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Show that $int_0^1 4 spaceoperatornameli(x)^3 space (x-1) space x^-3 dx = zeta(3) $
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My mentor tommy1729 wrote $int_0^1 4 space operatornameli(x)^3 space (x-1) space x^-3 dx = zeta(3) $
I wanted to prove it thus I looked at some methods for computing integrals and also representations of $zeta(3)$ that might be useful.
But nothing was very helpful to me.
In particular the fact that the RHS is so short - Just Apery’s constant - was surprising.
I expected it longer and more complicated.
So I tend to believe that either the integral computation requires many steps and Then Finally we Get a long expression but alot of cancellation until we are left with Apery’s constant only.
Or There is a simple way to get Apery’s constant directly with a trick I missed.
In either case it is amazing I would say.
So How to show that
$$int_0^1 4 space operatornameli(x)^3 space (x-1) space x^-3 dx = zeta(3) $$
I would like to see different ways to show it.
I assume real analysis methods are simpler than complex analysis methods ( on the complex plane like contour integration ).
I Also wondered If not knowing the RHS in advance would change the difficulty of this question.
Also I wonder about
$$ int_0^1 5 space operatornameli(x)^4 space (x-1) space x^-4 dx = ?? $$
calculus definite-integrals special-functions riemann-zeta conjectures
edited Jul 30 at 18:42
ComplexYetTrivial
2,637624
asked Jul 30 at 3:11
mick
4,78311961
add a comment |Â
up vote
3
down vote
favorite
My mentor tommy1729 wrote $int_0^1 4 space operatornameli(x)^3 space (x-1) space x^-3 dx = zeta(3) $
I wanted to prove it thus I looked at some methods for computing integrals and also representations of $zeta(3)$ that might be useful.
But nothing was very helpful to me.
In particular the fact that the RHS is so short - Just Apery’s constant - was surprising.
I expected it longer and more complicated.
So I tend to believe that either the integral computation requires many steps and Then Finally we Get a long expression but alot of cancellation until we are left with Apery’s constant only.
Or There is a simple way to get Apery’s constant directly with a trick I missed.
In either case it is amazing I would say.
So How to show that
$$int_0^1 4 space operatornameli(x)^3 space (x-1) space x^-3 dx = zeta(3) $$
I would like to see different ways to show it.
I assume real analysis methods are simpler than complex analysis methods ( on the complex plane like contour integration ).
I Also wondered If not knowing the RHS in advance would change the difficulty of this question.
Also I wonder about
$$ int_0^1 5 space operatornameli(x)^4 space (x-1) space x^-4 dx = ?? $$
calculus definite-integrals special-functions riemann-zeta conjectures
edited Jul 30 at 18:42
ComplexYetTrivial
2,637624
asked Jul 30 at 3:11
mick
4,78311961
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
My mentor tommy1729 wrote $int_0^1 4 space operatornameli(x)^3 space (x-1) space x^-3 dx = zeta(3) $
I wanted to prove it thus I looked at some methods for computing integrals and also representations of $zeta(3)$ that might be useful.
But nothing was very helpful to me.
In particular the fact that the RHS is so short - Just Apery’s constant - was surprising.
I expected it longer and more complicated.
So I tend to believe that either the integral computation requires many steps and Then Finally we Get a long expression but alot of cancellation until we are left with Apery’s constant only.
Or There is a simple way to get Apery’s constant directly with a trick I missed.
In either case it is amazing I would say.
So How to show that
$$int_0^1 4 space operatornameli(x)^3 space (x-1) space x^-3 dx = zeta(3) $$
I would like to see different ways to show it.
I assume real analysis methods are simpler than complex analysis methods ( on the complex plane like contour integration ).
I Also wondered If not knowing the RHS in advance would change the difficulty of this question.
Also I wonder about
$$ int_0^1 5 space operatornameli(x)^4 space (x-1) space x^-4 dx = ?? $$
calculus definite-integrals special-functions riemann-zeta conjectures
edited Jul 30 at 18:42
ComplexYetTrivial
2,637624
asked Jul 30 at 3:11
mick
4,78311961
My mentor tommy1729 wrote $int_0^1 4 space operatornameli(x)^3 space (x-1) space x^-3 dx = zeta(3) $
I wanted to prove it thus I looked at some methods for computing integrals and also representations of $zeta(3)$ that might be useful.
But nothing was very helpful to me.
In particular the fact that the RHS is so short - Just Apery’s constant - was surprising.
I expected it longer and more complicated.
So I tend to believe that either the integral computation requires many steps and Then Finally we Get a long expression but alot of cancellation until we are left with Apery’s constant only.
Or There is a simple way to get Apery’s constant directly with a trick I missed.
In either case it is amazing I would say.
So How to show that
$$int_0^1 4 space operatornameli(x)^3 space (x-1) space x^-3 dx = zeta(3) $$
I would like to see different ways to show it.
I assume real analysis methods are simpler than complex analysis methods ( on the complex plane like contour integration ).
