Zeta is involved? ∫01ln⁡(x)Li4(x)1−x dx=ζ(3)2−2512ζ(6)\displaystyle \int_0^1 \frac{\ln(x) \text{Li}_4(x)}{1-x} \, dx = \zeta(3)^2 - \frac{25}{12}\zeta(6)

Prove the following Integral:

∫01ln⁡(x)Li⁡4(x)1−xdx=ζ(3)2−2512ζ(6)\displaystyle \int\limits_{0}^{1} \dfrac{\ln (x) \operatorname{Li}_{4}(x)}{1-x} \mathrm{d} x = \zeta(3)^2 -\frac{25}{12}\zeta(6)

This is a part of the set Formidable Series and Integrals.

#Calculus

Note by Aman Rajput
5 years, 4 months ago

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Comments

A=∫01ln⁡(x)Li4(x)1−xdx\displaystyle A=\int _{ 0 }^{ 1 }{ \frac { \ln { \left( x \right) { Li }_{ 4 }\left( x \right) } }{ 1-x } dx }

A=∑k=1∞1k4∫01ln⁡(x)xk1−xdx\displaystyle A=\sum _{ k=1 }^{ \infty }{ \frac { 1 }{ { k }^{ 4 } } \int _{ 0 }^{ 1 }{ \frac { \ln { \left( x \right) { x }^{ k } } }{ 1-x } dx } }

Now, this is nothing but an integral representation of polygamma function.

A=−∑k=1∞1k4ψ1(k+1)\displaystyle A=-\sum _{ k=1 }^{ \infty }{ \frac { 1 }{ { k }^{ 4 } } { \psi }_{ 1 } } \left( k+1 \right)

Now we use the relation: ψa(b)=(−1)a+1a!∑n=0∞1(n+b)a+1\displaystyle { \psi }_{ a }\left( b \right) ={ \left( -1 \right) }^{ a+1 }a!\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ { \left( n+b \right) }^{ a+1 } } }

Therefore, we get:

A=−∑k=1∞∑n=0∞1k4(k+n)2\displaystyle A=-\sum _{ k=1 }^{ \infty }{ \sum _{ n=0 }^{ \infty }{ \frac { 1 }{ { k }^{ 4 }{ \left( k+n \right) }^{ 2 } } } }

Now, A=−T(4,0,2)A=-T\left( 4,0,2 \right)

Here, T=(a,b,c)T=\left( a,b,c \right) is Tornheim sum.

This can also be written as:

A=∑r1>r21r12r24\displaystyle A=\sum _{ { r }_{ 1 }>{ r }_{ 2 } }^{ }{ \frac { 1 }{ { { r }_{ 1 } }^{ 2 }{ { r }_{ 2 } }^{ 4 } } }

A=ζ(2,4)A=\zeta \left( 2,4 \right)

Where ζ(a,b)\zeta \left( a,b \right) is multi-zeta function and not Hurwitz-zeta function.

Now Tornheim sum can be evaluated as:

T(m,0,n)=(−1)m∑j=0⌊n−12⌋(m+n−2j−1m−1)ζ(2j)ζ(m+n−2j)+(−1)m∑j=0⌊m2⌋(m+n−2j−1n−1)ζ(2j)ζ(m+n−2j)−12ζ(m,n)\displaystyle T\left( m,0,n \right) ={ \left( -1 \right) }^{ m }\sum _{ j=0 }^{ \left\lfloor \frac { n-1 }{ 2 } \right\rfloor }{ \left( \begin{matrix} m+n-2j-1 \\ m-1 \end{matrix} \right) \zeta \left( 2j \right) \zeta \left( m+n-2j \right) + } { \left( -1 \right) }^{ m }\sum _{ j=0 }^{ \left\lfloor \frac { m }{ 2 } \right\rfloor }{ \left( \begin{matrix} m+n-2j-1 \\ n-1 \end{matrix} \right) \zeta \left( 2j \right) \zeta \left( m+n-2j \right) -\frac { 1 }{ 2 } \zeta \left( m,n \right) }

Now taking m=4m=4 and n=2n=2, we get:

A=(ζ(3))2−2512ζ(6)A={ \left( \zeta \left( 3 \right) \right) }^{ 2 }-\frac { 25 }{ 12 } \zeta \left( 6 \right)

∴∫01ln⁡(x)Li4(x)1−xdx=(ζ(3))2−2512ζ(6)\boxed {\displaystyle \therefore \int _{ 0 }^{ 1 }{ \frac { \ln { \left( x \right) { Li }_{ 4 }\left( x \right) } }{ 1-x } dx } ={ \left( \zeta \left( 3 \right) \right) }^{ 2 }-\frac { 25 }{ 12 } \zeta \left( 6 \right) }

Aditya Kumar - 5 years, 4 months ago

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Thanks Aman for introducing me to Tornheim sum.

