Feynman technique of integration for $int^infty_0 expleft(frac{-x^2}{y^2}-y^2right) dx$











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I've been learning a technique that Feynman describes in some of his books to integrate. The source can be found here:



http://ocw.mit.edu/courses/mathematics/18-304-undergraduate-seminar-in-discrete-mathematics-spring-2006/projects/integratnfeynman.pdf



The first few examples are integrals in $x$ and $y$ variables, and I can't see a good way to simplify them using differentiation, particularly the example:



$$int^infty_0 expleft(frac{-x^2}{y^2}-y^2right) dx$$










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  • 3




    Take out $exp(-y^2)$ and integrate directly, it is gaussian.
    – Rogelio Molina
    May 22 '15 at 19:01










  • I see another example in there that is just $int_0^infty exp(-x^2)dx$. Would love to see how this technique can be used to provide an alternate derivation.
    – muaddib
    May 22 '15 at 19:12










  • is there any way that it is "dy" instead of"dx"?
    – tired
    May 22 '15 at 19:26










  • If you scroll down to the bottom of the pdf linked it's the first example, unless they made a mistake, it's right.
    – Ryan Joseph
    May 22 '15 at 19:50






  • 1




    Just for note: There is similar, but harder problem, when you have $dy$ instead of $dx$. If you want to solve that problem (where $x$ is parameter, say $x>0$ (but you can do all for $x in mathbb{R}$)), first find $I'(x)$ and then use substitution $y mapsto frac{x}{t}$ for $I(x)$, where $I(x)$ is your integral. You should end with something like $I'(x)+2I(x)=0$, that is $I(x)=Ce^{-2x}$ (you can find that $C=frac{sqrt{pi}}{2}$ from $I(0)=lim_{x to 0^+} I(x)$).
    – Cortizol
    May 22 '15 at 20:19

















up vote
8
down vote

favorite
4












I've been learning a technique that Feynman describes in some of his books to integrate. The source can be found here:



http://ocw.mit.edu/courses/mathematics/18-304-undergraduate-seminar-in-discrete-mathematics-spring-2006/projects/integratnfeynman.pdf



The first few examples are integrals in $x$ and $y$ variables, and I can't see a good way to simplify them using differentiation, particularly the example:



$$int^infty_0 expleft(frac{-x^2}{y^2}-y^2right) dx$$










share|cite|improve this question




















  • 3




    Take out $exp(-y^2)$ and integrate directly, it is gaussian.
    – Rogelio Molina
    May 22 '15 at 19:01










  • I see another example in there that is just $int_0^infty exp(-x^2)dx$. Would love to see how this technique can be used to provide an alternate derivation.
    – muaddib
    May 22 '15 at 19:12










  • is there any way that it is "dy" instead of"dx"?
    – tired
    May 22 '15 at 19:26










  • If you scroll down to the bottom of the pdf linked it's the first example, unless they made a mistake, it's right.
    – Ryan Joseph
    May 22 '15 at 19:50






  • 1




    Just for note: There is similar, but harder problem, when you have $dy$ instead of $dx$. If you want to solve that problem (where $x$ is parameter, say $x>0$ (but you can do all for $x in mathbb{R}$)), first find $I'(x)$ and then use substitution $y mapsto frac{x}{t}$ for $I(x)$, where $I(x)$ is your integral. You should end with something like $I'(x)+2I(x)=0$, that is $I(x)=Ce^{-2x}$ (you can find that $C=frac{sqrt{pi}}{2}$ from $I(0)=lim_{x to 0^+} I(x)$).
    – Cortizol
    May 22 '15 at 20:19















up vote
8
down vote

favorite
4









up vote
8
down vote

favorite
4






4





I've been learning a technique that Feynman describes in some of his books to integrate. The source can be found here:



http://ocw.mit.edu/courses/mathematics/18-304-undergraduate-seminar-in-discrete-mathematics-spring-2006/projects/integratnfeynman.pdf



