(Parameter Choice) Using Differentiation under integral for $e^-x^2$ [duplicate]
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Integrating $int^infty_0 e^-x^2,dx$ using Feynman's parametrization trick
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I have the integral:
$$int_-infty^infty e^-x^2 , dx$$
And I’d like to solve it using differentiation under the integral sign. I understand that I must convert $e^-x^2$ to $e^-x^2g(x,t)$, where $g(x,t)$ is just some term involving a new parameter $t$. I tried multiplying by $e^-2xt$, which doesn’t really seem very logical, however, I just do not know what to put. How do I find the appropriate term to make the whole process work?
calculus integration improper-integrals
edited Sep 3 at 14:16
GoodDeeds
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asked Sep 3 at 9:58
Simplex1
183
marked as duplicate by Nosrati, Paul Frost, Adrian Keister, Theoretical Economist, Lord Shark the Unknown Sep 4 at 2:25
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
up vote
0
down vote
favorite
This question already has an answer here:
Integrating $int^infty_0 e^-x^2,dx$ using Feynman's parametrization trick
1 answer
I have the integral:
$$int_-infty^infty e^-x^2 , dx$$
And I’d like to solve it using differentiation under the integral sign. I understand that I must convert $e^-x^2$ to $e^-x^2g(x,t)$, where $g(x,t)$ is just some term involving a new parameter $t$. I tried multiplying by $e^-2xt$, which doesn’t really seem very logical, however, I just do not know what to put. How do I find the appropriate term to make the whole process work?
calculus integration improper-integrals
edited Sep 3 at 14:16
GoodDeeds
10.2k21335
asked Sep 3 at 9:58
Simplex1
183
marked as duplicate by Nosrati, Paul Frost, Adrian Keister, Theoretical Economist, Lord Shark the Unknown Sep 4 at 2:25
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
This Wikipedia page show lot of methods to solve this integral en.wikipedia.org/wiki/Gaussian_integral
– Deepesh Meena
Sep 3 at 14:38
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up vote
0
down vote
favorite
up vote
0
down vote
favorite
This question already has an answer here:
Integrating $int^infty_0 e^-x^2,dx$ using Feynman's parametrization trick
1 answer
I have the integral:
$$int_-infty^infty e^-x^2 , dx$$
And I’d like to solve it using differentiation under the integral sign. I understand that I must convert $e^-x^2$ to $e^-x^2g(x,t)$, where $g(x,t)$ is just some term involving a new parameter $t$. I tried multiplying by $e^-2xt$, which doesn’t really seem very logical, however, I just do not know what to put. How do I find the appropriate term to make the whole process work?
calculus integration improper-integrals
edited Sep 3 at 14:16
GoodDeeds
10.2k21335
asked Sep 3 at 9:58
Simplex1
183
This question already has an answer here:
Integrating $int^infty_0 e^-x^2,dx$ using Feynman's parametrization trick
1 answer
I have the integral:
$$int_-infty^infty e^-x^2 , dx$$
And I’d like to solve it using differentiation under the integral sign. I understand that I must convert $e^-x^2$ to $e^-x^2g(x,t)$, where $g(x,t)$ is just some term involving a new parameter $t$. I tried multiplying by $e^-2xt$, which doesn’t really seem very logical, however, I just do not know what to put. How do I find the appropriate term to make the whole process work?
This question already has an answer here:
Integrating $int^infty_0 e^-x^2,dx$ using Feynman's parametrization trick
1 answer
calculus integration improper-integrals
calculus integration improper-integrals
edited Sep 3 at 14:16
GoodDeeds
10.2k21335
asked Sep 3 at 9:58
Simplex1
183
edited Sep 3 at 14:16
GoodDeeds
10.2k21335
asked Sep 3 at 9:58
Simplex1
183
edited Sep 3 at 14:16
GoodDeeds
10.2k21335
edited Sep 3 at 14:16
GoodDeeds
10.2k21335
edited Sep 3 at 14:16
GoodDeeds
10.2k21335
10.2k21335
asked Sep 3 at 9:58
Simplex1
183
asked Sep 3 at 9:58
Simplex1
183
asked Sep 3 at 9:58
Simplex1
183
183
marked as duplicate by Nosrati, Paul Frost, Adrian Keister, Theoretical Economist, Lord Shark the Unknown Sep 4 at 2:25
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Nosrati, Paul Frost, Adrian Keister, Theoretical Economist, Lord Shark the Unknown Sep 4 at 2:25
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
This Wikipedia page show lot of methods to solve this integral en.wikipedia.org/wiki/Gaussian_integral
– Deepesh Meena
Sep 3 at 14:38
add a comment |Â
This Wikipedia page show lot of methods to solve this integral en.wikipedia.org/wiki/Gaussian_integral
– Deepesh Meena
Sep 3 at 14:38
This Wikipedia page show lot of methods to solve this integral en.wikipedia.org/wiki/Gaussian_integral
– Deepesh Meena
Sep 3 at 14:38
This Wikipedia page show lot of methods to solve this integral en.wikipedia.org/wiki/Gaussian_integral
– Deepesh Meena
Sep 3 at 14:38
add a comment |Â
2 Answers
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The easiest way to compute $int_-infty^inftye^-x²dx$ is to sqaure it and to use Fubini's theorem and polar coordinates:
$$left(int_-infty^inftye^-x²dxright )^2=int_mathbbR^2e^-(x^2+y^2)dlambda=int_0^inftye^-r^22pi rdr=pi$$
where $lambda$ denotes the Lebesgue measure on $mathbbR^2$.
