Integral involving the Associated Laguerre polynomials
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I'm trying to solve this integral
$int_{0}^{infty} L^n_p L^n_{p'} e^{-x} x^{n-1} dx = dfrac{1}{n} dfrac{(p!)^3}{(p-n)!} delta_{pp'}$
I started with integration by parts where
$u = $ $L^n_p L^n_{p'} e^{-x}$
$dv = x^{n-1} $
After some simplifications, I got the following:$\$
$ I = frac{-1}{n} int_{0}^{infty} frac{d}{dx} L^n_p L^n_{p'} e^{-x} x^{n} dx + frac{-1}{n} int_{0}^{infty} frac{d}{dx} L^n_{p'} L^n_p e^{-x} x^{n} dx + frac{1}{n} int_{0}^{infty} L^n_p L^n_{p'} e^{-x} x^{n} dx$
The integral
$frac{1}{n} int_{0}^{infty} L^n_p L^n_{p'} e^{-x} x^{n} dx$
is already known, which is equal to
$dfrac{(p!)^3}{(p-n)!} delta_{pp'}$
after multiply it with $frac{1}{n}$ it becomes
$dfrac{1}{n} dfrac{(p!)^3}{(p-n)!} delta_{pp'} $
All of that was okay, the issue here is with the rest integrals
$frac{-1}{n} int_{0}^{infty} frac{d}{dx} L^n_p L^n_{p'} e^{-x} x^{n} dx + frac{-1}{n} int_{0}^{infty} frac{d}{dx} L^n_{p'} L^n_p e^{-x} x^{n} dx $
I assumed that $p = p'$ so that this integral can be solved, so it becomes:
$frac{-2}{n} int_{0}^{infty} frac{d}{dx} L^n_p L^n_{p} e^{-x} x^{n} dx $
This integral must be equal to zero, so that the final result is $dfrac{1}{n} dfrac{(p!)^3}{(p-n)!} delta_{pp'} $
But I'm not sure with respect to what should I integrate? It's obvious not with respect to p nor n, there's some other substitution, but I have no idea what should that be.
definite-integrals special-functions orthogonal-polynomials
asked Nov 17 at 15:35
Lamyaa Hamad
61
add a comment |
up vote
1
down vote
favorite
I'm trying to solve this integral
$int_{0}^{infty} L^n_p L^n_{p'} e^{-x} x^{n-1} dx = dfrac{1}{n} dfrac{(p!)^3}{(p-n)!} delta_{pp'}$
I started with integration by parts where
$u = $ $L^n_p L^n_{p'} e^{-x}$
$dv = x^{n-1} $
After some simplifications, I got the following:$\$
$ I = frac{-1}{n} int_{0}^{infty} frac{d}{dx} L^n_p L^n_{p'} e^{-x} x^{n} dx + frac{-1}{n} int_{0}^{infty} frac{d}{dx} L^n_{p'} L^n_p e^{-x} x^{n} dx + frac{1}{n} int_{0}^{infty} L^n_p L^n_{p'} e^{-x} x^{n} dx$
The integral
$frac{1}{n} int_{0}^{infty} L^n_p L^n_{p'} e^{-x} x^{n} dx$
is already known, which is equal to
$dfrac{(p!)^3}{(p-n)!} delta_{pp'}$
after multiply it with $frac{1}{n}$ it becomes
$dfrac{1}{n} dfrac{(p!)^3}{(p-n)!} delta_{pp'} $
All of that was okay, the issue here is with the rest integrals
$frac{-1}{n} int_{0}^{infty} frac{d}{dx} L^n_p L^n_{p'} e^{-x} x^{n} dx + frac{-1}{n} int_{0}^{infty} frac{d}{dx} L^n_{p'} L^n_p e^{-x} x^{n} dx $
I assumed that $p = p'$ so that this integral can be solved, so it becomes:
$frac{-2}{n} int_{0}^{infty} frac{d}{dx} L^n_p L^n_{p} e^{-x} x^{n} dx $
This integral must be equal to zero, so that the final result is $dfrac{1}{n} dfrac{(p!)^3}{(p-n)!} delta_{pp'} $
But I'm not sure with respect to what should I integrate? It's obvious not with respect to p nor n, there's some other substitution, but I have no idea what should that be.
