Gaussian integral times polynomial
$begingroup$
I'm trying to calculate
$$
I = int_0^infty dq frac{e^{-qm^2 - R^2/(4q)}}{q^{d/2}}, quad R, m geq 0, d geq 1
tag1$$
To do it, I apply the change of variables,
$$
q = alpha^2, quad dq = 2alpha dalpha
tag2$$
This renders,
$$
I = int_0^infty dalpha alpha^{1 - d}e^{-alpha^2m^2 - R^2/(4alpha^2)}
tag3$$
I tryied to solve it and I looked for it in books and even Wolfram-Alpha, but I don't get any solutions. Any suggestions to solve Eq. (3)?
EDITION
Knowing the solution for $d = 0, d = 3$, would it be correct to use them and take derivatives respect to $R^2$ to generate the solutions to the rest $d$'s? For that pair of $d$-values, the solutions depend on $sqrt{R^2} = R$, so would it be right to use
$$
frac{partial}{partial(R^2)} = frac{partial R}{partial(R^2)}frac{partial}{partial R} = Big(frac{partial R^2}{partial R}Big)^{-1}frac{partial}{partial R} = frac{1}{2R}frac{partial}{partial R}
$$
or not? Probably it is a naive question, but I would like to have a second opinion. Actually, this is the same as integrating those results respect to $m^2$.
calculus integration definite-integrals improper-integrals
asked Dec 24 '18 at 0:03
VickyVicky
1457
$endgroup$
add a comment |
$begingroup$
I'm trying to calculate
$$
I = int_0^infty dq frac{e^{-qm^2 - R^2/(4q)}}{q^{d/2}}, quad R, m geq 0, d geq 1
tag1$$
To do it, I apply the change of variables,
$$
q = alpha^2, quad dq = 2alpha dalpha
tag2$$
This renders,
$$
I = int_0^infty dalpha alpha^{1 - d}e^{-alpha^2m^2 - R^2/(4alpha^2)}
tag3$$
I tryied to solve it and I looked for it in books and even Wolfram-Alpha, but I don't get any solutions. Any suggestions to solve Eq. (3)?
EDITION
Knowing the solution for $d = 0, d = 3$, would it be correct to use them and take derivatives respect to $R^2$ to generate the solutions to the rest $d$'s? For that pair of $d$-values, the solutions depend on $sqrt{R^2} = R$, so would it be right to use
$$
frac{partial}{partial(R^2)} = frac{partial R}{partial(R^2)}frac{partial}{partial R} = Big(frac{partial R^2}{partial R}Big)^{-1}frac{partial}{partial R} = frac{1}{2R}frac{partial}{partial R}
$$
or not? Probably it is a naive question, but I would like to have a second opinion. Actually, this is the same as integrating those results respect to $m^2$.
calculus integration definite-integrals improper-integrals
asked Dec 24 '18 at 0:03
VickyVicky
1457
$endgroup$
1
$begingroup$
For $d=1$, use Glasser's matter theorem. Then generalise by integrating $d-1$ times with respect to $m^2$.
$endgroup$
– J.G.
Dec 24 '18 at 0:20
$begingroup$
Matter or Master? By Matter I don't find anything. Could you tell me about the theorem? And thanks for the comment
$endgroup$
– Vicky
Dec 24 '18 at 0:35
$begingroup$
@J.G. What do you say about my edition to the post?
$endgroup$
– Vicky
Dec 24 '18 at 1:02
1
$begingroup$
Sorry, master, yes. Your addition to the question is a valid idea.
$endgroup$
– J.G.
Dec 24 '18 at 7:42
$begingroup$
I think you can use Schwinger's trick.
$endgroup$
– Dinesh Shankar
Dec 24 '18 at 13:34
add a comment |
$begingroup$
I'm trying to calculate
$$
I = int_0^infty dq frac{e^{-qm^2 - R^2/(4q)}}{q^{d/2}}, quad R, m geq 0, d geq 1
tag1$$
To do it, I apply the change of variables,
$$
q = alpha^2, quad dq = 2alpha dalpha
tag2$$
This renders,
$$
I = int_0^infty dalpha alpha^{1 - d}e^{-alpha^2m^2 - R^2/(4alpha^2)}
tag3$$
I tryied to solve it and I looked for it in books and even Wolfram-Alpha, but I don't get any solutions. Any suggestions to solve Eq. (3)?