The first few examples are integrals in $x$ and $y$ variables, and I can't see a good way to simplify them using differentiation, particularly the example:



$$int^infty_0 expleft(frac{-x^2}{y^2}-y^2right) dx$$










share|cite|improve this question















I've been learning a technique that Feynman describes in some of his books to integrate. The source can be found here:



http://ocw.mit.edu/courses/mathematics/18-304-undergraduate-seminar-in-discrete-mathematics-spring-2006/projects/integratnfeynman.pdf



The first few examples are integrals in $x$ and $y$ variables, and I can't see a good way to simplify them using differentiation, particularly the example:



$$int^infty_0 expleft(frac{-x^2}{y^2}-y^2right) dx$$







calculus integration






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share|cite|improve this question













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edited May 22 '15 at 18:59









Zev Chonoles

109k16223420




109k16223420










asked May 22 '15 at 18:56









Ryan Joseph

14018




14018








  • 3




    Take out $exp(-y^2)$ and integrate directly, it is gaussian.
    – Rogelio Molina
    May 22 '15 at 19:01










  • I see another example in there that is just $int_0^infty exp(-x^2)dx$. Would love to see how this technique can be used to provide an alternate derivation.
    – muaddib
    May 22 '15 at 19:12










  • is there any way that it is "dy" instead of"dx"?
    – tired
    May 22 '15 at 19:26










  • If you scroll down to the bottom of the pdf linked it's the first example, unless they made a mistake, it's right.
    – Ryan Joseph
    May 22 '15 at 19:50






  • 1




    Just for note: There is similar, but harder problem, when you have $dy$ instead of $dx$. If you want to solve that problem (where $x$ is parameter, say $x>0$ (but you can do all for $x in mathbb{R}$)), first find $I'(x)$ and then use substitution $y mapsto frac{x}{t}$ for $I(x)$, where $I(x)$ is your integral. You should end with something like $I'(x)+2I(x)=0$, that is $I(x)=Ce^{-2x}$ (you can find that $C=frac{sqrt{pi}}{2}$ from $I(0)=lim_{x to 0^+} I(x)$).
    – Cortizol
    May 22 '15 at 20:19
















  • 3




    Take out $exp(-y^2)$ and integrate directly, it is gaussian.
    – Rogelio Molina
    May 22 '15 at 19:01










  • I see another example in there that is just $int_0^infty exp(-x^2)dx$. Would love to see how this technique can be used to provide an alternate derivation.
    – muaddib
    May 22 '15 at 19:12










  • is there any way that it is "dy" instead of"dx"?
    – tired
    May 22 '15 at 19:26










  • If you scroll down to the bottom of the pdf linked it's the first example, unless they made a mistake, it's right.
    – Ryan Joseph
    May 22 '15 at 19:50






  • 1




    Just for note: There is similar, but harder problem, when you have $dy$ instead of $dx$. If you want to solve that problem (where $x$ is parameter, say $x>0$ (but you can do all for $x in mathbb{R}$)), first find $I'(x)$ and then use substitution $y mapsto frac{x}{t}$ for $I(x)$, where $I(x)$ is your integral. You should end with something like $I'(x)+2I(x)=0$, that is $I(x)=Ce^{-2x}$ (you can find that $C=frac{sqrt{pi}}{2}$ from $I(0)=lim_{x to 0^+} I(x)$).
    – Cortizol
    May 22 '15 at 20:19










3




3




Take out $exp(-y^2)$ and integrate directly, it is gaussian.
– Rogelio Molina
May 22 '15 at 19:01




Take out $exp(-y^2)$ and integrate directly, it is gaussian.
– Rogelio Molina
May 22 '15 at 19:01












I see another example in there that is just $int_0^infty exp(-x^2)dx$. Would love to see how this technique can be used to provide an alternate derivation.
– muaddib
May 22 '15 at 19:12




