Integral $int_0^infty dp , fracp^5 sin(p x) e^-b p^2p^4 + a^2$: any clever ideas?
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I am trying to solve the following integral, with $a>0,$ $b>0$:
$I equiv int_0^infty dp , fracp^5 sin(p x) e^-b p^2p^4 + a^2 $
By expanding the $sin$, I get
$I = sum_n=1^infty fracx^2n-1(2n-1)! int_0^infty dp , fracp^4+2n e^-b p^2p^4 + a^2 \
= sum_n=1^infty fracx^2n-1(2n-1)!Bigg
frac12 b^frac32-n Gamma (n-3/2) , _1F_2left(1;frac54-fracn2,frac74-fracn2;-frac14 a^2 b^2right)
+frac14 pi a^n-frac32
left[csc left((2 pi n+pi )/4right] cos (a b)
-sec left[ (2 pi n+pi )/4right] sin (a b)right)
Bigg.$
Here $_1F_2$ is a hypergeometric function. The first summation can be done, so that we are left with
$I= fracpi2a left[cos (a b) cos (xsqrta/2) sinh (xsqrta/2)+sin (a b) sin (xsqrta/2) cosh (xsqrta/2)right]
+
sum_n=1^infty fracx^2n-1(2n-1)!
frac12 b^frac32-n Gamma (n-3/2) , _1F_2left(1;frac54-fracn2,frac74-fracn2;-frac14 a^2 b^2right).$
I am unable to find a closed form for the second summation.
Is there an easier way to solve $I$? Presumably one can apply the residue theorem, but I did not find a quick way. Any ideas would be appreciated!
calculus definite-integrals residue-calculus hypergeometric-function
asked Sep 6 at 10:21
Can't integrate
547
add a comment |Â
up vote
1
down vote
favorite
I am trying to solve the following integral, with $a>0,$ $b>0$:
$I equiv int_0^infty dp , fracp^5 sin(p x) e^-b p^2p^4 + a^2 $
By expanding the $sin$, I get
$I = sum_n=1^infty fracx^2n-1(2n-1)! int_0^infty dp , fracp^4+2n e^-b p^2p^4 + a^2 \
= sum_n=1^infty fracx^2n-1(2n-1)!Bigg
frac12 b^frac32-n Gamma (n-3/2) , _1F_2left(1;frac54-fracn2,frac74-fracn2;-frac14 a^2 b^2right)
+frac14 pi a^n-frac32
left[csc left((2 pi n+pi )/4right] cos (a b)
-sec left[ (2 pi n+pi )/4right] sin (a b)right)
Bigg.$
Here $_1F_2$ is a hypergeometric function. The first summation can be done, so that we are left with
$I= fracpi2a left[cos (a b) cos (xsqrta/2) sinh (xsqrta/2)+sin (a b) sin (xsqrta/2) cosh (xsqrta/2)right]
+
sum_n=1^infty fracx^2n-1(2n-1)!
frac12 b^frac32-n Gamma (n-3/2) , _1F_2left(1;frac54-fracn2,frac74-fracn2;-frac14 a^2 b^2right).$
I am unable to find a closed form for the second summation.
Is there an easier way to solve $I$? Presumably one can apply the residue theorem, but I did not find a quick way. Any ideas would be appreciated!
calculus definite-integrals residue-calculus hypergeometric-function
asked Sep 6 at 10:21
Can't integrate
547
4
If I may ask, why is this integral interesting?
– Sobi
Sep 6 at 10:38
1
I'm trying to Fourier invert an expression, and this is the spatial (radially symmetric) part.
