Doubt about a known gaussian integral
$begingroup$
I've seen in books like "Mathematical Handbook of Formulas and tables" by Spiegel et al. (3rd. edition) that
$$
int_0^infty dx e^{-ax^2 - b/x^2} = frac{1}{2}sqrt{frac{pi}{a}}e^{-2sqrt{ab}}
tag1$$
Nevertheless, in order to solve a Physics problem I only get the right solution if I change $sqrt{frac{pi}{a}} leftrightarrow sqrt{frac{pi}{b}}$
Is there any chance of that formula to be an erratum? I've already computed with Wolfram-Apha and gives the same result as the book, but I'm pretty sure of not being wrong with the Physics and previous calculations.
Moreover, this integral appears when you try to compute
$$
I = int_{mathbb{R}^3}d^3p frac{e^{ivec{p}vec{x}}}{(2pi)^3(vec{p}^2 + m^2)} = frac{e^{-mx}}{4pi x}, quad x = |vec{x}|
tag2$$
By using
$$
frac{1}{vec{p}^2 + m^2} = int_0^infty dy e^{-y(m^2 + vec{p}^{ 2})}
tag3$$
So the only way Eq. (2) is right is if Eq. (1) needs the change I suggest or if I'm using it wrongly.
ADDENDUM
Let's see this last thing explicitly:
Introducing Eq. (3) in Eq. (2) renders,
$$
I = int_0^infty dyint_{mathbb{R}^3}frac{d^3p}{(2pi)^3} e^{-ycdot m^2}e^{-yvec{p}^{ 2} + ivec{p}vec{x}}
$$
Now, apply
$$
int_{mathbb{R}^3}frac{d^3p}{(2pi)^{3/2}}e^{-yvec{p}^{ 2} + ivec{p}vec{x}} = frac{1}{sqrt{2y}}e^{-x^2/(4y)}
$$
So there is just one remaining integral in $y$ inside $I$, let's call it $I_y$, that you can solve with Eq. (1) (without my suggestion and using the change of variable $y = alpha^2$):
$$
I_y = int_0^infty dy frac{e^{-ycdot m^2 - x^2/(4y)}}{sqrt{y}} = frac{sqrt{pi}}{m}e^{-mx}
$$
And the problem, the reasion why you don't get Eq. (2) is due to the factor $1/m$ that should be $2/x$ to obtain Eq. (2)
calculus integration multivariable-calculus definite-integrals multiple-integral
asked Dec 23 '18 at 8:23
VickyVicky
1557
$endgroup$
|
show 2 more comments
$begingroup$
I've seen in books like "Mathematical Handbook of Formulas and tables" by Spiegel et al. (3rd. edition) that
$$
int_0^infty dx e^{-ax^2 - b/x^2} = frac{1}{2}sqrt{frac{pi}{a}}e^{-2sqrt{ab}}
tag1$$
Nevertheless, in order to solve a Physics problem I only get the right solution if I change $sqrt{frac{pi}{a}} leftrightarrow sqrt{frac{pi}{b}}$
Is there any chance of that formula to be an erratum? I've already computed with Wolfram-Apha and gives the same result as the book, but I'm pretty sure of not being wrong with the Physics and previous calculations.
Moreover, this integral appears when you try to compute
$$
I = int_{mathbb{R}^3}d^3p frac{e^{ivec{p}vec{x}}}{(2pi)^3(vec{p}^2 + m^2)} = frac{e^{-mx}}{4pi x}, quad x = |vec{x}|
tag2$$
By using
$$
frac{1}{vec{p}^2 + m^2} = int_0^infty dy e^{-y(m^2 + vec{p}^{ 2})}
tag3$$
So the only way Eq. (2) is right is if Eq. (1) needs the change I suggest or if I'm using it wrongly.