I see another example in there that is just $int_0^infty exp(-x^2)dx$. Would love to see how this technique can be used to provide an alternate derivation.
– muaddib
May 22 '15 at 19:12












is there any way that it is "dy" instead of"dx"?
– tired
May 22 '15 at 19:26




is there any way that it is "dy" instead of"dx"?
– tired
May 22 '15 at 19:26












If you scroll down to the bottom of the pdf linked it's the first example, unless they made a mistake, it's right.
– Ryan Joseph
May 22 '15 at 19:50




If you scroll down to the bottom of the pdf linked it's the first example, unless they made a mistake, it's right.
– Ryan Joseph
May 22 '15 at 19:50




1




1




Just for note: There is similar, but harder problem, when you have $dy$ instead of $dx$. If you want to solve that problem (where $x$ is parameter, say $x>0$ (but you can do all for $x in mathbb{R}$)), first find $I'(x)$ and then use substitution $y mapsto frac{x}{t}$ for $I(x)$, where $I(x)$ is your integral. You should end with something like $I'(x)+2I(x)=0$, that is $I(x)=Ce^{-2x}$ (you can find that $C=frac{sqrt{pi}}{2}$ from $I(0)=lim_{x to 0^+} I(x)$).
– Cortizol
May 22 '15 at 20:19






Just for note: There is similar, but harder problem, when you have $dy$ instead of $dx$. If you want to solve that problem (where $x$ is parameter, say $x>0$ (but you can do all for $x in mathbb{R}$)), first find $I'(x)$ and then use substitution $y mapsto frac{x}{t}$ for $I(x)$, where $I(x)$ is your integral. You should end with something like $I'(x)+2I(x)=0$, that is $I(x)=Ce^{-2x}$ (you can find that $C=frac{sqrt{pi}}{2}$ from $I(0)=lim_{x to 0^+} I(x)$).
– Cortizol
May 22 '15 at 20:19












2 Answers
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5
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Suppose the integral were $I=int_0^{infty} e^{-y^2-frac{x^2}{y^2}}dy$. Then we note that $y^2+frac{x^2}{y^2}=left(y-frac{|x|}{y}right)^2+2|x|$.



Thus, we have



$$I=e^{-2|x|}int_0^{infty}e^{-left(y-frac{|x|}{y}right)^2}dy tag 1$$



Now, substitute $yto |x|/y$ so that $dyto -|x|dy/y^2$. Then,



$$I=e^{-2|x|}int_0^{infty}frac{|x|}{y^2}e^{-left(y-frac{|x|}{y}right)^2}dy tag 2$$



If we add $(1)$ and (2), we find



$$begin{align}
I&=frac12,e^{-2|x|}int_0^{infty}left(1+frac{|x|}{y^2}right)e^{-left(y-frac{|x|}{y}right)^2}dy \\
&=frac12,e^{-2|x|}int_{-infty}^{infty}e^{-y^2}dy\\
&=e^{-2|x|}frac{sqrt{pi}}{2}
end{align}$$



So, while not quite a "Feynmann" trick, it is an effective way of evaluation.






share|cite|improve this answer






























    up vote
    4
    down vote













    I looked through the paper and see that the "Feynman Technique" is really just a clever application of Leibniz's rule for taking derivatives under integrals. As is pointed out in the comments, the interesting part of this is evaluating $$int_0^infty e^{-x^2} dx$$ using the technique. The paper is a bit misleading for this one since all the other examples added a function of $b$ into the integrand, but that doesn't seem to be the right way to do this one. Instead consider:
    $$I(b) = left(int_0^b e^{-x^2} dxright)^2$$
    Ultimately you want to evaluate that at $b = infty$ and take its square root.



    The full derivation of the result can be found here. That paper discusses this in terms of the Leibniz rule and gives several other interesting derivations of different integrals.






    share|cite|improve this answer























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      2 Answers
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      up vote
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      down vote













      Suppose the integral were $I=int_0^{infty} e^{-y^2-frac{x^2}{y^2}}dy$. Then we note that $y^2+frac{x^2}{y^2}=left(y-frac{|x|}{y}right)^2+2|x|$.