I Also wondered If not knowing the RHS in advance would change the difficulty of this question.
Also I wonder about
$$ int_0^1 5 space operatornameli(x)^4 space (x-1) space x^-4 dx = ?? $$
calculus definite-integrals special-functions riemann-zeta conjectures
edited Jul 30 at 18:42
ComplexYetTrivial
2,637624
asked Jul 30 at 3:11
mick
4,78311961
edited Jul 30 at 18:42
ComplexYetTrivial
2,637624
edited Jul 30 at 18:42
ComplexYetTrivial
2,637624
edited Jul 30 at 18:42
ComplexYetTrivial
2,637624
2,637624
asked Jul 30 at 3:11
mick
4,78311961
asked Jul 30 at 3:11
mick
4,78311961
asked Jul 30 at 3:11
mick
4,78311961
4,78311961
add a comment |Â
add a comment |Â
2 Answers
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This solution reduces the expression to an integral that Mathematica knows how to solve. Use a shift in the integrand of Gradshteyn & Rhyzhik 4.351.3 and rewrite the Ei function in terms of li:
$$ frac12 Big( fracli(x)x Big)^2 = int_0^infty x^t log(1+t)fracdt2+t .$$
Use the expansion (B) as given by ComplexYetTrivial. Interchange sums and integral to find
$$int_0^1 Big( fracli(x)x Big)^2 Big( fracli(x)x Big) (x-1) dx= 2int_0^infty dt , fraclog(1+t)2+tsum_k=0^inftyfrac1k+1int_0^1 x^t L_k(-logx)(x-1)dx
$$
Use the well-known evaluation, below, twice
$$int_0^1 x^t L_k(-logx) = t^k/(1+t)^k+1$$
to get
$$int_0^1 Big( fracli(x)x Big)^3(x-1)dx = 2int_0^infty dt , fraclog(1+t)2+t sum_k=0^inftyfrac1k+1Big(frac1t+1 big(fract+1t+2big)^k+1 - frac1t big(fractt+1big)^k+1 Big)$$
The sums have closed-forms in terms of log. Collect and simplify the integrand to get
$$int_0^1 Big( fracli(x)x Big)^3(x-1)dx =
2int_0^infty dt , fraclog(1+t)2+tBig(fraclog(t+1)t
-fraclog(t+2)t+1Big)=fraczeta(3)4
$$
where the single integral has been performed by Mathematica.
answered Jul 30 at 23:40
skbmoore
1,00026
1
Nice answer! Note that by the substitution $1 + t = mathrme^s$ the last integral can be reduced to $$ int_0^infty fracs^2sinh(s) , mathrmd s - 2 int_0^infty fracs^2mathrme^s +1 , mathrmd s - 2 sum_k,n=1^infty frac(-1)^n+kk (n+k)^2 = frac72 zeta(3) - 3 zeta(3) - frac14 zeta(3) , . $$ The value of the double series can be deduced from robjohn's answer to this question, so we don't even need Mathematica.
– ComplexYetTrivial
Jul 31 at 13:21
@ComplexYetTrivial Thanks for completing the proof. ( I have a tendency to rely on Mathematica but also do the integral numerically to make sure the analytic answer pans out.)
– skbmoore
Jul 31 at 17:01
add a comment |Â
up vote
0
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This is not a complete answer, but just a description of two ideas that might help with the evaluation of the integral
$$ I equiv 4 int limits_0^1 left(fracoperatornameli(x)xright)^3 (x-1) , mathrmd x , . $$
They are based on methods that can be applied to find the easier integral
$$ J equiv int limits_0^1 left(fracoperatornameli(x)xright)^2 , mathrmd x , . $$
The first approach relies on integration by parts and the series
$$ x-1 = sum limits_k=1^infty frac1k! ln^k (x) , , , x > 0 , .$$
In order to evaluate $J$ we can use the antiderivative $x mapsto 1-frac1x$ of $x mapsto frac1x^2$ to avoid problems with the singularity of $operatornameli(x)$ at $x = 1$ . We get
beginalign
J &= 2 int limits_0^1 fracoperatornameli(x)x frac1-xln(x) , mathrmd x = - 2 sum limits_k=1^infty frac1k! int limits_0^1 fracoperatornameli(x)x ln^k-1 (x) , mathrmd x\
&= 2 sum limits_k=1^infty frac1k! k int limits_0^1 ln^k-1 (x) , mathrmd x = 2 sum limits_k=1^infty frac1k! k (-1)^k-1 (k-1)! \
&= 2 sum limits_k=1^infty frac(-1)^k-1k^2 = 2 eta (2) = zeta(2) = fracpi^26 , .
endalign
Similarly, we can use the antiderivative $x mapsto frac(x-1)^22 x^2$ of $x mapsto fracx-1x^3$ to find
beginalign
I &= - frac32 int limits_0^1 left(fracoperatornameli(x)xright)^2 frac(x-1)^2ln(x) , mathrmd x \
&= frac32 sum limits_k=0^infty frac1k! int limits_0^1 operatornameli^2 (x) frac1-xx fracln^k-1 (x)x , mathrmd x , . tagA
endalign
We can now integrate by parts once more to obtain at least one term that reduces to a multiple of $zeta(3)$ as in the simpler case. However, I have not managed to solve the remaining integrals yet. We could of course use the series again to express the remaining $1-x$ in terms of logarithm powers, but that does not seem to solve the problem.