EDITION
Knowing the solution for $d = 0, d = 3$, would it be correct to use them and take derivatives respect to $R^2$ to generate the solutions to the rest $d$'s? For that pair of $d$-values, the solutions depend on $sqrt{R^2} = R$, so would it be right to use
$$
frac{partial}{partial(R^2)} = frac{partial R}{partial(R^2)}frac{partial}{partial R} = Big(frac{partial R^2}{partial R}Big)^{-1}frac{partial}{partial R} = frac{1}{2R}frac{partial}{partial R}
$$
or not? Probably it is a naive question, but I would like to have a second opinion. Actually, this is the same as integrating those results respect to $m^2$.
calculus integration definite-integrals improper-integrals
asked Dec 24 '18 at 0:03
VickyVicky
1457
$endgroup$
I'm trying to calculate
$$
I = int_0^infty dq frac{e^{-qm^2 - R^2/(4q)}}{q^{d/2}}, quad R, m geq 0, d geq 1
tag1$$
To do it, I apply the change of variables,
$$
q = alpha^2, quad dq = 2alpha dalpha
tag2$$
This renders,
$$
I = int_0^infty dalpha alpha^{1 - d}e^{-alpha^2m^2 - R^2/(4alpha^2)}
tag3$$
I tryied to solve it and I looked for it in books and even Wolfram-Alpha, but I don't get any solutions. Any suggestions to solve Eq. (3)?
EDITION
Knowing the solution for $d = 0, d = 3$, would it be correct to use them and take derivatives respect to $R^2$ to generate the solutions to the rest $d$'s? For that pair of $d$-values, the solutions depend on $sqrt{R^2} = R$, so would it be right to use
$$
frac{partial}{partial(R^2)} = frac{partial R}{partial(R^2)}frac{partial}{partial R} = Big(frac{partial R^2}{partial R}Big)^{-1}frac{partial}{partial R} = frac{1}{2R}frac{partial}{partial R}
$$
or not? Probably it is a naive question, but I would like to have a second opinion. Actually, this is the same as integrating those results respect to $m^2$.
calculus integration definite-integrals improper-integrals
calculus integration definite-integrals improper-integrals
asked Dec 24 '18 at 0:03
VickyVicky
1457
asked Dec 24 '18 at 0:03
VickyVicky
1457
edited Dec 24 '18 at 1:10
Vicky
asked Dec 24 '18 at 0:03
VickyVicky
1457
asked Dec 24 '18 at 0:03
VickyVicky
1457
asked Dec 24 '18 at 0:03
VickyVicky
1457
1457
1
$begingroup$
For $d=1$, use Glasser's matter theorem. Then generalise by integrating $d-1$ times with respect to $m^2$.
$endgroup$
– J.G.
Dec 24 '18 at 0:20
$begingroup$
Matter or Master? By Matter I don't find anything. Could you tell me about the theorem? And thanks for the comment
$endgroup$
– Vicky
Dec 24 '18 at 0:35
$begingroup$
@J.G. What do you say about my edition to the post?
$endgroup$
– Vicky
Dec 24 '18 at 1:02
1
$begingroup$
Sorry, master, yes. Your addition to the question is a valid idea.
$endgroup$
– J.G.
Dec 24 '18 at 7:42
$begingroup$
I think you can use Schwinger's trick.
$endgroup$
– Dinesh Shankar
Dec 24 '18 at 13:34
add a comment |
1
$begingroup$
For $d=1$, use Glasser's matter theorem. Then generalise by integrating $d-1$ times with respect to $m^2$.
$endgroup$
– J.G.
Dec 24 '18 at 0:20
$begingroup$
Matter or Master? By Matter I don't find anything. Could you tell me about the theorem? And thanks for the comment
$endgroup$
– Vicky
Dec 24 '18 at 0:35
$begingroup$
@J.G. What do you say about my edition to the post?
$endgroup$
– Vicky
Dec 24 '18 at 1:02
1
$begingroup$
Sorry, master, yes. Your addition to the question is a valid idea.
$endgroup$
– J.G.
Dec 24 '18 at 7:42
$begingroup$
I think you can use Schwinger's trick.
$endgroup$
– Dinesh Shankar
Dec 24 '18 at 13:34
1
1
$begingroup$
For $d=1$, use Glasser's matter theorem. Then generalise by integrating $d-1$ times with respect to $m^2$.
$endgroup$
– J.G.
Dec 24 '18 at 0:20
$begingroup$
For $d=1$, use Glasser's matter theorem. Then generalise by integrating $d-1$ times with respect to $m^2$.
$endgroup$
– J.G.
Dec 24 '18 at 0:20
$begingroup$
Matter or Master? By Matter I don't find anything. Could you tell me about the theorem? And thanks for the comment
$endgroup$
– Vicky
Dec 24 '18 at 0:35
$begingroup$
Matter or Master? By Matter I don't find anything. Could you tell me about the theorem? And thanks for the comment
$endgroup$
– Vicky
Dec 24 '18 at 0:35
$begingroup$
@J.G. What do you say about my edition to the post?