answered Sep 3 at 10:12
Peter Melech
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let $u=x^2$, $du=2xdx$
$$int_-infty^infty e^-x^2 , dx$$
$$int_-infty^infty e^-x^2dx=int_0^inftyu^-frac12e^-udu=Gammaleft(frac12right).$$
Here $Gamma (s)$ is the Gamma function.
$$Gamma(1-s)Gamma(s)=fracpisinpi s$$
$$Gammaleft(frac12right)=sqrtpi.$$
$$int_-infty^infty e^-x^2 , dx=sqrtpi$$
answered Sep 3 at 14:33
Deepesh Meena
3,6482824
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The easiest way to compute $int_-infty^inftye^-x²dx$ is to sqaure it and to use Fubini's theorem and polar coordinates:
$$left(int_-infty^inftye^-x²dxright )^2=int_mathbbR^2e^-(x^2+y^2)dlambda=int_0^inftye^-r^22pi rdr=pi$$
where $lambda$ denotes the Lebesgue measure on $mathbbR^2$.
answered Sep 3 at 10:12
Peter Melech
2,065711
add a comment |Â
up vote
0
down vote
The easiest way to compute $int_-infty^inftye^-x²dx$ is to sqaure it and to use Fubini's theorem and polar coordinates:
$$left(int_-infty^inftye^-x²dxright )^2=int_mathbbR^2e^-(x^2+y^2)dlambda=int_0^inftye^-r^22pi rdr=pi$$
where $lambda$ denotes the Lebesgue measure on $mathbbR^2$.
answered Sep 3 at 10:12
Peter Melech
2,065711
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The easiest way to compute $int_-infty^inftye^-x²dx$ is to sqaure it and to use Fubini's theorem and polar coordinates:
$$left(int_-infty^inftye^-x²dxright )^2=int_mathbbR^2e^-(x^2+y^2)dlambda=int_0^inftye^-r^22pi rdr=pi$$
where $lambda$ denotes the Lebesgue measure on $mathbbR^2$.
answered Sep 3 at 10:12
Peter Melech
2,065711
The easiest way to compute $int_-infty^inftye^-x²dx$ is to sqaure it and to use Fubini's theorem and polar coordinates:
$$left(int_-infty^inftye^-x²dxright )^2=int_mathbbR^2e^-(x^2+y^2)dlambda=int_0^inftye^-r^22pi rdr=pi$$
where $lambda$ denotes the Lebesgue measure on $mathbbR^2$.
answered Sep 3 at 10:12
Peter Melech
2,065711
answered Sep 3 at 10:12
Peter Melech
2,065711
answered Sep 3 at 10:12
Peter Melech
2,065711
answered Sep 3 at 10:12
Peter Melech
2,065711
2,065711
add a comment |Â
add a comment |Â
up vote
0
down vote
let $u=x^2$, $du=2xdx$
$$int_-infty^infty e^-x^2 , dx$$
$$int_-infty^infty e^-x^2dx=int_0^inftyu^-frac12e^-udu=Gammaleft(frac12right).$$
Here $Gamma (s)$ is the Gamma function.
$$Gamma(1-s)Gamma(s)=fracpisinpi s$$
$$Gammaleft(frac12right)=sqrtpi.$$
$$int_-infty^infty e^-x^2 , dx=sqrtpi$$
answered Sep 3 at 14:33
Deepesh Meena
3,6482824
add a comment |Â
up vote
0
down vote
let $u=x^2$, $du=2xdx$
$$int_-infty^infty e^-x^2 , dx$$
$$int_-infty^infty e^-x^2dx=int_0^inftyu^-frac12e^-udu=Gammaleft(frac12right).$$
Here $Gamma (s)$ is the Gamma function.
$$Gamma(1-s)Gamma(s)=fracpisinpi s$$
$$Gammaleft(frac12right)=sqrtpi.$$
$$int_-infty^infty e^-x^2 , dx=sqrtpi$$
answered Sep 3 at 14:33
Deepesh Meena
3,6482824
add a comment |Â
up vote
0
down vote
up vote
0
down vote
let $u=x^2$, $du=2xdx$
$$int_-infty^infty e^-x^2 , dx$$
$$int_-infty^infty e^-x^2dx=int_0^inftyu^-frac12e^-udu=Gammaleft(frac12right).$$
Here $Gamma (s)$ is the Gamma function.
$$Gamma(1-s)Gamma(s)=fracpisinpi s$$
$$Gammaleft(frac12right)=sqrtpi.$$
$$int_-infty^infty e^-x^2 , dx=sqrtpi$$
answered Sep 3 at 14:33
Deepesh Meena
3,6482824
let $u=x^2$, $du=2xdx$
$$int_-infty^infty e^-x^2 , dx$$
$$int_-infty^infty e^-x^2dx=int_0^inftyu^-frac12e^-udu=Gammaleft(frac12right).$$
Here $Gamma (s)$ is the Gamma function.
$$Gamma(1-s)Gamma(s)=fracpisinpi s$$
$$Gammaleft(frac12right)=sqrtpi.$$
$$int_-infty^infty e^-x^2 , dx=sqrtpi$$
answered Sep 3 at 14:33
Deepesh Meena
3,6482824
answered Sep 3 at 14:33
Deepesh Meena
3,6482824
answered Sep 3 at 14:33
Deepesh Meena
3,6482824
answered Sep 3 at 14:33
Deepesh Meena
3,6482824
3,6482824
add a comment |Â
add a comment |Â
This Wikipedia page show lot of methods to solve this integral en.wikipedia.org/wiki/Gaussian_integral
– Deepesh Meena
Sep 3 at 14:38