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Asymptotics to $f(n) = int_0^1bigl ( frac operatornameli(x)x bigr)^2n + 1 ,(x-1) , dx $
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Consider
$$f(n)=int_0^1Bigl(fracoperatornameli(x)xBigr)^2n + 1,(x-1),dx $$
Where $n$ is a positive integer.
(I know that $f(1) = zeta(3) $ but I already made a Question about proving that)
If I am not mistaken $f(6) = 123 482 $ or about that value. (I assume not exactly that integer!)
So I started to wonder.
How fast does $f(n)$ grow?
What are very good asymptotics for $f(n) $?
I had the idea to investigate $ f ‘ (t) $ (with respect to $t$) and could “ express “ that simply by differentiation under the integral sign.
(Feyman loved that btw)
But then we are still left with integral(s) , I'm not sure how that helps.
Similarly I could make a Taylor series of $f(t)$ but that would still not help me I guess?
Also the behaviour at half-integers is radically different!
How to Find very good asymptotics then !?
calculus real-analysis definite-integrals asymptotics exponentiation
edited Jul 30 at 15:15
Bernard
110k635102
asked Jul 30 at 14:27
mick
4,78311961
add a comment |Â
up vote
0
down vote
favorite
Consider
$$f(n)=int_0^1Bigl(fracoperatornameli(x)xBigr)^2n + 1,(x-1),dx $$
Where $n$ is a positive integer.
(I know that $f(1) = zeta(3) $ but I already made a Question about proving that)
If I am not mistaken $f(6) = 123 482 $ or about that value. (I assume not exactly that integer!)
So I started to wonder.
How fast does $f(n)$ grow?
What are very good asymptotics for $f(n) $?
I had the idea to investigate $ f ‘ (t) $ (with respect to $t$) and could “ express “ that simply by differentiation under the integral sign.
(Feyman loved that btw)
But then we are still left with integral(s) , I'm not sure how that helps.
Similarly I could make a Taylor series of $f(t)$ but that would still not help me I guess?
Also the behaviour at half-integers is radically different!
How to Find very good asymptotics then !?
calculus real-analysis definite-integrals asymptotics exponentiation
edited Jul 30 at 15:15
Bernard
110k635102
asked Jul 30 at 14:27
mick
4,78311961
Slightly related as mentioned in the OP is the zeta(3) question : math.stackexchange.com/questions/2866629/…
– mick
Jul 30 at 14:30
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Consider
$$f(n)=int_0^1Bigl(fracoperatornameli(x)xBigr)^2n + 1,(x-1),dx $$
Where $n$ is a positive integer.
(I know that $f(1) = zeta(3) $ but I already made a Question about proving that)
If I am not mistaken $f(6) = 123 482 $ or about that value. (I assume not exactly that integer!)
So I started to wonder.
How fast does $f(n)$ grow?
What are very good asymptotics for $f(n) $?
I had the idea to investigate $ f ‘ (t) $ (with respect to $t$) and could “ express “ that simply by differentiation under the integral sign.
(Feyman loved that btw)
But then we are still left with integral(s) , I'm not sure how that helps.
Similarly I could make a Taylor series of $f(t)$ but that would still not help me I guess?
Also the behaviour at half-integers is radically different!
How to Find very good asymptotics then !?
calculus real-analysis definite-integrals asymptotics exponentiation
edited Jul 30 at 15:15
Bernard
110k635102
asked Jul 30 at 14:27
mick
4,78311961
Consider
$$f(n)=int_0^1Bigl(fracoperatornameli(x)xBigr)^2n + 1,(x-1),dx $$
Where $n$ is a positive integer.
(I know that $f(1) = zeta(3) $ but I already made a Question about proving that)
If I am not mistaken $f(6) = 123 482 $ or about that value. (I assume not exactly that integer!)
So I started to wonder.
How fast does $f(n)$ grow?
What are very good asymptotics for $f(n) $?
I had the idea to investigate $ f ‘ (t) $ (with respect to $t$) and could “ express “ that simply by differentiation under the integral sign.
(Feyman loved that btw)
But then we are still left with integral(s) , I'm not sure how that helps.
Similarly I could make a Taylor series of $f(t)$ but that would still not help me I guess?
Also the behaviour at half-integers is radically different!