ADDENDUM
Let's see this last thing explicitly:
Introducing Eq. (3) in Eq. (2) renders,
$$
I = int_0^infty dyint_{mathbb{R}^3}frac{d^3p}{(2pi)^3} e^{-ycdot m^2}e^{-yvec{p}^{ 2} + ivec{p}vec{x}}
$$
Now, apply
$$
int_{mathbb{R}^3}frac{d^3p}{(2pi)^{3/2}}e^{-yvec{p}^{ 2} + ivec{p}vec{x}} = frac{1}{sqrt{2y}}e^{-x^2/(4y)}
$$
So there is just one remaining integral in $y$ inside $I$, let's call it $I_y$, that you can solve with Eq. (1) (without my suggestion and using the change of variable $y = alpha^2$):
$$
I_y = int_0^infty dy frac{e^{-ycdot m^2 - x^2/(4y)}}{sqrt{y}} = frac{sqrt{pi}}{m}e^{-mx}
$$
And the problem, the reasion why you don't get Eq. (2) is due to the factor $1/m$ that should be $2/x$ to obtain Eq. (2)
calculus integration multivariable-calculus definite-integrals multiple-integral
asked Dec 23 '18 at 8:23
VickyVicky
1557
$endgroup$
$begingroup$
I'm sure the $a$ is in the right place.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 8:32
$begingroup$
Then how do you get Eq. (2) via the last formula (Eq. (3))? I don't obtain it as I explained
$endgroup$
– Vicky
Dec 23 '18 at 8:33
$begingroup$
The formula is correct
$endgroup$
– Claude Leibovici
Dec 23 '18 at 8:40
$begingroup$
@Vicky Maybe Physics is wrong?
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 8:43
$begingroup$
Forget Physics, if you don't mind, could you take a look to my ADDENDUM where I prove the inconsistency?
$endgroup$
– Vicky
Dec 23 '18 at 9:32
|
show 2 more comments
$begingroup$
I've seen in books like "Mathematical Handbook of Formulas and tables" by Spiegel et al. (3rd. edition) that
$$
int_0^infty dx e^{-ax^2 - b/x^2} = frac{1}{2}sqrt{frac{pi}{a}}e^{-2sqrt{ab}}
tag1$$
Nevertheless, in order to solve a Physics problem I only get the right solution if I change $sqrt{frac{pi}{a}} leftrightarrow sqrt{frac{pi}{b}}$
Is there any chance of that formula to be an erratum? I've already computed with Wolfram-Apha and gives the same result as the book, but I'm pretty sure of not being wrong with the Physics and previous calculations.
Moreover, this integral appears when you try to compute
$$
I = int_{mathbb{R}^3}d^3p frac{e^{ivec{p}vec{x}}}{(2pi)^3(vec{p}^2 + m^2)} = frac{e^{-mx}}{4pi x}, quad x = |vec{x}|
tag2$$
By using
$$
frac{1}{vec{p}^2 + m^2} = int_0^infty dy e^{-y(m^2 + vec{p}^{ 2})}
tag3$$
So the only way Eq. (2) is right is if Eq. (1) needs the change I suggest or if I'm using it wrongly.
ADDENDUM
Let's see this last thing explicitly:
Introducing Eq. (3) in Eq. (2) renders,
$$
I = int_0^infty dyint_{mathbb{R}^3}frac{d^3p}{(2pi)^3} e^{-ycdot m^2}e^{-yvec{p}^{ 2} + ivec{p}vec{x}}
$$
Now, apply
$$
int_{mathbb{R}^3}frac{d^3p}{(2pi)^{3/2}}e^{-yvec{p}^{ 2} + ivec{p}vec{x}} = frac{1}{sqrt{2y}}e^{-x^2/(4y)}
$$
So there is just one remaining integral in $y$ inside $I$, let's call it $I_y$, that you can solve with Eq. (1) (without my suggestion and using the change of variable $y = alpha^2$):
$$
I_y = int_0^infty dy frac{e^{-ycdot m^2 - x^2/(4y)}}{sqrt{y}} = frac{sqrt{pi}}{m}e^{-mx}
$$
And the problem, the reasion why you don't get Eq. (2) is due to the factor $1/m$ that should be $2/x$ to obtain Eq. (2)
calculus integration multivariable-calculus definite-integrals multiple-integral
asked Dec 23 '18 at 8:23
VickyVicky
1557
$endgroup$
I've seen in books like "Mathematical Handbook of Formulas and tables" by Spiegel et al. (3rd. edition) that
$$
int_0^infty dx e^{-ax^2 - b/x^2} = frac{1}{2}sqrt{frac{pi}{a}}e^{-2sqrt{ab}}
tag1$$
Nevertheless, in order to solve a Physics problem I only get the right solution if I change $sqrt{frac{pi}{a}} leftrightarrow sqrt{frac{pi}{b}}$
Is there any chance of that formula to be an erratum? I've already computed with Wolfram-Apha and gives the same result as the book, but I'm pretty sure of not being wrong with the Physics and previous calculations.
