# 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
2 years, 11 months ago

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

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

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:

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

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

- 2 years, 11 months ago

Thanks Aman for introducing me to Tornheim sum.

- 2 years, 11 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 $$\zeta(2,4)$$

- 2 years, 11 months ago

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

- 2 years, 11 months ago

I have edited it .. Check again

- 2 years, 11 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

- 2 years, 11 months ago

Can u provide a formula to evaluate it?

- 2 years, 11 months ago

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,

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

- 2 years, 11 months ago