The second suggestion employs the Fourier-Laguerre series
$$ operatornameli (x) = - x sum_n=0^infty fracmathrmL_n (-ln(x))n+1 , , , x in (0,1) , , tagB$$
of the logarithmic integral. It can be proved by deriving a recurrence relation for the coefficients from that of the Laguerre polynomials.
Using the substitution $x = mathrme^-t$ and the orthogonality relation of the Laguerre polynomials we immediately obtain
$$ J = sum limits_p=0^infty sum limits_q=0^infty frac1(p+1)(q+1) int limits_0^infty mathrmL_p (t) mathrmL_q (t) mathrme^-t , mathrmd t = sum limits_p=0^infty frac1(n+1)^2 = zeta(2) = fracpi^26 , .$$
Similarly, we have
$$ I = 4sum limits_p=0^infty sum limits_q=0^infty sum limits_r=0^infty frac1(p+1)(q+1)(r+1) int limits_0^infty mathrmL_p (t) mathrmL_q (t) mathrmL_r (t) (1- mathrme^-t) mathrme^-t , mathrmd t , .$$
General formulas for integrals involving three Laguerre polynomials appear to be known (see this paper or this one for a generalisation). I do not know whether they are nice enough to reduce the triple series to a representation of $zeta(3)$ though.
Remark: After doing some numerical calculations I now suspect that the triple series diverges. This is probably due to the fact the original series $(mathrmB)$ only converges in $L^2$, so it cannot be used here. For the simpler integral everything works out though.
It is of course possible to combine the two methods by applying the Laguerre series $(mathrmB)$ in equation $(mathrmA)$. I do not know if these ideas can be used to get the final result, but maybe they can help someone else to find a way.
answered Jul 30 at 18:49
ComplexYetTrivial
2,637624
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
This solution reduces the expression to an integral that Mathematica knows how to solve. Use a shift in the integrand of Gradshteyn & Rhyzhik 4.351.3 and rewrite the Ei function in terms of li:
$$ frac12 Big( fracli(x)x Big)^2 = int_0^infty x^t log(1+t)fracdt2+t .$$
Use the expansion (B) as given by ComplexYetTrivial. Interchange sums and integral to find
$$int_0^1 Big( fracli(x)x Big)^2 Big( fracli(x)x Big) (x-1) dx= 2int_0^infty dt , fraclog(1+t)2+tsum_k=0^inftyfrac1k+1int_0^1 x^t L_k(-logx)(x-1)dx
$$
Use the well-known evaluation, below, twice
$$int_0^1 x^t L_k(-logx) = t^k/(1+t)^k+1$$
to get
$$int_0^1 Big( fracli(x)x Big)^3(x-1)dx = 2int_0^infty dt , fraclog(1+t)2+t sum_k=0^inftyfrac1k+1Big(frac1t+1 big(fract+1t+2big)^k+1 - frac1t big(fractt+1big)^k+1 Big)$$
The sums have closed-forms in terms of log. Collect and simplify the integrand to get
$$int_0^1 Big( fracli(x)x Big)^3(x-1)dx =
2int_0^infty dt , fraclog(1+t)2+tBig(fraclog(t+1)t
-fraclog(t+2)t+1Big)=fraczeta(3)4
$$
where the single integral has been performed by Mathematica.
answered Jul 30 at 23:40
skbmoore
1,00026
1
Nice answer! Note that by the substitution $1 + t = mathrme^s$ the last integral can be reduced to $$ int_0^infty fracs^2sinh(s) , mathrmd s - 2 int_0^infty fracs^2mathrme^s +1 , mathrmd s - 2 sum_k,n=1^infty frac(-1)^n+kk (n+k)^2 = frac72 zeta(3) - 3 zeta(3) - frac14 zeta(3) , . $$ The value of the double series can be deduced from robjohn's answer to this question, so we don't even need Mathematica.
– ComplexYetTrivial
Jul 31 at 13:21
@ComplexYetTrivial Thanks for completing the proof. ( I have a tendency to rely on Mathematica but also do the integral numerically to make sure the analytic answer pans out.)