Moreover, this integral appears when you try to compute
$$
I = int_{mathbb{R}^3}d^3p frac{e^{ivec{p}vec{x}}}{(2pi)^3(vec{p}^2 + m^2)} = frac{e^{-mx}}{4pi x}, quad x = |vec{x}|
tag2$$
By using
$$
frac{1}{vec{p}^2 + m^2} = int_0^infty dy e^{-y(m^2 + vec{p}^{ 2})}
tag3$$
So the only way Eq. (2) is right is if Eq. (1) needs the change I suggest or if I'm using it wrongly.
ADDENDUM
Let's see this last thing explicitly:
Introducing Eq. (3) in Eq. (2) renders,
$$
I = int_0^infty dyint_{mathbb{R}^3}frac{d^3p}{(2pi)^3} e^{-ycdot m^2}e^{-yvec{p}^{ 2} + ivec{p}vec{x}}
$$
Now, apply
$$
int_{mathbb{R}^3}frac{d^3p}{(2pi)^{3/2}}e^{-yvec{p}^{ 2} + ivec{p}vec{x}} = frac{1}{sqrt{2y}}e^{-x^2/(4y)}
$$
So there is just one remaining integral in $y$ inside $I$, let's call it $I_y$, that you can solve with Eq. (1) (without my suggestion and using the change of variable $y = alpha^2$):
$$
I_y = int_0^infty dy frac{e^{-ycdot m^2 - x^2/(4y)}}{sqrt{y}} = frac{sqrt{pi}}{m}e^{-mx}
$$
And the problem, the reasion why you don't get Eq. (2) is due to the factor $1/m$ that should be $2/x$ to obtain Eq. (2)
calculus integration multivariable-calculus definite-integrals multiple-integral
calculus integration multivariable-calculus definite-integrals multiple-integral
asked Dec 23 '18 at 8:23
VickyVicky
1557
asked Dec 23 '18 at 8:23
VickyVicky
1557
edited Dec 23 '18 at 9:19
Vicky
asked Dec 23 '18 at 8:23
VickyVicky
1557
asked Dec 23 '18 at 8:23
VickyVicky
1557
asked Dec 23 '18 at 8:23
VickyVicky
1557
1557
$begingroup$
I'm sure the $a$ is in the right place.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 8:32
$begingroup$
Then how do you get Eq. (2) via the last formula (Eq. (3))? I don't obtain it as I explained
$endgroup$
– Vicky
Dec 23 '18 at 8:33
$begingroup$
The formula is correct
$endgroup$
– Claude Leibovici
Dec 23 '18 at 8:40
$begingroup$
@Vicky Maybe Physics is wrong?
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 8:43
$begingroup$
Forget Physics, if you don't mind, could you take a look to my ADDENDUM where I prove the inconsistency?
$endgroup$
– Vicky
Dec 23 '18 at 9:32
|
show 2 more comments
$begingroup$
I'm sure the $a$ is in the right place.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 8:32
$begingroup$
Then how do you get Eq. (2) via the last formula (Eq. (3))? I don't obtain it as I explained
$endgroup$
– Vicky
Dec 23 '18 at 8:33
$begingroup$
The formula is correct
$endgroup$
– Claude Leibovici
Dec 23 '18 at 8:40
$begingroup$
@Vicky Maybe Physics is wrong?