      Thus, we have



      $$I=e^{-2|x|}int_0^{infty}e^{-left(y-frac{|x|}{y}right)^2}dy tag 1$$



      Now, substitute $yto |x|/y$ so that $dyto -|x|dy/y^2$. Then,



      $$I=e^{-2|x|}int_0^{infty}frac{|x|}{y^2}e^{-left(y-frac{|x|}{y}right)^2}dy tag 2$$



      If we add $(1)$ and (2), we find



      $$begin{align}
      I&=frac12,e^{-2|x|}int_0^{infty}left(1+frac{|x|}{y^2}right)e^{-left(y-frac{|x|}{y}right)^2}dy \\
      &=frac12,e^{-2|x|}int_{-infty}^{infty}e^{-y^2}dy\\
      &=e^{-2|x|}frac{sqrt{pi}}{2}
      end{align}$$



      So, while not quite a "Feynmann" trick, it is an effective way of evaluation.






      share|cite|improve this answer



























        up vote
        5
        down vote













        Suppose the integral were $I=int_0^{infty} e^{-y^2-frac{x^2}{y^2}}dy$. Then we note that $y^2+frac{x^2}{y^2}=left(y-frac{|x|}{y}right)^2+2|x|$.



        Thus, we have



        $$I=e^{-2|x|}int_0^{infty}e^{-left(y-frac{|x|}{y}right)^2}dy tag 1$$



        Now, substitute $yto |x|/y$ so that $dyto -|x|dy/y^2$. Then,



        $$I=e^{-2|x|}int_0^{infty}frac{|x|}{y^2}e^{-left(y-frac{|x|}{y}right)^2}dy tag 2$$



        If we add $(1)$ and (2), we find



        $$begin{align}
        I&=frac12,e^{-2|x|}int_0^{infty}left(1+frac{|x|}{y^2}right)e^{-left(y-frac{|x|}{y}right)^2}dy \\
        &=frac12,e^{-2|x|}int_{-infty}^{infty}e^{-y^2}dy\\
        &=e^{-2|x|}frac{sqrt{pi}}{2}
        end{align}$$



        So, while not quite a "Feynmann" trick, it is an effective way of evaluation.






        share|cite|improve this answer

























          up vote
          5
          down vote










          up vote
          5
          down vote









          Suppose the integral were $I=int_0^{infty} e^{-y^2-frac{x^2}{y^2}}dy$. Then we note that $y^2+frac{x^2}{y^2}=left(y-frac{|x|}{y}right)^2+2|x|$.



          Thus, we have



          $$I=e^{-2|x|}int_0^{infty}e^{-left(y-frac{|x|}{y}right)^2}dy tag 1$$



          Now, substitute $yto |x|/y$ so that $dyto -|x|dy/y^2$. Then,



          $$I=e^{-2|x|}int_0^{infty}frac{|x|}{y^2}e^{-left(y-frac{|x|}{y}right)^2}dy tag 2$$



          If we add $(1)$ and (2), we find



          $$begin{align}
          I&=frac12,e^{-2|x|}int_0^{infty}left(1+frac{|x|}{y^2}right)e^{-left(y-frac{|x|}{y}right)^2}dy \\
          &=frac12,e^{-2|x|}int_{-infty}^{infty}e^{-y^2}dy\\
          &=e^{-2|x|}frac{sqrt{pi}}{2}
          end{align}$$



          So, while not quite a "Feynmann" trick, it is an effective way of evaluation.






          share|cite|improve this answer














          Suppose the integral were $I=int_0^{infty} e^{-y^2-frac{x^2}{y^2}}dy$. Then we note that $y^2+frac{x^2}{y^2}=left(y-frac{|x|}{y}right)^2+2|x|$.