– skbmoore
Jul 31 at 17:01
add a comment |Â
up vote
3
down vote
This solution reduces the expression to an integral that Mathematica knows how to solve. Use a shift in the integrand of Gradshteyn & Rhyzhik 4.351.3 and rewrite the Ei function in terms of li:
$$ frac12 Big( fracli(x)x Big)^2 = int_0^infty x^t log(1+t)fracdt2+t .$$
Use the expansion (B) as given by ComplexYetTrivial. Interchange sums and integral to find
$$int_0^1 Big( fracli(x)x Big)^2 Big( fracli(x)x Big) (x-1) dx= 2int_0^infty dt , fraclog(1+t)2+tsum_k=0^inftyfrac1k+1int_0^1 x^t L_k(-logx)(x-1)dx
$$
Use the well-known evaluation, below, twice
$$int_0^1 x^t L_k(-logx) = t^k/(1+t)^k+1$$
to get
$$int_0^1 Big( fracli(x)x Big)^3(x-1)dx = 2int_0^infty dt , fraclog(1+t)2+t sum_k=0^inftyfrac1k+1Big(frac1t+1 big(fract+1t+2big)^k+1 - frac1t big(fractt+1big)^k+1 Big)$$
The sums have closed-forms in terms of log. Collect and simplify the integrand to get
$$int_0^1 Big( fracli(x)x Big)^3(x-1)dx =
2int_0^infty dt , fraclog(1+t)2+tBig(fraclog(t+1)t
-fraclog(t+2)t+1Big)=fraczeta(3)4
$$
where the single integral has been performed by Mathematica.
answered Jul 30 at 23:40
skbmoore
1,00026
1
Nice answer! Note that by the substitution $1 + t = mathrme^s$ the last integral can be reduced to $$ int_0^infty fracs^2sinh(s) , mathrmd s - 2 int_0^infty fracs^2mathrme^s +1 , mathrmd s - 2 sum_k,n=1^infty frac(-1)^n+kk (n+k)^2 = frac72 zeta(3) - 3 zeta(3) - frac14 zeta(3) , . $$ The value of the double series can be deduced from robjohn's answer to this question, so we don't even need Mathematica.
– ComplexYetTrivial
Jul 31 at 13:21
@ComplexYetTrivial Thanks for completing the proof. ( I have a tendency to rely on Mathematica but also do the integral numerically to make sure the analytic answer pans out.)
– skbmoore
Jul 31 at 17:01
add a comment |Â
up vote
3
down vote
up vote
3
down vote
This solution reduces the expression to an integral that Mathematica knows how to solve. Use a shift in the integrand of Gradshteyn & Rhyzhik 4.351.3 and rewrite the Ei function in terms of li:
$$ frac12 Big( fracli(x)x Big)^2 = int_0^infty x^t log(1+t)fracdt2+t .$$
Use the expansion (B) as given by ComplexYetTrivial. Interchange sums and integral to find
$$int_0^1 Big( fracli(x)x Big)^2 Big( fracli(x)x Big) (x-1) dx= 2int_0^infty dt , fraclog(1+t)2+tsum_k=0^inftyfrac1k+1int_0^1 x^t L_k(-logx)(x-1)dx
$$
Use the well-known evaluation, below, twice
$$int_0^1 x^t L_k(-logx) = t^k/(1+t)^k+1$$
to get
$$int_0^1 Big( fracli(x)x Big)^3(x-1)dx = 2int_0^infty dt , fraclog(1+t)2+t sum_k=0^inftyfrac1k+1Big(frac1t+1 big(fract+1t+2big)^k+1 - frac1t big(fractt+1big)^k+1 Big)$$
The sums have closed-forms in terms of log. Collect and simplify the integrand to get
$$int_0^1 Big( fracli(x)x Big)^3(x-1)dx =
2int_0^infty dt , fraclog(1+t)2+tBig(fraclog(t+1)t
-fraclog(t+2)t+1Big)=fraczeta(3)4
$$
where the single integral has been performed by Mathematica.
answered Jul 30 at 23:40
skbmoore
1,00026
This solution reduces the expression to an integral that Mathematica knows how to solve. Use a shift in the integrand of Gradshteyn & Rhyzhik 4.351.3 and rewrite the Ei function in terms of li:
$$ frac12 Big( fracli(x)x Big)^2 = int_0^infty x^t log(1+t)fracdt2+t .$$
Use the expansion (B) as given by ComplexYetTrivial. Interchange sums and integral to find
$$int_0^1 Big( fracli(x)x Big)^2 Big( fracli(x)x Big) (x-1) dx= 2int_0^infty dt , fraclog(1+t)2+tsum_k=0^inftyfrac1k+1int_0^1 x^t L_k(-logx)(x-1)dx
$$
Use the well-known evaluation, below, twice
$$int_0^1 x^t L_k(-logx) = t^k/(1+t)^k+1$$
to get
$$int_0^1 Big( fracli(x)x Big)^3(x-1)dx = 2int_0^infty dt , fraclog(1+t)2+t sum_k=0^inftyfrac1k+1Big(frac1t+1 big(fract+1t+2big)^k+1 - frac1t big(fractt+1big)^k+1 Big)$$
The sums have closed-forms in terms of log. Collect and simplify the integrand to get
$$int_0^1 Big( fracli(x)x Big)^3(x-1)dx =
2int_0^infty dt , fraclog(1+t)2+tBig(fraclog(t+1)t
-fraclog(t+2)t+1Big)=fraczeta(3)4
$$
where the single integral has been performed by Mathematica.