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 8:43
$begingroup$
Forget Physics, if you don't mind, could you take a look to my ADDENDUM where I prove the inconsistency?
$endgroup$
– Vicky
Dec 23 '18 at 9:32
$begingroup$
I'm sure the $a$ is in the right place.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 8:32
$begingroup$
I'm sure the $a$ is in the right place.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 8:32
$begingroup$
Then how do you get Eq. (2) via the last formula (Eq. (3))? I don't obtain it as I explained
$endgroup$
– Vicky
Dec 23 '18 at 8:33
$begingroup$
Then how do you get Eq. (2) via the last formula (Eq. (3))? I don't obtain it as I explained
$endgroup$
– Vicky
Dec 23 '18 at 8:33
$begingroup$
The formula is correct
$endgroup$
– Claude Leibovici
Dec 23 '18 at 8:40
$begingroup$
The formula is correct
$endgroup$
– Claude Leibovici
Dec 23 '18 at 8:40
$begingroup$
@Vicky Maybe Physics is wrong?
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 8:43
$begingroup$
@Vicky Maybe Physics is wrong?
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 8:43
$begingroup$
Forget Physics, if you don't mind, could you take a look to my ADDENDUM where I prove the inconsistency?
$endgroup$
– Vicky
Dec 23 '18 at 9:32
$begingroup$
Forget Physics, if you don't mind, could you take a look to my ADDENDUM where I prove the inconsistency?
$endgroup$
– Vicky
Dec 23 '18 at 9:32
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
It's convenient to put $a=r^2$ and $b=s^2$ and consider
$$I=e^{2rs}int_0^inftyexp(-r^2x^2-s^2/x^2),dx
=int_0^inftyexp(-(rx-s/x)^2),dx.$$
Letting $x'=s/(rx)$ gives
$$I=frac srint_0^inftyexp(-(s/x'-rx')^2),frac{dx'}{x'^2}$$
and so $$2I=int_0^inftyleft(1+frac{s}{rx^2}right)exp(-(rx-s/x)^2),dx.$$
Now define $y=rx-s/x$, to get
$$2I=frac1rint_{-infty}^infty exp(-y^2),dy=frac{sqrtpi}{r}.$$
answered Dec 23 '18 at 8:42
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
$begingroup$
Ok, I get this demostration of Eq. (1), but then where am I failing in the ADDENDUM (see edition to my post)? Surely it's a naive issue but I don't see it
$endgroup$
– Vicky
Dec 23 '18 at 9:16
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3050161%2fdoubt-about-a-known-gaussian-integral%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It's convenient to put $a=r^2$ and $b=s^2$ and consider
$$I=e^{2rs}int_0^inftyexp(-r^2x^2-s^2/x^2),dx
=int_0^inftyexp(-(rx-s/x)^2),dx.$$
Letting $x'=s/(rx)$ gives
$$I=frac srint_0^inftyexp(-(s/x'-rx')^2),frac{dx'}{x'^2}$$
and so $$2I=int_0^inftyleft(1+frac{s}{rx^2}right)exp(-(rx-s/x)^2),dx.$$
Now define $y=rx-s/x$, to get
$$2I=frac1rint_{-infty}^infty exp(-y^2),dy=frac{sqrtpi}{r}.$$
answered Dec 23 '18 at 8:42
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
$begingroup$
Ok, I get this demostration of Eq. (1), but then where am I failing in the ADDENDUM (see edition to my post)? Surely it's a naive issue but I don't see it
$endgroup$
– Vicky
Dec 23 '18 at 9:16
add a comment |
$begingroup$
It's convenient to put $a=r^2$ and $b=s^2$ and consider
$$I=e^{2rs}int_0^inftyexp(-r^2x^2-s^2/x^2),dx
=int_0^inftyexp(-(rx-s/x)^2),dx.$$
Letting $x'=s/(rx)$ gives
$$I=frac srint_0^inftyexp(-(s/x'-rx')^2),frac{dx'}{x'^2}$$