          Thus, we have



          $$I=e^{-2|x|}int_0^{infty}e^{-left(y-frac{|x|}{y}right)^2}dy tag 1$$



          Now, substitute $yto |x|/y$ so that $dyto -|x|dy/y^2$. Then,



          $$I=e^{-2|x|}int_0^{infty}frac{|x|}{y^2}e^{-left(y-frac{|x|}{y}right)^2}dy tag 2$$



          If we add $(1)$ and (2), we find



          $$begin{align}
          I&=frac12,e^{-2|x|}int_0^{infty}left(1+frac{|x|}{y^2}right)e^{-left(y-frac{|x|}{y}right)^2}dy \\
          &=frac12,e^{-2|x|}int_{-infty}^{infty}e^{-y^2}dy\\
          &=e^{-2|x|}frac{sqrt{pi}}{2}
          end{align}$$



          So, while not quite a "Feynmann" trick, it is an effective way of evaluation.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited May 23 '15 at 15:10

























          answered May 22 '15 at 21:48









          Mark Viola

          129k1273170




          129k1273170






















              up vote
              4
              down vote













              I looked through the paper and see that the "Feynman Technique" is really just a clever application of Leibniz's rule for taking derivatives under integrals. As is pointed out in the comments, the interesting part of this is evaluating $$int_0^infty e^{-x^2} dx$$ using the technique. The paper is a bit misleading for this one since all the other examples added a function of $b$ into the integrand, but that doesn't seem to be the right way to do this one. Instead consider:
              $$I(b) = left(int_0^b e^{-x^2} dxright)^2$$
              Ultimately you want to evaluate that at $b = infty$ and take its square root.



              The full derivation of the result can be found here. That paper discusses this in terms of the Leibniz rule and gives several other interesting derivations of different integrals.






              share|cite|improve this answer



























                up vote
                4
                down vote













                I looked through the paper and see that the "Feynman Technique" is really just a clever application of Leibniz's rule for taking derivatives under integrals. As is pointed out in the comments, the interesting part of this is evaluating $$int_0^infty e^{-x^2} dx$$ using the technique. The paper is a bit misleading for this one since all the other examples added a function of $b$ into the integrand, but that doesn't seem to be the right way to do this one. Instead consider:
                $$I(b) = left(int_0^b e^{-x^2} dxright)^2$$
                Ultimately you want to evaluate that at $b = infty$ and take its square root.



                The full derivation of the result can be found here. That paper discusses this in terms of the Leibniz rule and gives several other interesting derivations of different integrals.






                share|cite|improve this answer

























                  up vote
                  4
                  down vote










                  up vote
                  4
                  down vote









                  I looked through the paper and see that the "Feynman Technique" is really just a clever application of Leibniz's rule for taking derivatives under integrals. As is pointed out in the comments, the interesting part of this is evaluating $$int_0^infty e^{-x^2} dx$$ using the technique. The paper is a bit misleading for this one since all the other examples added a function of $b$ into the integrand, but that doesn't seem to be the right way to do this one. Instead consider:
                  $$I(b) = left(int_0^b e^{-x^2} dxright)^2$$
                  Ultimately you want to evaluate that at $b = infty$ and take its square root.



                  The full derivation of the result can be found here. That paper discusses this in terms of the Leibniz rule and gives several other interesting derivations of different integrals.






                  share|cite|improve this answer














                  I looked through the paper and see that the "Feynman Technique" is really just a clever application of Leibniz's rule for taking derivatives under integrals. As is pointed out in the comments, the interesting part of this is evaluating $$int_0^infty e^{-x^2} dx$$ using the technique. The paper is a bit misleading for this one since all the other examples added a function of $b$ into the integrand, but that doesn't seem to be the right way to do this one. Instead consider:
                  $$I(b) = left(int_0^b e^{-x^2} dxright)^2$$
                  Ultimately you want to evaluate that at $b = infty$ and take its square root.



                  The full derivation of the result can be found here. That paper discusses this in terms of the Leibniz rule and gives several other interesting derivations of different integrals.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 16 at 23:34

























                  answered May 22 '15 at 20:02









                  muaddib

                  6,90121135




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