Aditya Kumar - 5 years, 4 months ago

Good aditya... Why you are going for polygamma here... If you want to use tornheim sum.. Directly apply this

11−x=∑n>0xn−1\displaystyle \frac{1}{1-x}=\sum_{n>0}x^{n-1}

And in last formula.. At the end there is a term ζ(m,n)\zeta(m,n) which will become ζ(4,2)\zeta(4,2) . now how will you evaluate this value to get T(4,0,2)T(4,0,2) . But you are trying to find ζ(2,4)\zeta(2,4)

Aman Rajput - 5 years, 4 months ago

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Ahahahaha. I'm a fool. Since this problem involves challenging ideas I went through polygamma.

Aditya Kumar - 5 years, 4 months ago

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@Aditya Kumar – I have edited it .. Check again

Aman Rajput - 5 years, 4 months ago

And I think the last formula you are using.. Will not gonna work.. Check it again . I don't how you get this from that ??? This formula is valid for m+nm+n odd.

Here , 4+2 =6 (even) Formula not valid @Aditya Kumar

Aman Rajput - 5 years, 4 months ago

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Can u provide a formula to evaluate it?

Aditya Kumar - 5 years, 4 months ago

Aliter:

−A=∑m=1∞∑n=0∞1(m+n)2n4-\text{A} = \displaystyle \sum_{m=1}^{\infty}\sum_{n=0}^{\infty} \dfrac{1}{(m+n)^2 n^4}

=ζ(6)+∑m=1∞∑n=1∞1(m+n)2n4 \displaystyle = \zeta(6) + \sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \dfrac{1}{(m+n)^2n^4}

=ζ(6)+12(∑m=1∞∑n=1∞1(m+n)2n4+∑m=1∞∑n=1∞1(m+n)2m4) \displaystyle = \zeta(6) + \dfrac{1}{2} \left( \sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \dfrac{1}{(m+n)^2n^4} + \sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \dfrac{1}{(m+n)^2m^4}\right) (Interchanging variables and averaging)

=ζ(6)+12(∑m=1∞∑n=1∞m4+n4(m+n)2m4n4) \displaystyle = \zeta(6) + \dfrac{1}{2} \left(\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \dfrac{m^4+n^4}{(m+n)^2 m^4 n^4}\right)

=ζ(6)+12∑m=1∞∑n=1∞[1m2n4+1n2m4−2m3n3+2m2n2(m+n)]\displaystyle = \zeta(6) + \dfrac{1}{2} \sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \left[\dfrac{1}{m^2 n^4} + \dfrac{1}{n^2 m^4} -\dfrac{2}{m^3n^3} + \dfrac{2}{m^2 n^2 (m+n)} \right] (Expressing m4+n4m^4+n^4 as powers of (m+n)(m+n) and mnmn )

=ζ(6)+ζ(2)ζ(4)−[ζ(3)]2+S=\displaystyle \zeta(6) + \zeta(2)\zeta(4) - \left[\zeta(3)\right]^2 + \text{S}

where S=∑m=1∞∑n=1∞1m2n2(m+n)2\displaystyle \text{S} = \sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \dfrac{1}{m^2 n^2 (m+n)^2}

Since S=ζ(6)3\text{S} = \dfrac{\zeta(6)}{3}, we have,

−A=43ζ(6)+ζ(2)ζ(4)−[ζ(3)]2-\text{A} = \dfrac{4}{3} \zeta(6) + \zeta(2)\zeta(4) - \left[\zeta(3)\right]^2

Expressing ζ(2)ζ(4)\zeta(2)\zeta(4) in terms of ζ(6)\zeta(6), we have,

A=[ζ(3)]2−2512ζ(6)\text{A} = \boxed{\left[\zeta(3)\right]^2 - \dfrac{25}{12} \zeta(6)}

Ishan Singh - 5 years, 4 months ago
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