definite-integrals special-functions orthogonal-polynomials
asked Nov 17 at 15:35
Lamyaa Hamad
61
$int_0^{infty } L_p^n(x) L_r^n(x) exp (-x) x^{n-1} , dx=left( begin{array}{cc} { & begin{array}{cc} frac{Gamma (n+p+1)}{n p!} & pleq r \ frac{Gamma (n+r+1)}{n r!} & p>r \ end{array} \ end{array} right)$ for:n > 0, p >= 0, r >= 0and $nin mathbb{Z}$,$pin mathbb{Z}$,$rin mathbb{Z}$
– Mariusz Iwaniuk
Nov 17 at 18:02
From where did that assumption comes?
– Lamyaa Hamad
Nov 17 at 18:33
My solution(CAS me helped to solve) is true only with this assumption what to write in the comment above.
– Mariusz Iwaniuk
Nov 17 at 20:10
Well yeah, I know it's true only with this assumption. But what I'm trying to say is what steps were taken to get this conclusion? I didn't understand where did these conditions come from. Kindly, I need more explanation.
– Lamyaa Hamad
Nov 17 at 21:31
No steps was taken,I'm only putt random values to my solution and on the basis of this I have drawn conclusions.
– Mariusz Iwaniuk
Nov 17 at 21:49
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm trying to solve this integral
$int_{0}^{infty} L^n_p L^n_{p'} e^{-x} x^{n-1} dx = dfrac{1}{n} dfrac{(p!)^3}{(p-n)!} delta_{pp'}$
I started with integration by parts where
$u = $ $L^n_p L^n_{p'} e^{-x}$
$dv = x^{n-1} $
After some simplifications, I got the following:$\$
$ I = frac{-1}{n} int_{0}^{infty} frac{d}{dx} L^n_p L^n_{p'} e^{-x} x^{n} dx + frac{-1}{n} int_{0}^{infty} frac{d}{dx} L^n_{p'} L^n_p e^{-x} x^{n} dx + frac{1}{n} int_{0}^{infty} L^n_p L^n_{p'} e^{-x} x^{n} dx$
The integral
$frac{1}{n} int_{0}^{infty} L^n_p L^n_{p'} e^{-x} x^{n} dx$
is already known, which is equal to
$dfrac{(p!)^3}{(p-n)!} delta_{pp'}$
after multiply it with $frac{1}{n}$ it becomes
$dfrac{1}{n} dfrac{(p!)^3}{(p-n)!} delta_{pp'} $
All of that was okay, the issue here is with the rest integrals
$frac{-1}{n} int_{0}^{infty} frac{d}{dx} L^n_p L^n_{p'} e^{-x} x^{n} dx + frac{-1}{n} int_{0}^{infty} frac{d}{dx} L^n_{p'} L^n_p e^{-x} x^{n} dx $
I assumed that $p = p'$ so that this integral can be solved, so it becomes:
$frac{-2}{n} int_{0}^{infty} frac{d}{dx} L^n_p L^n_{p} e^{-x} x^{n} dx $
This integral must be equal to zero, so that the final result is $dfrac{1}{n} dfrac{(p!)^3}{(p-n)!} delta_{pp'} $
But I'm not sure with respect to what should I integrate? It's obvious not with respect to p nor n, there's some other substitution, but I have no idea what should that be.