answered Jul 30 at 23:40
skbmoore
1,00026
answered Jul 30 at 23:40
skbmoore
1,00026
answered Jul 30 at 23:40
skbmoore
1,00026
answered Jul 30 at 23:40
skbmoore
1,00026
1,00026
1
Nice answer! Note that by the substitution $1 + t = mathrme^s$ the last integral can be reduced to $$ int_0^infty fracs^2sinh(s) , mathrmd s - 2 int_0^infty fracs^2mathrme^s +1 , mathrmd s - 2 sum_k,n=1^infty frac(-1)^n+kk (n+k)^2 = frac72 zeta(3) - 3 zeta(3) - frac14 zeta(3) , . $$ The value of the double series can be deduced from robjohn's answer to this question, so we don't even need Mathematica.
– ComplexYetTrivial
Jul 31 at 13:21
@ComplexYetTrivial Thanks for completing the proof. ( I have a tendency to rely on Mathematica but also do the integral numerically to make sure the analytic answer pans out.)
– skbmoore
Jul 31 at 17:01
add a comment |Â
1
Nice answer! Note that by the substitution $1 + t = mathrme^s$ the last integral can be reduced to $$ int_0^infty fracs^2sinh(s) , mathrmd s - 2 int_0^infty fracs^2mathrme^s +1 , mathrmd s - 2 sum_k,n=1^infty frac(-1)^n+kk (n+k)^2 = frac72 zeta(3) - 3 zeta(3) - frac14 zeta(3) , . $$ The value of the double series can be deduced from robjohn's answer to this question, so we don't even need Mathematica.
– ComplexYetTrivial
Jul 31 at 13:21
@ComplexYetTrivial Thanks for completing the proof. ( I have a tendency to rely on Mathematica but also do the integral numerically to make sure the analytic answer pans out.)
– skbmoore
Jul 31 at 17:01
1
1
Nice answer! Note that by the substitution $1 + t = mathrme^s$ the last integral can be reduced to $$ int_0^infty fracs^2sinh(s) , mathrmd s - 2 int_0^infty fracs^2mathrme^s +1 , mathrmd s - 2 sum_k,n=1^infty frac(-1)^n+kk (n+k)^2 = frac72 zeta(3) - 3 zeta(3) - frac14 zeta(3) , . $$ The value of the double series can be deduced from robjohn's answer to this question, so we don't even need Mathematica.
– ComplexYetTrivial
Jul 31 at 13:21
Nice answer! Note that by the substitution $1 + t = mathrme^s$ the last integral can be reduced to $$ int_0^infty fracs^2sinh(s) , mathrmd s - 2 int_0^infty fracs^2mathrme^s +1 , mathrmd s - 2 sum_k,n=1^infty frac(-1)^n+kk (n+k)^2 = frac72 zeta(3) - 3 zeta(3) - frac14 zeta(3) , . $$ The value of the double series can be deduced from robjohn's answer to this question, so we don't even need Mathematica.
– ComplexYetTrivial
Jul 31 at 13:21
@ComplexYetTrivial Thanks for completing the proof. ( I have a tendency to rely on Mathematica but also do the integral numerically to make sure the analytic answer pans out.)
– skbmoore
Jul 31 at 17:01
@ComplexYetTrivial Thanks for completing the proof. ( I have a tendency to rely on Mathematica but also do the integral numerically to make sure the analytic answer pans out.)
– skbmoore
Jul 31 at 17:01
add a comment |Â
up vote
0
down vote
This is not a complete answer, but just a description of two ideas that might help with the evaluation of the integral
$$ I equiv 4 int limits_0^1 left(fracoperatornameli(x)xright)^3 (x-1) , mathrmd x , . $$
They are based on methods that can be applied to find the easier integral
$$ J equiv int limits_0^1 left(fracoperatornameli(x)xright)^2 , mathrmd x , . $$
The first approach relies on integration by parts and the series
$$ x-1 = sum limits_k=1^infty frac1k! ln^k (x) , , , x > 0 , .$$
In order to evaluate $J$ we can use the antiderivative $x mapsto 1-frac1x$ of $x mapsto frac1x^2$ to avoid problems with the singularity of $operatornameli(x)$ at $x = 1$ . We get
beginalign
J &= 2 int limits_0^1 fracoperatornameli(x)x frac1-xln(x) , mathrmd x = - 2 sum limits_k=1^infty frac1k! int limits_0^1 fracoperatornameli(x)x ln^k-1 (x) , mathrmd x\
&= 2 sum limits_k=1^infty frac1k! k int limits_0^1 ln^k-1 (x) , mathrmd x = 2 sum limits_k=1^infty frac1k! k (-1)^k-1 (k-1)! \
&= 2 sum limits_k=1^infty frac(-1)^k-1k^2 = 2 eta (2) = zeta(2) = fracpi^26 , .