– Can't integrate
Sep 6 at 12:12
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am trying to solve the following integral, with $a>0,$ $b>0$:
$I equiv int_0^infty dp , fracp^5 sin(p x) e^-b p^2p^4 + a^2 $
By expanding the $sin$, I get
$I = sum_n=1^infty fracx^2n-1(2n-1)! int_0^infty dp , fracp^4+2n e^-b p^2p^4 + a^2 \
= sum_n=1^infty fracx^2n-1(2n-1)!Bigg
frac12 b^frac32-n Gamma (n-3/2) , _1F_2left(1;frac54-fracn2,frac74-fracn2;-frac14 a^2 b^2right)
+frac14 pi a^n-frac32
left[csc left((2 pi n+pi )/4right] cos (a b)
-sec left[ (2 pi n+pi )/4right] sin (a b)right)
Bigg.$
Here $_1F_2$ is a hypergeometric function. The first summation can be done, so that we are left with
$I= fracpi2a left[cos (a b) cos (xsqrta/2) sinh (xsqrta/2)+sin (a b) sin (xsqrta/2) cosh (xsqrta/2)right]
+
sum_n=1^infty fracx^2n-1(2n-1)!
frac12 b^frac32-n Gamma (n-3/2) , _1F_2left(1;frac54-fracn2,frac74-fracn2;-frac14 a^2 b^2right).$
I am unable to find a closed form for the second summation.
Is there an easier way to solve $I$? Presumably one can apply the residue theorem, but I did not find a quick way. Any ideas would be appreciated!
calculus definite-integrals residue-calculus hypergeometric-function
asked Sep 6 at 10:21
Can't integrate
547
I am trying to solve the following integral, with $a>0,$ $b>0$:
$I equiv int_0^infty dp , fracp^5 sin(p x) e^-b p^2p^4 + a^2 $
By expanding the $sin$, I get
$I = sum_n=1^infty fracx^2n-1(2n-1)! int_0^infty dp , fracp^4+2n e^-b p^2p^4 + a^2 \
= sum_n=1^infty fracx^2n-1(2n-1)!Bigg
frac12 b^frac32-n Gamma (n-3/2) , _1F_2left(1;frac54-fracn2,frac74-fracn2;-frac14 a^2 b^2right)
+frac14 pi a^n-frac32
left[csc left((2 pi n+pi )/4right] cos (a b)
-sec left[ (2 pi n+pi )/4right] sin (a b)right)
Bigg.$
Here $_1F_2$ is a hypergeometric function. The first summation can be done, so that we are left with
$I= fracpi2a left[cos (a b) cos (xsqrta/2) sinh (xsqrta/2)+sin (a b) sin (xsqrta/2) cosh (xsqrta/2)right]
+
sum_n=1^infty fracx^2n-1(2n-1)!
frac12 b^frac32-n Gamma (n-3/2) , _1F_2left(1;frac54-fracn2,frac74-fracn2;-frac14 a^2 b^2right).$
I am unable to find a closed form for the second summation.
Is there an easier way to solve $I$? Presumably one can apply the residue theorem, but I did not find a quick way. Any ideas would be appreciated!
calculus definite-integrals residue-calculus hypergeometric-function
calculus definite-integrals residue-calculus hypergeometric-function
asked Sep 6 at 10:21
Can't integrate
547
asked Sep 6 at 10:21
Can't integrate
547
asked Sep 6 at 10:21
Can't integrate
547
asked Sep 6 at 10:21
Can't integrate
547
asked Sep 6 at 10:21
Can't integrate
547
547
4
If I may ask, why is this integral interesting?
– Sobi
Sep 6 at 10:38
1
I'm trying to Fourier invert an expression, and this is the spatial (radially symmetric) part.
– Can't integrate
Sep 6 at 12:12
add a comment |Â
4
If I may ask, why is this integral interesting?
– Sobi
Sep 6 at 10:38
1
I'm trying to Fourier invert an expression, and this is the spatial (radially symmetric) part.
– Can't integrate
Sep 6 at 12:12
4
4
If I may ask, why is this integral interesting?
– Sobi
Sep 6 at 10:38
If I may ask, why is this integral interesting?
– Sobi
Sep 6 at 10:38
1
1
I'm trying to Fourier invert an expression, and this is the spatial (radially symmetric) part.
– Can't integrate
Sep 6 at 12:12
I'm trying to Fourier invert an expression, and this is the spatial (radially symmetric) part.