and so $$2I=int_0^inftyleft(1+frac{s}{rx^2}right)exp(-(rx-s/x)^2),dx.$$
Now define $y=rx-s/x$, to get
$$2I=frac1rint_{-infty}^infty exp(-y^2),dy=frac{sqrtpi}{r}.$$
answered Dec 23 '18 at 8:42
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
$begingroup$
Ok, I get this demostration of Eq. (1), but then where am I failing in the ADDENDUM (see edition to my post)? Surely it's a naive issue but I don't see it
$endgroup$
– Vicky
Dec 23 '18 at 9:16
add a comment |
$begingroup$
It's convenient to put $a=r^2$ and $b=s^2$ and consider
$$I=e^{2rs}int_0^inftyexp(-r^2x^2-s^2/x^2),dx
=int_0^inftyexp(-(rx-s/x)^2),dx.$$
Letting $x'=s/(rx)$ gives
$$I=frac srint_0^inftyexp(-(s/x'-rx')^2),frac{dx'}{x'^2}$$
and so $$2I=int_0^inftyleft(1+frac{s}{rx^2}right)exp(-(rx-s/x)^2),dx.$$
Now define $y=rx-s/x$, to get
$$2I=frac1rint_{-infty}^infty exp(-y^2),dy=frac{sqrtpi}{r}.$$
answered Dec 23 '18 at 8:42
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
It's convenient to put $a=r^2$ and $b=s^2$ and consider
$$I=e^{2rs}int_0^inftyexp(-r^2x^2-s^2/x^2),dx
=int_0^inftyexp(-(rx-s/x)^2),dx.$$
Letting $x'=s/(rx)$ gives
$$I=frac srint_0^inftyexp(-(s/x'-rx')^2),frac{dx'}{x'^2}$$
and so $$2I=int_0^inftyleft(1+frac{s}{rx^2}right)exp(-(rx-s/x)^2),dx.$$
Now define $y=rx-s/x$, to get
$$2I=frac1rint_{-infty}^infty exp(-y^2),dy=frac{sqrtpi}{r}.$$
answered Dec 23 '18 at 8:42
Lord Shark the UnknownLord Shark the Unknown
105k1160133
answered Dec 23 '18 at 8:42
Lord Shark the UnknownLord Shark the Unknown
105k1160133
answered Dec 23 '18 at 8:42
Lord Shark the UnknownLord Shark the Unknown
105k1160133
answered Dec 23 '18 at 8:42
Lord Shark the UnknownLord Shark the Unknown
105k1160133
105k1160133
$begingroup$
Ok, I get this demostration of Eq. (1), but then where am I failing in the ADDENDUM (see edition to my post)? Surely it's a naive issue but I don't see it
$endgroup$
– Vicky
Dec 23 '18 at 9:16
add a comment |
$begingroup$
Ok, I get this demostration of Eq. (1), but then where am I failing in the ADDENDUM (see edition to my post)? Surely it's a naive issue but I don't see it
$endgroup$
– Vicky
Dec 23 '18 at 9:16
$begingroup$
Ok, I get this demostration of Eq. (1), but then where am I failing in the ADDENDUM (see edition to my post)? Surely it's a naive issue but I don't see it
$endgroup$
– Vicky
Dec 23 '18 at 9:16
$begingroup$
Ok, I get this demostration of Eq. (1), but then where am I failing in the ADDENDUM (see edition to my post)? Surely it's a naive issue but I don't see it
$endgroup$
– Vicky
Dec 23 '18 at 9:16
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3050161%2fdoubt-about-a-known-gaussian-integral%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
I'm sure the $a$ is in the right place.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 8:32
$begingroup$
Then how do you get Eq. (2) via the last formula (Eq. (3))? I don't obtain it as I explained
$endgroup$
– Vicky
Dec 23 '18 at 8:33
$begingroup$
The formula is correct
$endgroup$
– Claude Leibovici
Dec 23 '18 at 8:40
$begingroup$
@Vicky Maybe Physics is wrong?
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 8:43
$begingroup$
Forget Physics, if you don't mind, could you take a look to my ADDENDUM where I prove the inconsistency?
$endgroup$
– Vicky
Dec 23 '18 at 9:32