Gaussian integral times polynomial












0












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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


















0












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
















0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 24 '18 at 1:10







Vicky

















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
















  • 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












1 Answer
1






active

oldest

votes


















3












$begingroup$

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



$$ int_{0}^{infty}frac{e^{-m^{2}q-R^{2}/4q}}{q^{d/2}},mathrm{d}q = 2left(frac{2m}{R}right)^{d/2-1}K_{d/2-1}(Rm). $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











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1 Answer
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1 Answer
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active

oldest

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active

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active

oldest

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3












$begingroup$

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



$$ int_{0}^{infty}frac{e^{-m^{2}q-R^{2}/4q}}{q^{d/2}},mathrm{d}q = 2left(frac{2m}{R}right)^{d/2-1}K_{d/2-1}(Rm). $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















3












$begingroup$

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



$$ int_{0}^{infty}frac{e^{-m^{2}q-R^{2}/4q}}{q^{d/2}},mathrm{d}q = 2left(frac{2m}{R}right)^{d/2-1}K_{d/2-1}(Rm). $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














3












3








3





$begingroup$

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



$$ int_{0}^{infty}frac{e^{-m^{2}q-R^{2}/4q}}{q^{d/2}},mathrm{d}q = 2left(frac{2m}{R}right)^{d/2-1}K_{d/2-1}(Rm). $$






share|cite|improve this answer









$endgroup$



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



$$ int_{0}^{infty}frac{e^{-m^{2}q-R^{2}/4q}}{q^{d/2}},mathrm{d}q = 2left(frac{2m}{R}right)^{d/2-1}K_{d/2-1}(Rm). $$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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