– Can't integrate
Sep 6 at 12:12
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
We have
$$frac p^5 e^-b p^2 sin x p p^4 + a^2 =
p e^-b p^2 sin x p left( 1 +
frac i a 2 (p^2 - i a) - frac i a 2 (p^2 + i a) right), \
frac d db left( e^i a b int_0^infty
frac p e^-b p^2 sin x p p^2 - i a dp right) =
-e^i a b int_0^infty p e^-b p^2 sin x p ,dp,$$
and, after some calculations,
$$int_0^infty frac p^5 e^-b p^2 sin x p p^4 + a^2 dp =
F(a) + F(-a) + frac sqrt pi x e^-x^2/(4 b) 4 b^3/2, \
F(a) = frac i pi a e^i a b 8 left(
e^x sqrt i a operatornameerfc frac 2 b sqrti a + x 2 sqrt b -
e^-x sqrt i a operatornameerfc frac 2 b sqrti a - x 2 sqrt b
right).$$
answered Sep 13 at 9:04
Maxim
2,595113
This is great. Thank you so much. What an elegant and creative solution. I've checked it, and it's correct. May I ask how you came up with the idea? I guess partial fraction expansion of the main integral is the key.
– Can't integrate
Sep 14 at 15:02
Right, when there is a polynomial in the exponent and a polynomial in the denominator, differentiation under the integral sign often works, and, since we have $p^2$ in the exponent and $p^4$ in the denominator, that suggests applying partial fraction decomposition.
– Maxim
Sep 15 at 7:03
add a comment |Â
up vote
2
down vote
Amazingly this integral can be solved in terms of well-known functions of mathematical physics. It is easy to see that
$$I(a,b,x):=int_0^infty fracp^5,sin(px)p^4+a^2,e^-b,p^2 dp =frac12 fracddb fracddx underbraceint_-infty^infty fracp^2,cos(px)p^4+a^2,e^-b,p^2 dp_:=J(a,b,x).$$
I won't do the derivatives, but present the formula for $J(a,b,x).$ Let $c=x/sqrt4b.$ Then
$$ J(a,b,x)=fracpi2sqrtb,e^-c^2,ReBig[frac1sqrti,a,b
Big( expbig( (sqrti,a,b - c)^2 big) , texterfcbig(sqrti,a,b - cbig) + $$
$$+ expbig( (sqrti,a,b + c)^2 big) , texterfcbig(sqrti,a,b + cbig) Big)Big] $$
The 'erfc' is the complimentary error function. My proof is long and not rigorous so we'll wait a few days to see if anyone will present a proof. If not, I may return to it. However, I've tested the numerical integration vs. the closed form for a total of 1000 evaluations for 10 different $a,, b, , x.$ The differences were 0 to within machine precision. The tests were over positive $a,, b, , x$ each from 0.1 to 6.5 by increments of 0.6.
answered Sep 6 at 21:30
skbmoore
1,52429
Wow, I've looked at it numerically - and can't find any disagreements! :) Is your reasoning based on the residue theorem, or some other trick?
– Can't integrate
Sep 6 at 22:08
1
Briefly, I came up with another integral relationship that I started to estimate asymptotically. However, I recognized the asymptotic series could be summed in terms of the exp * erfc terms; hence the non-rigor. I'd be happy if someone would attempt a rigorous proof. One of the variables can be eliminated and probably a partial differential equation in two variables can be established that is satisfied by the integral form of J(a,b,x) and the closed form.