$endgroup$
– Vicky
Dec 24 '18 at 1:02
$begingroup$
@J.G. What do you say about my edition to the post?
$endgroup$
– Vicky
Dec 24 '18 at 1:02
1
1
$begingroup$
Sorry, master, yes. Your addition to the question is a valid idea.
$endgroup$
– J.G.
Dec 24 '18 at 7:42
$begingroup$
Sorry, master, yes. Your addition to the question is a valid idea.
$endgroup$
– J.G.
Dec 24 '18 at 7:42
$begingroup$
I think you can use Schwinger's trick.
$endgroup$
– Dinesh Shankar
Dec 24 '18 at 13:34
$begingroup$
I think you can use Schwinger's trick.
$endgroup$
– Dinesh Shankar
Dec 24 '18 at 13:34
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It is known that the modified Bessel function of the second kind can be written in an integral form
$$ K_{nu}(z) = frac{z^{nu}}{2^{nu+1}}int_{0}^{infty}frac{e^{-t-z^{2}/4t}}{t^{nu+1}},mathrm{d}t. $$
This post has some more information on its derivation. Identifying $nu = d/2-1$ and $z = Rm$, one can write the integral as
$$ int_{0}^{infty}frac{e^{-m^{2}q-R^{2}/4q}}{q^{d/2}},mathrm{d}q = 2left(frac{2m}{R}right)^{d/2-1}K_{d/2-1}(Rm). $$
answered Dec 25 '18 at 5:15
IninterrompueIninterrompue
67519
$endgroup$
$begingroup$
In this Bessel function, $nu$ can be any number, either positive, negative, integer or non-integer, true?
$endgroup$
– Vicky
Jan 2 at 22:08
1
$begingroup$
Yes, however, there will always be a singularity at $x = 0$, so cases where $R = 0$ and $m = 0$ for your integral need to be handled separately. Both can be written in terms of the gamma function, with additional restrictions on $d$ because the gamma function has singularities at the non-positive integers.
$endgroup$
– Ininterrompue
Jan 2 at 23:02
add a comment |
Your Answer
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1 Answer
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1 Answer
1
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$begingroup$
It is known that the modified Bessel function of the second kind can be written in an integral form
$$ K_{nu}(z) = frac{z^{nu}}{2^{nu+1}}int_{0}^{infty}frac{e^{-t-z^{2}/4t}}{t^{nu+1}},mathrm{d}t. $$
This post has some more information on its derivation. Identifying $nu = d/2-1$ and $z = Rm$, one can write the integral as
$$ int_{0}^{infty}frac{e^{-m^{2}q-R^{2}/4q}}{q^{d/2}},mathrm{d}q = 2left(frac{2m}{R}right)^{d/2-1}K_{d/2-1}(Rm). $$
answered Dec 25 '18 at 5:15
IninterrompueIninterrompue
67519
$endgroup$
$begingroup$
In this Bessel function, $nu$ can be any number, either positive, negative, integer or non-integer, true?
$endgroup$
– Vicky
Jan 2 at 22:08
1
$begingroup$
Yes, however, there will always be a singularity at $x = 0$, so cases where $R = 0$ and $m = 0$ for your integral need to be handled separately. Both can be written in terms of the gamma function, with additional restrictions on $d$ because the gamma function has singularities at the non-positive integers.
$endgroup$
– Ininterrompue
Jan 2 at 23:02
add a comment |
$begingroup$
It is known that the modified Bessel function of the second kind can be written in an integral form
$$ K_{nu}(z) = frac{z^{nu}}{2^{nu+1}}int_{0}^{infty}frac{e^{-t-z^{2}/4t}}{t^{nu+1}},mathrm{d}t. $$
This post has some more information on its derivation. Identifying $nu = d/2-1$ and $z = Rm$, one can write the integral as
$$ int_{0}^{infty}frac{e^{-m^{2}q-R^{2}/4q}}{q^{d/2}},mathrm{d}q = 2left(frac{2m}{R}right)^{d/2-1}K_{d/2-1}(Rm). $$
answered Dec 25 '18 at 5:15
IninterrompueIninterrompue
67519
$endgroup$
$begingroup$
In this Bessel function, $nu$ can be any number, either positive, negative, integer or non-integer, true?
$endgroup$
– Vicky
Jan 2 at 22:08
1
$begingroup$
Yes, however, there will always be a singularity at $x = 0$, so cases where $R = 0$ and $m = 0$ for your integral need to be handled separately. Both can be written in terms of the gamma function, with additional restrictions on $d$ because the gamma function has singularities at the non-positive integers.
