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:
$$int^infty_0 expleft(frac{-x^2}{y^2}-y^2right) dx$$
calculus integration
|
show 1 more comment
up vote
8
down vote
favorite
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
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
|
show 1 more comment
up vote
8
down vote
favorite
up vote
8
down vote
favorite
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
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
calculus integration
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
|
show 1 more comment
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
|
show 1 more comment
2 Answers
2
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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.
add a comment |
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.
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
add a comment |
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.
add a comment |
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.
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.
edited May 23 '15 at 15:10
answered May 22 '15 at 21:48
Mark Viola
129k1273170
129k1273170
add a comment |
add a comment |
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.
add a comment |
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.
add a comment |
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.
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.
edited Nov 16 at 23:34
answered May 22 '15 at 20:02
muaddib
6,90121135
6,90121135
add a comment |
add a comment |
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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