– skbmoore
Sep 6 at 22:44
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
We have
$$frac p^5 e^-b p^2 sin x p p^4 + a^2 =
p e^-b p^2 sin x p left( 1 +
frac i a 2 (p^2 - i a) - frac i a 2 (p^2 + i a) right), \
frac d db left( e^i a b int_0^infty
frac p e^-b p^2 sin x p p^2 - i a dp right) =
-e^i a b int_0^infty p e^-b p^2 sin x p ,dp,$$
and, after some calculations,
$$int_0^infty frac p^5 e^-b p^2 sin x p p^4 + a^2 dp =
F(a) + F(-a) + frac sqrt pi x e^-x^2/(4 b) 4 b^3/2, \
F(a) = frac i pi a e^i a b 8 left(
e^x sqrt i a operatornameerfc frac 2 b sqrti a + x 2 sqrt b -
e^-x sqrt i a operatornameerfc frac 2 b sqrti a - x 2 sqrt b
right).$$
answered Sep 13 at 9:04
Maxim
2,595113
This is great. Thank you so much. What an elegant and creative solution. I've checked it, and it's correct. May I ask how you came up with the idea? I guess partial fraction expansion of the main integral is the key.
– Can't integrate
Sep 14 at 15:02
Right, when there is a polynomial in the exponent and a polynomial in the denominator, differentiation under the integral sign often works, and, since we have $p^2$ in the exponent and $p^4$ in the denominator, that suggests applying partial fraction decomposition.
– Maxim
Sep 15 at 7:03
add a comment |Â
up vote
1
down vote
accepted
We have
$$frac p^5 e^-b p^2 sin x p p^4 + a^2 =
p e^-b p^2 sin x p left( 1 +
frac i a 2 (p^2 - i a) - frac i a 2 (p^2 + i a) right), \
frac d db left( e^i a b int_0^infty
frac p e^-b p^2 sin x p p^2 - i a dp right) =
-e^i a b int_0^infty p e^-b p^2 sin x p ,dp,$$
and, after some calculations,
$$int_0^infty frac p^5 e^-b p^2 sin x p p^4 + a^2 dp =
F(a) + F(-a) + frac sqrt pi x e^-x^2/(4 b) 4 b^3/2, \
F(a) = frac i pi a e^i a b 8 left(
e^x sqrt i a operatornameerfc frac 2 b sqrti a + x 2 sqrt b -
e^-x sqrt i a operatornameerfc frac 2 b sqrti a - x 2 sqrt b
right).$$
answered Sep 13 at 9:04
Maxim
2,595113
This is great. Thank you so much. What an elegant and creative solution. I've checked it, and it's correct. May I ask how you came up with the idea? I guess partial fraction expansion of the main integral is the key.
– Can't integrate
Sep 14 at 15:02
Right, when there is a polynomial in the exponent and a polynomial in the denominator, differentiation under the integral sign often works, and, since we have $p^2$ in the exponent and $p^4$ in the denominator, that suggests applying partial fraction decomposition.
– Maxim
Sep 15 at 7:03
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
We have
$$frac p^5 e^-b p^2 sin x p p^4 + a^2 =
p e^-b p^2 sin x p left( 1 +
frac i a 2 (p^2 - i a) - frac i a 2 (p^2 + i a) right), \
frac d db left( e^i a b int_0^infty
frac p e^-b p^2 sin x p p^2 - i a dp right) =
-e^i a b int_0^infty p e^-b p^2 sin x p ,dp,$$
and, after some calculations,
$$int_0^infty frac p^5 e^-b p^2 sin x p p^4 + a^2 dp =
F(a) + F(-a) + frac sqrt pi x e^-x^2/(4 b) 4 b^3/2, \
F(a) = frac i pi a e^i a b 8 left(
e^x sqrt i a operatornameerfc frac 2 b sqrti a + x 2 sqrt b -
e^-x sqrt i a operatornameerfc frac 2 b sqrti a - x 2 sqrt b
right).$$
answered Sep 13 at 9:04
Maxim
2,595113
We have
$$frac p^5 e^-b p^2 sin x p p^4 + a^2 =
p e^-b p^2 sin x p left( 1 +
frac i a 2 (p^2 - i a) - frac i a 2 (p^2 + i a) right), \
frac d db left( e^i a b int_0^infty
frac p e^-b p^2 sin x p p^2 - i a dp right) =
-e^i a b int_0^infty p e^-b p^2 sin x p ,dp,$$
and, after some calculations,
$$int_0^infty frac p^5 e^-b p^2 sin x p p^4 + a^2 dp =
F(a) + F(-a) + frac sqrt pi x e^-x^2/(4 b) 4 b^3/2, \
F(a) = frac i pi a e^i a b 8 left(
e^x sqrt i a operatornameerfc frac 2 b sqrti a + x 2 sqrt b -
e^-x sqrt i a operatornameerfc frac 2 b sqrti a - x 2 sqrt b
right).$$
answered Sep 13 at 9:04
Maxim
2,595113
answered Sep 13 at 9:04
Maxim
2,595113
answered Sep 13 at 9:04
Maxim
2,595113
answered Sep 13 at 9:04
Maxim
2,595113
2,595113
This is great. Thank you so much. What an elegant and creative solution. I've checked it, and it's correct. May I ask how you came up with the idea? I guess partial fraction expansion of the main integral is the key.
