Rama Murthy
      Apr 9 at 23:49














    8












    8








    8





    $begingroup$

    $s=-r^{2}$ 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 Apr 9 at 23:44









    Kavi Rama MurthyKavi Rama Murthy

    74.9k53270




    74.9k53270








    • 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
      Apr 9 at 23:45






    • 1




      $begingroup$
      @JohnDoe Right. I didn't mention it explicitly but that is what I meant.
      $endgroup$
      – Kavi Rama Murthy
      Apr 9 at 23:49














    • 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
      Apr 9 at 23:45






    • 1




      $begingroup$
      @JohnDoe Right. I didn't mention it explicitly but that is what I meant.
      $endgroup$
      – Kavi Rama Murthy
      Apr 9 at 23:49








    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
    Apr 9 at 23:45




    $begingroup$
    It should be noted that the minus sign from the substitution is then used to reverse the order of the limits.
    $endgroup$
    – John Doe
    Apr 9 at 23:45




    1




    1




    $begingroup$
    @JohnDoe Right. I didn't mention it explicitly but that is what I meant.
    $endgroup$
    – Kavi Rama Murthy
    Apr 9 at 23:49




    $begingroup$
    @JohnDoe Right. I didn't mention it explicitly but that is what I meant.
    $endgroup$
    – Kavi Rama Murthy
    Apr 9 at 23:49











    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 Apr 9 at 23:47

























        answered Apr 9 at 23:45









        HAMIDINE SOUMAREHAMIDINE SOUMARE

        2,403215




        2,403215






























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