How to Find very good asymptotics then !?
calculus real-analysis definite-integrals asymptotics exponentiation
edited Jul 30 at 15:15
Bernard
110k635102
asked Jul 30 at 14:27
mick
4,78311961
edited Jul 30 at 15:15
Bernard
110k635102
edited Jul 30 at 15:15
Bernard
110k635102
edited Jul 30 at 15:15
Bernard
110k635102
110k635102
asked Jul 30 at 14:27
mick
4,78311961
asked Jul 30 at 14:27
mick
4,78311961
asked Jul 30 at 14:27
mick
4,78311961
4,78311961
Slightly related as mentioned in the OP is the zeta(3) question : math.stackexchange.com/questions/2866629/…
– mick
Jul 30 at 14:30
add a comment |Â
Slightly related as mentioned in the OP is the zeta(3) question : math.stackexchange.com/questions/2866629/…
– mick
Jul 30 at 14:30
Slightly related as mentioned in the OP is the zeta(3) question : math.stackexchange.com/questions/2866629/…
– mick
Jul 30 at 14:30
Slightly related as mentioned in the OP is the zeta(3) question : math.stackexchange.com/questions/2866629/…
– mick
Jul 30 at 14:30
add a comment |Â
1 Answer
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Let
$$phi_n (x) = left(fracoperatornameli(x)xright)^2n+1 (x-1)$$
for $n in mathbbN_0$ and $x in (0,1)$. Note that for large $n$ the function $phi_n$ is extremely small everywhere except for one sharp peak close to $x = 1$ (at about $x approx 1 - mathrme^-(2n+1)$) .
Therefore we can approximate it by its asymptotic form near $x = 1$ if $n$ is large. Using $x sim 1$ and $operatornameli(x) sim ln(1-x) + gamma$ with the Euler-Mascheroni constant $gamma$ as $x nearrow 1$ , we obtain
$$ phi_n (x) sim [ln(1-x) + gamma]^2n+1 (x-1) , , , x nearrow 1 , . $$
Plugging this result into the integral and using the substitution $1 - x = mathrme^-t$ we get
beginalign
f(n) & sim int limits_0^1 [ln(1-x) + gamma]^2n+1 (x-1) , mathrmd x \
&= int limits_0^infty (t-gamma)^2n+1 mathrme^-2 t , mathrmd t \
&= mathrme^-2gamma int limits_-gamma^infty s^2n+1 mathrme^-2 s , mathrmd s \
&sim mathrme^-2gamma fracGamma(2n+2)2^2n+2 \
&= mathrme^-2gamma frac(2n+1)!4^n+1
endalign
as $n to infty$ .
This derivation is not quite rigorous yet, as I did not prove that the corrections to these approximations are asymptotically negligible compared to the above leading term. However, numerical calculations seem to agree with this result and confirm the value of the prefactor.
answered Jul 30 at 16:58
ComplexYetTrivial
2,637624
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
Let
$$phi_n (x) = left(fracoperatornameli(x)xright)^2n+1 (x-1)$$
for $n in mathbbN_0$ and $x in (0,1)$. Note that for large $n$ the function $phi_n$ is extremely small everywhere except for one sharp peak close to $x = 1$ (at about $x approx 1 - mathrme^-(2n+1)$) .
Therefore we can approximate it by its asymptotic form near $x = 1$ if $n$ is large. Using $x sim 1$ and $operatornameli(x) sim ln(1-x) + gamma$ with the Euler-Mascheroni constant $gamma$ as $x nearrow 1$ , we obtain
$$ phi_n (x) sim [ln(1-x) + gamma]^2n+1 (x-1) , , , x nearrow 1 , . $$
Plugging this result into the integral and using the substitution $1 - x = mathrme^-t$ we get
beginalign
f(n) & sim int limits_0^1 [ln(1-x) + gamma]^2n+1 (x-1) , mathrmd x \
&= int limits_0^infty (t-gamma)^2n+1 mathrme^-2 t , mathrmd t \
&= mathrme^-2gamma int limits_-gamma^infty s^2n+1 mathrme^-2 s , mathrmd s \
&sim mathrme^-2gamma fracGamma(2n+2)2^2n+2 \
&= mathrme^-2gamma frac(2n+1)!4^n+1
endalign
as $n to infty$ .
This derivation is not quite rigorous yet, as I did not prove that the corrections to these approximations are asymptotically negligible compared to the above leading term. However, numerical calculations seem to agree with this result and confirm the value of the prefactor.
answered Jul 30 at 16:58
ComplexYetTrivial
2,637624
add a comment |Â
up vote
5
down vote
Let
$$phi_n (x) = left(fracoperatornameli(x)xright)^2n+1 (x-1)$$
for $n in mathbbN_0$ and $x in (0,1)$. Note that for large $n$ the function $phi_n$ is extremely small everywhere except for one sharp peak close to $x = 1$ (at about $x approx 1 - mathrme^-(2n+1)$) .
