Zeta is involved? $\displaystyle \int_0^1 \frac{\ln(x) \text{Li}_4(x)}{1-x} \, dx = \zeta(3)^2 - \frac{25}{12}\zeta(6)$

Prove the following Integral:

$\displaystyle \int\limits_{0}^{1} \dfrac{\ln (x) \operatorname{Li}_{4}(x)}{1-x} \mathrm{d} x = \zeta(3)^2 -\frac{25}{12}\zeta(6)$

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

Note by Aman Rajput
5 years, 2 months ago

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Aliter:

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

$\displaystyle = \zeta(6) + \sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \dfrac{1}{(m+n)^2n^4}$

$\displaystyle = \zeta(6) + \dfrac{1}{2} \left( \sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \dfrac{1}{(m+n)^2n^4} + \sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \dfrac{1}{(m+n)^2m^4}\right)$ (Interchanging variables and averaging)

$\displaystyle = \zeta(6) + \dfrac{1}{2} \left(\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \dfrac{m^4+n^4}{(m+n)^2 m^4 n^4}\right)$

$\displaystyle = \zeta(6) + \dfrac{1}{2} \sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \left[\dfrac{1}{m^2 n^4} + \dfrac{1}{n^2 m^4} -\dfrac{2}{m^3n^3} + \dfrac{2}{m^2 n^2 (m+n)} \right]$ (Expressing $m^4+n^4$ as powers of $(m+n)$ and $mn$ )

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

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

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

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

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

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

- 5 years, 2 months ago

$\displaystyle A=\int _{ 0 }^{ 1 }{ \frac { \ln { \left( x \right) { Li }_{ 4 }\left( x \right) } }{ 1-x } dx }$

$\displaystyle A=\sum _{ k=1 }^{ \infty }{ \frac { 1 }{ { k }^{ 4 } } \int _{ 0 }^{ 1 }{ \frac { \ln { \left( x \right) { x }^{ k } } }{ 1-x } dx } }$

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

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

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

Therefore, we get:

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

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

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

This can also be written as:

$\displaystyle A=\sum _{ { r }_{ 1 }>{ r }_{ 2 } }^{ }{ \frac { 1 }{ { { r }_{ 1 } }^{ 2 }{ { r }_{ 2 } }^{ 4 } } }$

$A=\zeta \left( 2,4 \right)$

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

Now Tornheim sum can be evaluated as:

$\displaystyle T\left( m,0,n \right) ={ \left( -1 \right) }^{ m }\sum _{ j=0 }^{ \left\lfloor \frac { n-1 }{ 2 } \right\rfloor }{ \left( \begin{matrix} m+n-2j-1 \\ m-1 \end{matrix} \right) \zeta \left( 2j \right) \zeta \left( m+n-2j \right) + } { \left( -1 \right) }^{ m }\sum _{ j=0 }^{ \left\lfloor \frac { m }{ 2 } \right\rfloor }{ \left( \begin{matrix} m+n-2j-1 \\ n-1 \end{matrix} \right) \zeta \left( 2j \right) \zeta \left( m+n-2j \right) -\frac { 1 }{ 2 } \zeta \left( m,n \right) }$

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

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

$\boxed {\displaystyle \therefore \int _{ 0 }^{ 1 }{ \frac { \ln { \left( x \right) { Li }_{ 4 }\left( x \right) } }{ 1-x } dx } ={ \left( \zeta \left( 3 \right) \right) }^{ 2 }-\frac { 25 }{ 12 } \zeta \left( 6 \right) }$

- 5 years, 2 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 $m+n$ odd.

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

- 5 years, 2 months ago

Can u provide a formula to evaluate it?

- 5 years, 2 months ago

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

$\displaystyle \frac{1}{1-x}=\sum_{n>0}x^{n-1}$

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

- 5 years, 2 months ago

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

- 5 years, 2 months ago

I have edited it .. Check again

- 5 years, 2 months ago

Thanks Aman for introducing me to Tornheim sum.

- 5 years, 2 months ago