– Can't integrate
Sep 14 at 15:02
Right, when there is a polynomial in the exponent and a polynomial in the denominator, differentiation under the integral sign often works, and, since we have $p^2$ in the exponent and $p^4$ in the denominator, that suggests applying partial fraction decomposition.
– Maxim
Sep 15 at 7:03
add a comment |Â
This is great. Thank you so much. What an elegant and creative solution. I've checked it, and it's correct. May I ask how you came up with the idea? I guess partial fraction expansion of the main integral is the key.
– Can't integrate
Sep 14 at 15:02
Right, when there is a polynomial in the exponent and a polynomial in the denominator, differentiation under the integral sign often works, and, since we have $p^2$ in the exponent and $p^4$ in the denominator, that suggests applying partial fraction decomposition.
– Maxim
Sep 15 at 7:03
This is great. Thank you so much. What an elegant and creative solution. I've checked it, and it's correct. May I ask how you came up with the idea? I guess partial fraction expansion of the main integral is the key.
– Can't integrate
Sep 14 at 15:02
This is great. Thank you so much. What an elegant and creative solution. I've checked it, and it's correct. May I ask how you came up with the idea? I guess partial fraction expansion of the main integral is the key.
– Can't integrate
Sep 14 at 15:02
Right, when there is a polynomial in the exponent and a polynomial in the denominator, differentiation under the integral sign often works, and, since we have $p^2$ in the exponent and $p^4$ in the denominator, that suggests applying partial fraction decomposition.
– Maxim
Sep 15 at 7:03
Right, when there is a polynomial in the exponent and a polynomial in the denominator, differentiation under the integral sign often works, and, since we have $p^2$ in the exponent and $p^4$ in the denominator, that suggests applying partial fraction decomposition.
– Maxim
Sep 15 at 7:03
add a comment |Â
up vote
2
down vote
Amazingly this integral can be solved in terms of well-known functions of mathematical physics. It is easy to see that
$$I(a,b,x):=int_0^infty fracp^5,sin(px)p^4+a^2,e^-b,p^2 dp =frac12 fracddb fracddx underbraceint_-infty^infty fracp^2,cos(px)p^4+a^2,e^-b,p^2 dp_:=J(a,b,x).$$
I won't do the derivatives, but present the formula for $J(a,b,x).$ Let $c=x/sqrt4b.$ Then
$$ J(a,b,x)=fracpi2sqrtb,e^-c^2,ReBig[frac1sqrti,a,b
Big( expbig( (sqrti,a,b - c)^2 big) , texterfcbig(sqrti,a,b - cbig) + $$
$$+ expbig( (sqrti,a,b + c)^2 big) , texterfcbig(sqrti,a,b + cbig) Big)Big] $$
The 'erfc' is the complimentary error function. My proof is long and not rigorous so we'll wait a few days to see if anyone will present a proof. If not, I may return to it. However, I've tested the numerical integration vs. the closed form for a total of 1000 evaluations for 10 different $a,, b, , x.$ The differences were 0 to within machine precision. The tests were over positive $a,, b, , x$ each from 0.1 to 6.5 by increments of 0.6.
answered Sep 6 at 21:30
skbmoore
1,52429
Wow, I've looked at it numerically - and can't find any disagreements! :) Is your reasoning based on the residue theorem, or some other trick?