endalign
Similarly, we can use the antiderivative $x mapsto frac(x-1)^22 x^2$ of $x mapsto fracx-1x^3$ to find
beginalign
I &= - frac32 int limits_0^1 left(fracoperatornameli(x)xright)^2 frac(x-1)^2ln(x) , mathrmd x \
&= frac32 sum limits_k=0^infty frac1k! int limits_0^1 operatornameli^2 (x) frac1-xx fracln^k-1 (x)x , mathrmd x , . tagA
endalign
We can now integrate by parts once more to obtain at least one term that reduces to a multiple of $zeta(3)$ as in the simpler case. However, I have not managed to solve the remaining integrals yet. We could of course use the series again to express the remaining $1-x$ in terms of logarithm powers, but that does not seem to solve the problem.
The second suggestion employs the Fourier-Laguerre series
$$ operatornameli (x) = - x sum_n=0^infty fracmathrmL_n (-ln(x))n+1 , , , x in (0,1) , , tagB$$
of the logarithmic integral. It can be proved by deriving a recurrence relation for the coefficients from that of the Laguerre polynomials.
Using the substitution $x = mathrme^-t$ and the orthogonality relation of the Laguerre polynomials we immediately obtain
$$ J = sum limits_p=0^infty sum limits_q=0^infty frac1(p+1)(q+1) int limits_0^infty mathrmL_p (t) mathrmL_q (t) mathrme^-t , mathrmd t = sum limits_p=0^infty frac1(n+1)^2 = zeta(2) = fracpi^26 , .$$
Similarly, we have
$$ I = 4sum limits_p=0^infty sum limits_q=0^infty sum limits_r=0^infty frac1(p+1)(q+1)(r+1) int limits_0^infty mathrmL_p (t) mathrmL_q (t) mathrmL_r (t) (1- mathrme^-t) mathrme^-t , mathrmd t , .$$
General formulas for integrals involving three Laguerre polynomials appear to be known (see this paper or this one for a generalisation). I do not know whether they are nice enough to reduce the triple series to a representation of $zeta(3)$ though.
Remark: After doing some numerical calculations I now suspect that the triple series diverges. This is probably due to the fact the original series $(mathrmB)$ only converges in $L^2$, so it cannot be used here. For the simpler integral everything works out though.
It is of course possible to combine the two methods by applying the Laguerre series $(mathrmB)$ in equation $(mathrmA)$. I do not know if these ideas can be used to get the final result, but maybe they can help someone else to find a way.
answered Jul 30 at 18:49
ComplexYetTrivial
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This is not a complete answer, but just a description of two ideas that might help with the evaluation of the integral
$$ I equiv 4 int limits_0^1 left(fracoperatornameli(x)xright)^3 (x-1) , mathrmd x , . $$
They are based on methods that can be applied to find the easier integral
$$ J equiv int limits_0^1 left(fracoperatornameli(x)xright)^2 , mathrmd x , . $$
The first approach relies on integration by parts and the series
$$ x-1 = sum limits_k=1^infty frac1k! ln^k (x) , , , x > 0 , .$$
In order to evaluate $J$ we can use the antiderivative $x mapsto 1-frac1x$ of $x mapsto frac1x^2$ to avoid problems with the singularity of $operatornameli(x)$ at $x = 1$ . We get
beginalign
J &= 2 int limits_0^1 fracoperatornameli(x)x frac1-xln(x) , mathrmd x = - 2 sum limits_k=1^infty frac1k! int limits_0^1 fracoperatornameli(x)x ln^k-1 (x) , mathrmd x\
&= 2 sum limits_k=1^infty frac1k! k int limits_0^1 ln^k-1 (x) , mathrmd x = 2 sum limits_k=1^infty frac1k! k (-1)^k-1 (k-1)! \
&= 2 sum limits_k=1^infty frac(-1)^k-1k^2 = 2 eta (2) = zeta(2) = fracpi^26 , .
endalign
Similarly, we can use the antiderivative $x mapsto frac(x-1)^22 x^2$ of $x mapsto fracx-1x^3$ to find
beginalign
I &= - frac32 int limits_0^1 left(fracoperatornameli(x)xright)^2 frac(x-1)^2ln(x) , mathrmd x \
&= frac32 sum limits_k=0^infty frac1k! int limits_0^1 operatornameli^2 (x) frac1-xx fracln^k-1 (x)x , mathrmd x , . tagA
endalign
We can now integrate by parts once more to obtain at least one term that reduces to a multiple of $zeta(3)$ as in the simpler case. However, I have not managed to solve the remaining integrals yet. We could of course use the series again to express the remaining $1-x$ in terms of logarithm powers, but that does not seem to solve the problem.