$endgroup$
– Ininterrompue
Jan 2 at 23:02
add a comment |
$begingroup$
It is known that the modified Bessel function of the second kind can be written in an integral form
$$ K_{nu}(z) = frac{z^{nu}}{2^{nu+1}}int_{0}^{infty}frac{e^{-t-z^{2}/4t}}{t^{nu+1}},mathrm{d}t. $$
This post has some more information on its derivation. Identifying $nu = d/2-1$ and $z = Rm$, one can write the integral as
$$ int_{0}^{infty}frac{e^{-m^{2}q-R^{2}/4q}}{q^{d/2}},mathrm{d}q = 2left(frac{2m}{R}right)^{d/2-1}K_{d/2-1}(Rm). $$
answered Dec 25 '18 at 5:15
IninterrompueIninterrompue
67519
$endgroup$
It is known that the modified Bessel function of the second kind can be written in an integral form
$$ K_{nu}(z) = frac{z^{nu}}{2^{nu+1}}int_{0}^{infty}frac{e^{-t-z^{2}/4t}}{t^{nu+1}},mathrm{d}t. $$
This post has some more information on its derivation. Identifying $nu = d/2-1$ and $z = Rm$, one can write the integral as
$$ int_{0}^{infty}frac{e^{-m^{2}q-R^{2}/4q}}{q^{d/2}},mathrm{d}q = 2left(frac{2m}{R}right)^{d/2-1}K_{d/2-1}(Rm). $$
answered Dec 25 '18 at 5:15
IninterrompueIninterrompue
67519
answered Dec 25 '18 at 5:15
IninterrompueIninterrompue
67519
answered Dec 25 '18 at 5:15
IninterrompueIninterrompue
67519
answered Dec 25 '18 at 5:15
IninterrompueIninterrompue
67519
67519
$begingroup$
In this Bessel function, $nu$ can be any number, either positive, negative, integer or non-integer, true?
$endgroup$
– Vicky
Jan 2 at 22:08
1
$begingroup$
Yes, however, there will always be a singularity at $x = 0$, so cases where $R = 0$ and $m = 0$ for your integral need to be handled separately. Both can be written in terms of the gamma function, with additional restrictions on $d$ because the gamma function has singularities at the non-positive integers.
$endgroup$
– Ininterrompue
Jan 2 at 23:02
add a comment |
$begingroup$
In this Bessel function, $nu$ can be any number, either positive, negative, integer or non-integer, true?
$endgroup$
– Vicky
Jan 2 at 22:08
1
$begingroup$
Yes, however, there will always be a singularity at $x = 0$, so cases where $R = 0$ and $m = 0$ for your integral need to be handled separately. Both can be written in terms of the gamma function, with additional restrictions on $d$ because the gamma function has singularities at the non-positive integers.
$endgroup$
– Ininterrompue
Jan 2 at 23:02
$begingroup$
In this Bessel function, $nu$ can be any number, either positive, negative, integer or non-integer, true?
$endgroup$
– Vicky
Jan 2 at 22:08
$begingroup$
In this Bessel function, $nu$ can be any number, either positive, negative, integer or non-integer, true?
$endgroup$
– Vicky
Jan 2 at 22:08
1
1
$begingroup$
Yes, however, there will always be a singularity at $x = 0$, so cases where $R = 0$ and $m = 0$ for your integral need to be handled separately. Both can be written in terms of the gamma function, with additional restrictions on $d$ because the gamma function has singularities at the non-positive integers.
$endgroup$
– Ininterrompue
Jan 2 at 23:02
$begingroup$
Yes, however, there will always be a singularity at $x = 0$, so cases where $R = 0$ and $m = 0$ for your integral need to be handled separately. Both can be written in terms of the gamma function, with additional restrictions on $d$ because the gamma function has singularities at the non-positive integers.
$endgroup$
– Ininterrompue
Jan 2 at 23:02
add a comment |
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1
$begingroup$
For $d=1$, use Glasser's matter theorem. Then generalise by integrating $d-1$ times with respect to $m^2$.
$endgroup$
– J.G.
Dec 24 '18 at 0:20
$begingroup$
Matter or Master? By Matter I don't find anything. Could you tell me about the theorem? And thanks for the comment
$endgroup$
– Vicky
Dec 24 '18 at 0:35
$begingroup$
@J.G. What do you say about my edition to the post?
$endgroup$
– Vicky
Dec 24 '18 at 1:02
1
$begingroup$
Sorry, master, yes. Your addition to the question is a valid idea.
$endgroup$
– J.G.
Dec 24 '18 at 7:42
$begingroup$
I think you can use Schwinger's trick.
$endgroup$
– Dinesh Shankar
Dec 24 '18 at 13:34