Therefore we can approximate it by its asymptotic form near $x = 1$ if $n$ is large. Using $x sim 1$ and $operatornameli(x) sim ln(1-x) + gamma$ with the Euler-Mascheroni constant $gamma$ as $x nearrow 1$ , we obtain
$$ phi_n (x) sim [ln(1-x) + gamma]^2n+1 (x-1) , , , x nearrow 1 , . $$
Plugging this result into the integral and using the substitution $1 - x = mathrme^-t$ we get
beginalign
f(n) & sim int limits_0^1 [ln(1-x) + gamma]^2n+1 (x-1) , mathrmd x \
&= int limits_0^infty (t-gamma)^2n+1 mathrme^-2 t , mathrmd t \
&= mathrme^-2gamma int limits_-gamma^infty s^2n+1 mathrme^-2 s , mathrmd s \
&sim mathrme^-2gamma fracGamma(2n+2)2^2n+2 \
&= mathrme^-2gamma frac(2n+1)!4^n+1
endalign
as $n to infty$ .
This derivation is not quite rigorous yet, as I did not prove that the corrections to these approximations are asymptotically negligible compared to the above leading term. However, numerical calculations seem to agree with this result and confirm the value of the prefactor.
answered Jul 30 at 16:58
ComplexYetTrivial
2,637624
add a comment |Â
up vote
5
down vote
up vote
5
down vote
Let
$$phi_n (x) = left(fracoperatornameli(x)xright)^2n+1 (x-1)$$
for $n in mathbbN_0$ and $x in (0,1)$. Note that for large $n$ the function $phi_n$ is extremely small everywhere except for one sharp peak close to $x = 1$ (at about $x approx 1 - mathrme^-(2n+1)$) .
Therefore we can approximate it by its asymptotic form near $x = 1$ if $n$ is large. Using $x sim 1$ and $operatornameli(x) sim ln(1-x) + gamma$ with the Euler-Mascheroni constant $gamma$ as $x nearrow 1$ , we obtain
$$ phi_n (x) sim [ln(1-x) + gamma]^2n+1 (x-1) , , , x nearrow 1 , . $$
Plugging this result into the integral and using the substitution $1 - x = mathrme^-t$ we get
beginalign
f(n) & sim int limits_0^1 [ln(1-x) + gamma]^2n+1 (x-1) , mathrmd x \
&= int limits_0^infty (t-gamma)^2n+1 mathrme^-2 t , mathrmd t \
&= mathrme^-2gamma int limits_-gamma^infty s^2n+1 mathrme^-2 s , mathrmd s \
&sim mathrme^-2gamma fracGamma(2n+2)2^2n+2 \
&= mathrme^-2gamma frac(2n+1)!4^n+1
endalign
as $n to infty$ .
This derivation is not quite rigorous yet, as I did not prove that the corrections to these approximations are asymptotically negligible compared to the above leading term. However, numerical calculations seem to agree with this result and confirm the value of the prefactor.
answered Jul 30 at 16:58
ComplexYetTrivial
2,637624
Let
$$phi_n (x) = left(fracoperatornameli(x)xright)^2n+1 (x-1)$$
for $n in mathbbN_0$ and $x in (0,1)$. Note that for large $n$ the function $phi_n$ is extremely small everywhere except for one sharp peak close to $x = 1$ (at about $x approx 1 - mathrme^-(2n+1)$) .
Therefore we can approximate it by its asymptotic form near $x = 1$ if $n$ is large. Using $x sim 1$ and $operatornameli(x) sim ln(1-x) + gamma$ with the Euler-Mascheroni constant $gamma$ as $x nearrow 1$ , we obtain
$$ phi_n (x) sim [ln(1-x) + gamma]^2n+1 (x-1) , , , x nearrow 1 , . $$
Plugging this result into the integral and using the substitution $1 - x = mathrme^-t$ we get
beginalign
f(n) & sim int limits_0^1 [ln(1-x) + gamma]^2n+1 (x-1) , mathrmd x \
&= int limits_0^infty (t-gamma)^2n+1 mathrme^-2 t , mathrmd t \
&= mathrme^-2gamma int limits_-gamma^infty s^2n+1 mathrme^-2 s , mathrmd s \
&sim mathrme^-2gamma fracGamma(2n+2)2^2n+2 \
&= mathrme^-2gamma frac(2n+1)!4^n+1
endalign
as $n to infty$ .
This derivation is not quite rigorous yet, as I did not prove that the corrections to these approximations are asymptotically negligible compared to the above leading term. However, numerical calculations seem to agree with this result and confirm the value of the prefactor.
answered Jul 30 at 16:58
ComplexYetTrivial
2,637624
answered Jul 30 at 16:58
ComplexYetTrivial
2,637624
answered Jul 30 at 16:58
ComplexYetTrivial
2,637624
answered Jul 30 at 16:58
ComplexYetTrivial
2,637624
2,637624
add a comment |Â
add a comment |Â
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Slightly related as mentioned in the OP is the zeta(3) question : math.stackexchange.com/questions/2866629/…
– mick
Jul 30 at 14:30