The second suggestion employs the Fourier-Laguerre series
$$ operatornameli (x) = - x sum_n=0^infty fracmathrmL_n (-ln(x))n+1 , , , x in (0,1) , , tagB$$
of the logarithmic integral. It can be proved by deriving a recurrence relation for the coefficients from that of the Laguerre polynomials.
Using the substitution $x = mathrme^-t$ and the orthogonality relation of the Laguerre polynomials we immediately obtain
$$ J = sum limits_p=0^infty sum limits_q=0^infty frac1(p+1)(q+1) int limits_0^infty mathrmL_p (t) mathrmL_q (t) mathrme^-t , mathrmd t = sum limits_p=0^infty frac1(n+1)^2 = zeta(2) = fracpi^26 , .$$
Similarly, we have
$$ I = 4sum limits_p=0^infty sum limits_q=0^infty sum limits_r=0^infty frac1(p+1)(q+1)(r+1) int limits_0^infty mathrmL_p (t) mathrmL_q (t) mathrmL_r (t) (1- mathrme^-t) mathrme^-t , mathrmd t , .$$
General formulas for integrals involving three Laguerre polynomials appear to be known (see this paper or this one for a generalisation). I do not know whether they are nice enough to reduce the triple series to a representation of $zeta(3)$ though.
Remark: After doing some numerical calculations I now suspect that the triple series diverges. This is probably due to the fact the original series $(mathrmB)$ only converges in $L^2$, so it cannot be used here. For the simpler integral everything works out though.
It is of course possible to combine the two methods by applying the Laguerre series $(mathrmB)$ in equation $(mathrmA)$. I do not know if these ideas can be used to get the final result, but maybe they can help someone else to find a way.
answered Jul 30 at 18:49
ComplexYetTrivial
2,637624
add a comment |Â
up vote
0
down vote
up vote
0
down vote
This is not a complete answer, but just a description of two ideas that might help with the evaluation of the integral
$$ I equiv 4 int limits_0^1 left(fracoperatornameli(x)xright)^3 (x-1) , mathrmd x , . $$
They are based on methods that can be applied to find the easier integral
$$ J equiv int limits_0^1 left(fracoperatornameli(x)xright)^2 , mathrmd x , . $$
The first approach relies on integration by parts and the series
$$ x-1 = sum limits_k=1^infty frac1k! ln^k (x) , , , x > 0 , .$$
In order to evaluate $J$ we can use the antiderivative $x mapsto 1-frac1x$ of $x mapsto frac1x^2$ to avoid problems with the singularity of $operatornameli(x)$ at $x = 1$ . We get
beginalign
J &= 2 int limits_0^1 fracoperatornameli(x)x frac1-xln(x) , mathrmd x = - 2 sum limits_k=1^infty frac1k! int limits_0^1 fracoperatornameli(x)x ln^k-1 (x) , mathrmd x\
&= 2 sum limits_k=1^infty frac1k! k int limits_0^1 ln^k-1 (x) , mathrmd x = 2 sum limits_k=1^infty frac1k! k (-1)^k-1 (k-1)! \
&= 2 sum limits_k=1^infty frac(-1)^k-1k^2 = 2 eta (2) = zeta(2) = fracpi^26 , .
endalign
Similarly, we can use the antiderivative $x mapsto frac(x-1)^22 x^2$ of $x mapsto fracx-1x^3$ to find
beginalign
I &= - frac32 int limits_0^1 left(fracoperatornameli(x)xright)^2 frac(x-1)^2ln(x) , mathrmd x \
&= frac32 sum limits_k=0^infty frac1k! int limits_0^1 operatornameli^2 (x) frac1-xx fracln^k-1 (x)x , mathrmd x , . tagA
endalign
We can now integrate by parts once more to obtain at least one term that reduces to a multiple of $zeta(3)$ as in the simpler case. However, I have not managed to solve the remaining integrals yet. We could of course use the series again to express the remaining $1-x$ in terms of logarithm powers, but that does not seem to solve the problem.
The second suggestion employs the Fourier-Laguerre series
$$ operatornameli (x) = - x sum_n=0^infty fracmathrmL_n (-ln(x))n+1 , , , x in (0,1) , , tagB$$
of the logarithmic integral. It can be proved by deriving a recurrence relation for the coefficients from that of the Laguerre polynomials.
Using the substitution $x = mathrme^-t$ and the orthogonality relation of the Laguerre polynomials we immediately obtain
$$ J = sum limits_p=0^infty sum limits_q=0^infty frac1(p+1)(q+1) int limits_0^infty mathrmL_p (t) mathrmL_q (t) mathrme^-t , mathrmd t = sum limits_p=0^infty frac1(n+1)^2 = zeta(2) = fracpi^26 , .$$
Similarly, we have
$$ I = 4sum limits_p=0^infty sum limits_q=0^infty sum limits_r=0^infty frac1(p+1)(q+1)(r+1) int limits_0^infty mathrmL_p (t) mathrmL_q (t) mathrmL_r (t) (1- mathrme^-t) mathrme^-t , mathrmd t , .$$
General formulas for integrals involving three Laguerre polynomials appear to be known (see this paper or this one for a generalisation). I do not know whether they are nice enough to reduce the triple series to a representation of $zeta(3)$ though.