– Can't integrate
Sep 6 at 22:08
1
Briefly, I came up with another integral relationship that I started to estimate asymptotically. However, I recognized the asymptotic series could be summed in terms of the exp * erfc terms; hence the non-rigor. I'd be happy if someone would attempt a rigorous proof. One of the variables can be eliminated and probably a partial differential equation in two variables can be established that is satisfied by the integral form of J(a,b,x) and the closed form.
– skbmoore
Sep 6 at 22:44
add a comment |Â
up vote
2
down vote
Amazingly this integral can be solved in terms of well-known functions of mathematical physics. It is easy to see that
$$I(a,b,x):=int_0^infty fracp^5,sin(px)p^4+a^2,e^-b,p^2 dp =frac12 fracddb fracddx underbraceint_-infty^infty fracp^2,cos(px)p^4+a^2,e^-b,p^2 dp_:=J(a,b,x).$$
I won't do the derivatives, but present the formula for $J(a,b,x).$ Let $c=x/sqrt4b.$ Then
$$ J(a,b,x)=fracpi2sqrtb,e^-c^2,ReBig[frac1sqrti,a,b
Big( expbig( (sqrti,a,b - c)^2 big) , texterfcbig(sqrti,a,b - cbig) + $$
$$+ expbig( (sqrti,a,b + c)^2 big) , texterfcbig(sqrti,a,b + cbig) Big)Big] $$
The 'erfc' is the complimentary error function. My proof is long and not rigorous so we'll wait a few days to see if anyone will present a proof. If not, I may return to it. However, I've tested the numerical integration vs. the closed form for a total of 1000 evaluations for 10 different $a,, b, , x.$ The differences were 0 to within machine precision. The tests were over positive $a,, b, , x$ each from 0.1 to 6.5 by increments of 0.6.
answered Sep 6 at 21:30
skbmoore
1,52429
Wow, I've looked at it numerically - and can't find any disagreements! :) Is your reasoning based on the residue theorem, or some other trick?
– Can't integrate
Sep 6 at 22:08
1
Briefly, I came up with another integral relationship that I started to estimate asymptotically. However, I recognized the asymptotic series could be summed in terms of the exp * erfc terms; hence the non-rigor. I'd be happy if someone would attempt a rigorous proof. One of the variables can be eliminated and probably a partial differential equation in two variables can be established that is satisfied by the integral form of J(a,b,x) and the closed form.
– skbmoore
Sep 6 at 22:44
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Amazingly this integral can be solved in terms of well-known functions of mathematical physics. It is easy to see that
$$I(a,b,x):=int_0^infty fracp^5,sin(px)p^4+a^2,e^-b,p^2 dp =frac12 fracddb fracddx underbraceint_-infty^infty fracp^2,cos(px)p^4+a^2,e^-b,p^2 dp_:=J(a,b,x).$$
I won't do the derivatives, but present the formula for $J(a,b,x).$ Let $c=x/sqrt4b.$ Then
$$ J(a,b,x)=fracpi2sqrtb,e^-c^2,ReBig[frac1sqrti,a,b
Big( expbig( (sqrti,a,b - c)^2 big) , texterfcbig(sqrti,a,b - cbig) + $$
$$+ expbig( (sqrti,a,b + c)^2 big) , texterfcbig(sqrti,a,b + cbig) Big)Big] $$
The 'erfc' is the complimentary error function. My proof is long and not rigorous so we'll wait a few days to see if anyone will present a proof. If not, I may return to it. However, I've tested the numerical integration vs. the closed form for a total of 1000 evaluations for 10 different $a,, b, , x.$ The differences were 0 to within machine precision. The tests were over positive $a,, b, , x$ each from 0.1 to 6.5 by increments of 0.6.