definite-integrals special-functions orthogonal-polynomials
asked Nov 17 at 15:35
Lamyaa Hamad
61
I'm trying to solve this integral
$int_{0}^{infty} L^n_p L^n_{p'} e^{-x} x^{n-1} dx = dfrac{1}{n} dfrac{(p!)^3}{(p-n)!} delta_{pp'}$
I started with integration by parts where
$u = $ $L^n_p L^n_{p'} e^{-x}$
$dv = x^{n-1} $
After some simplifications, I got the following:$\$
$ I = frac{-1}{n} int_{0}^{infty} frac{d}{dx} L^n_p L^n_{p'} e^{-x} x^{n} dx + frac{-1}{n} int_{0}^{infty} frac{d}{dx} L^n_{p'} L^n_p e^{-x} x^{n} dx + frac{1}{n} int_{0}^{infty} L^n_p L^n_{p'} e^{-x} x^{n} dx$
The integral
$frac{1}{n} int_{0}^{infty} L^n_p L^n_{p'} e^{-x} x^{n} dx$
is already known, which is equal to
$dfrac{(p!)^3}{(p-n)!} delta_{pp'}$
after multiply it with $frac{1}{n}$ it becomes
$dfrac{1}{n} dfrac{(p!)^3}{(p-n)!} delta_{pp'} $
All of that was okay, the issue here is with the rest integrals
$frac{-1}{n} int_{0}^{infty} frac{d}{dx} L^n_p L^n_{p'} e^{-x} x^{n} dx + frac{-1}{n} int_{0}^{infty} frac{d}{dx} L^n_{p'} L^n_p e^{-x} x^{n} dx $
I assumed that $p = p'$ so that this integral can be solved, so it becomes:
$frac{-2}{n} int_{0}^{infty} frac{d}{dx} L^n_p L^n_{p} e^{-x} x^{n} dx $
This integral must be equal to zero, so that the final result is $dfrac{1}{n} dfrac{(p!)^3}{(p-n)!} delta_{pp'} $
But I'm not sure with respect to what should I integrate? It's obvious not with respect to p nor n, there's some other substitution, but I have no idea what should that be.
definite-integrals special-functions orthogonal-polynomials
definite-integrals special-functions orthogonal-polynomials
asked Nov 17 at 15:35
Lamyaa Hamad
61
asked Nov 17 at 15:35
Lamyaa Hamad
61
asked Nov 17 at 15:35
Lamyaa Hamad
61
asked Nov 17 at 15:35
Lamyaa Hamad
61
asked Nov 17 at 15:35
Lamyaa Hamad
61
61
$int_0^{infty } L_p^n(x) L_r^n(x) exp (-x) x^{n-1} , dx=left( begin{array}{cc} { & begin{array}{cc} frac{Gamma (n+p+1)}{n p!} & pleq r \ frac{Gamma (n+r+1)}{n r!} & p>r \ end{array} \ end{array} right)$ for:n > 0, p >= 0, r >= 0and $nin mathbb{Z}$,$pin mathbb{Z}$,$rin mathbb{Z}$
– Mariusz Iwaniuk
Nov 17 at 18:02
From where did that assumption comes?
– Lamyaa Hamad
Nov 17 at 18:33
My solution(CAS me helped to solve) is true only with this assumption what to write in the comment above.
– Mariusz Iwaniuk
Nov 17 at 20:10
Well yeah, I know it's true only with this assumption. But what I'm trying to say is what steps were taken to get this conclusion? I didn't understand where did these conditions come from. Kindly, I need more explanation.
– Lamyaa Hamad
Nov 17 at 21:31
No steps was taken,I'm only putt random values to my solution and on the basis of this I have drawn conclusions.
– Mariusz Iwaniuk
Nov 17 at 21:49
add a comment |
$int_0^{infty } L_p^n(x) L_r^n(x) exp (-x) x^{n-1} , dx=left( begin{array}{cc} { & begin{array}{cc} frac{Gamma (n+p+1)}{n p!} & pleq r \ frac{Gamma (n+r+1)}{n r!} & p>r \ end{array} \ end{array} right)$ for:n > 0, p >= 0, r >= 0and $nin mathbb{Z}$,$pin mathbb{Z}$,$rin mathbb{Z}$
– Mariusz Iwaniuk
Nov 17 at 18:02
From where did that assumption comes?
– Lamyaa Hamad
Nov 17 at 18:33
My solution(CAS me helped to solve) is true only with this assumption what to write in the comment above.