Remark: After doing some numerical calculations I now suspect that the triple series diverges. This is probably due to the fact the original series $(mathrmB)$ only converges in $L^2$, so it cannot be used here. For the simpler integral everything works out though.
It is of course possible to combine the two methods by applying the Laguerre series $(mathrmB)$ in equation $(mathrmA)$. I do not know if these ideas can be used to get the final result, but maybe they can help someone else to find a way.
answered Jul 30 at 18:49
ComplexYetTrivial
2,637624
This is not a complete answer, but just a description of two ideas that might help with the evaluation of the integral
$$ I equiv 4 int limits_0^1 left(fracoperatornameli(x)xright)^3 (x-1) , mathrmd x , . $$
They are based on methods that can be applied to find the easier integral
$$ J equiv int limits_0^1 left(fracoperatornameli(x)xright)^2 , mathrmd x , . $$
The first approach relies on integration by parts and the series
$$ x-1 = sum limits_k=1^infty frac1k! ln^k (x) , , , x > 0 , .$$
In order to evaluate $J$ we can use the antiderivative $x mapsto 1-frac1x$ of $x mapsto frac1x^2$ to avoid problems with the singularity of $operatornameli(x)$ at $x = 1$ . We get
beginalign
J &= 2 int limits_0^1 fracoperatornameli(x)x frac1-xln(x) , mathrmd x = - 2 sum limits_k=1^infty frac1k! int limits_0^1 fracoperatornameli(x)x ln^k-1 (x) , mathrmd x\
&= 2 sum limits_k=1^infty frac1k! k int limits_0^1 ln^k-1 (x) , mathrmd x = 2 sum limits_k=1^infty frac1k! k (-1)^k-1 (k-1)! \
&= 2 sum limits_k=1^infty frac(-1)^k-1k^2 = 2 eta (2) = zeta(2) = fracpi^26 , .
endalign
Similarly, we can use the antiderivative $x mapsto frac(x-1)^22 x^2$ of $x mapsto fracx-1x^3$ to find
beginalign
I &= - frac32 int limits_0^1 left(fracoperatornameli(x)xright)^2 frac(x-1)^2ln(x) , mathrmd x \
&= frac32 sum limits_k=0^infty frac1k! int limits_0^1 operatornameli^2 (x) frac1-xx fracln^k-1 (x)x , mathrmd x , . tagA
endalign
We can now integrate by parts once more to obtain at least one term that reduces to a multiple of $zeta(3)$ as in the simpler case. However, I have not managed to solve the remaining integrals yet. We could of course use the series again to express the remaining $1-x$ in terms of logarithm powers, but that does not seem to solve the problem.
The second suggestion employs the Fourier-Laguerre series
$$ operatornameli (x) = - x sum_n=0^infty fracmathrmL_n (-ln(x))n+1 , , , x in (0,1) , , tagB$$
of the logarithmic integral. It can be proved by deriving a recurrence relation for the coefficients from that of the Laguerre polynomials.
Using the substitution $x = mathrme^-t$ and the orthogonality relation of the Laguerre polynomials we immediately obtain
$$ J = sum limits_p=0^infty sum limits_q=0^infty frac1(p+1)(q+1) int limits_0^infty mathrmL_p (t) mathrmL_q (t) mathrme^-t , mathrmd t = sum limits_p=0^infty frac1(n+1)^2 = zeta(2) = fracpi^26 , .$$
Similarly, we have
$$ I = 4sum limits_p=0^infty sum limits_q=0^infty sum limits_r=0^infty frac1(p+1)(q+1)(r+1) int limits_0^infty mathrmL_p (t) mathrmL_q (t) mathrmL_r (t) (1- mathrme^-t) mathrme^-t , mathrmd t , .$$
General formulas for integrals involving three Laguerre polynomials appear to be known (see this paper or this one for a generalisation). I do not know whether they are nice enough to reduce the triple series to a representation of $zeta(3)$ though.
Remark: After doing some numerical calculations I now suspect that the triple series diverges. This is probably due to the fact the original series $(mathrmB)$ only converges in $L^2$, so it cannot be used here. For the simpler integral everything works out though.
It is of course possible to combine the two methods by applying the Laguerre series $(mathrmB)$ in equation $(mathrmA)$. I do not know if these ideas can be used to get the final result, but maybe they can help someone else to find a way.
answered Jul 30 at 18:49
ComplexYetTrivial
2,637624
edited Jul 30 at 19:25
answered Jul 30 at 18:49
ComplexYetTrivial
2,637624
answered Jul 30 at 18:49
ComplexYetTrivial
2,637624
answered Jul 30 at 18:49
ComplexYetTrivial
2,637624
2,637624
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