answered Sep 6 at 21:30
skbmoore
1,52429
Amazingly this integral can be solved in terms of well-known functions of mathematical physics. It is easy to see that
$$I(a,b,x):=int_0^infty fracp^5,sin(px)p^4+a^2,e^-b,p^2 dp =frac12 fracddb fracddx underbraceint_-infty^infty fracp^2,cos(px)p^4+a^2,e^-b,p^2 dp_:=J(a,b,x).$$
I won't do the derivatives, but present the formula for $J(a,b,x).$ Let $c=x/sqrt4b.$ Then
$$ J(a,b,x)=fracpi2sqrtb,e^-c^2,ReBig[frac1sqrti,a,b
Big( expbig( (sqrti,a,b - c)^2 big) , texterfcbig(sqrti,a,b - cbig) + $$
$$+ expbig( (sqrti,a,b + c)^2 big) , texterfcbig(sqrti,a,b + cbig) Big)Big] $$
The 'erfc' is the complimentary error function. My proof is long and not rigorous so we'll wait a few days to see if anyone will present a proof. If not, I may return to it. However, I've tested the numerical integration vs. the closed form for a total of 1000 evaluations for 10 different $a,, b, , x.$ The differences were 0 to within machine precision. The tests were over positive $a,, b, , x$ each from 0.1 to 6.5 by increments of 0.6.
answered Sep 6 at 21:30
skbmoore
1,52429
answered Sep 6 at 21:30
skbmoore
1,52429
answered Sep 6 at 21:30
skbmoore
1,52429
answered Sep 6 at 21:30
skbmoore
1,52429
1,52429
Wow, I've looked at it numerically - and can't find any disagreements! :) Is your reasoning based on the residue theorem, or some other trick?
– Can't integrate
Sep 6 at 22:08
1
Briefly, I came up with another integral relationship that I started to estimate asymptotically. However, I recognized the asymptotic series could be summed in terms of the exp * erfc terms; hence the non-rigor. I'd be happy if someone would attempt a rigorous proof. One of the variables can be eliminated and probably a partial differential equation in two variables can be established that is satisfied by the integral form of J(a,b,x) and the closed form.
– skbmoore
Sep 6 at 22:44
add a comment |Â
Wow, I've looked at it numerically - and can't find any disagreements! :) Is your reasoning based on the residue theorem, or some other trick?
– Can't integrate
Sep 6 at 22:08
1
Briefly, I came up with another integral relationship that I started to estimate asymptotically. However, I recognized the asymptotic series could be summed in terms of the exp * erfc terms; hence the non-rigor. I'd be happy if someone would attempt a rigorous proof. One of the variables can be eliminated and probably a partial differential equation in two variables can be established that is satisfied by the integral form of J(a,b,x) and the closed form.
– skbmoore
Sep 6 at 22:44
Wow, I've looked at it numerically - and can't find any disagreements! :) Is your reasoning based on the residue theorem, or some other trick?
– Can't integrate
Sep 6 at 22:08
Wow, I've looked at it numerically - and can't find any disagreements! :) Is your reasoning based on the residue theorem, or some other trick?
– Can't integrate
Sep 6 at 22:08
1
1
Briefly, I came up with another integral relationship that I started to estimate asymptotically. However, I recognized the asymptotic series could be summed in terms of the exp * erfc terms; hence the non-rigor. I'd be happy if someone would attempt a rigorous proof. One of the variables can be eliminated and probably a partial differential equation in two variables can be established that is satisfied by the integral form of J(a,b,x) and the closed form.
– skbmoore
Sep 6 at 22:44
Briefly, I came up with another integral relationship that I started to estimate asymptotically. However, I recognized the asymptotic series could be summed in terms of the exp * erfc terms; hence the non-rigor. I'd be happy if someone would attempt a rigorous proof. One of the variables can be eliminated and probably a partial differential equation in two variables can be established that is satisfied by the integral form of J(a,b,x) and the closed form.
– skbmoore
Sep 6 at 22:44
add a comment |Â
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4
If I may ask, why is this integral interesting?
– Sobi
Sep 6 at 10:38
1
I'm trying to Fourier invert an expression, and this is the spatial (radially symmetric) part.
– Can't integrate
Sep 6 at 12:12