– Mariusz Iwaniuk
Nov 17 at 20:10
Well yeah, I know it's true only with this assumption. But what I'm trying to say is what steps were taken to get this conclusion? I didn't understand where did these conditions come from. Kindly, I need more explanation.
– Lamyaa Hamad
Nov 17 at 21:31
No steps was taken,I'm only putt random values to my solution and on the basis of this I have drawn conclusions.
– Mariusz Iwaniuk
Nov 17 at 21:49
$int_0^{infty } L_p^n(x) L_r^n(x) exp (-x) x^{n-1} , dx=left( begin{array}{cc} { & begin{array}{cc} frac{Gamma (n+p+1)}{n p!} & pleq r \ frac{Gamma (n+r+1)}{n r!} & p>r \ end{array} \ end{array} right)$ for:
n > 0, p >= 0, r >= 0 and $nin mathbb{Z}$,$pin mathbb{Z}$,$rin mathbb{Z}$– Mariusz Iwaniuk
Nov 17 at 18:02
$int_0^{infty } L_p^n(x) L_r^n(x) exp (-x) x^{n-1} , dx=left( begin{array}{cc} { & begin{array}{cc} frac{Gamma (n+p+1)}{n p!} & pleq r \ frac{Gamma (n+r+1)}{n r!} & p>r \ end{array} \ end{array} right)$ for:
n > 0, p >= 0, r >= 0 and $nin mathbb{Z}$,$pin mathbb{Z}$,$rin mathbb{Z}$– Mariusz Iwaniuk
Nov 17 at 18:02
From where did that assumption comes?
– Lamyaa Hamad
Nov 17 at 18:33
From where did that assumption comes?
– Lamyaa Hamad
Nov 17 at 18:33
My solution(CAS me helped to solve) is true only with this assumption what to write in the comment above.
– Mariusz Iwaniuk
Nov 17 at 20:10
My solution(CAS me helped to solve) is true only with this assumption what to write in the comment above.
– Mariusz Iwaniuk
Nov 17 at 20:10
Well yeah, I know it's true only with this assumption. But what I'm trying to say is what steps were taken to get this conclusion? I didn't understand where did these conditions come from. Kindly, I need more explanation.
– Lamyaa Hamad
Nov 17 at 21:31
Well yeah, I know it's true only with this assumption. But what I'm trying to say is what steps were taken to get this conclusion? I didn't understand where did these conditions come from. Kindly, I need more explanation.
– Lamyaa Hamad
Nov 17 at 21:31
No steps was taken,I'm only putt random values to my solution and on the basis of this I have drawn conclusions.
– Mariusz Iwaniuk
Nov 17 at 21:49
No steps was taken,I'm only putt random values to my solution and on the basis of this I have drawn conclusions.
– Mariusz Iwaniuk
Nov 17 at 21:49
add a comment |
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$int_0^{infty } L_p^n(x) L_r^n(x) exp (-x) x^{n-1} , dx=left( begin{array}{cc} { & begin{array}{cc} frac{Gamma (n+p+1)}{n p!} & pleq r \ frac{Gamma (n+r+1)}{n r!} & p>r \ end{array} \ end{array} right)$ for:
n > 0, p >= 0, r >= 0and $nin mathbb{Z}$,$pin mathbb{Z}$,$rin mathbb{Z}$– Mariusz Iwaniuk
Nov 17 at 18:02
From where did that assumption comes?
– Lamyaa Hamad
Nov 17 at 18:33
My solution(CAS me helped to solve) is true only with this assumption what to write in the comment above.
– Mariusz Iwaniuk
Nov 17 at 20:10
Well yeah, I know it's true only with this assumption. But what I'm trying to say is what steps were taken to get this conclusion? I didn't understand where did these conditions come from. Kindly, I need more explanation.
– Lamyaa Hamad
Nov 17 at 21:31
No steps was taken,I'm only putt random values to my solution and on the basis of this I have drawn conclusions.
– Mariusz Iwaniuk
Nov 17 at 21:49