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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
10 months, 1 week 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:

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

\[\boxed {\displaystyle \therefore \int _{ 0 }^{ 1 }{ \frac { \ln { \left( x \right) { Li }_{ 4 }\left( x \right) } }{ 1-x } dx } ={ \left( \zeta \left( 3 \right) \right) }^{ 2 }-\frac { 25 }{ 12 } \zeta \left( 6 \right) } \] Aditya Kumar · 10 months, 1 week ago

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@Aditya Kumar Thanks Aman for introducing me to Tornheim sum. Aditya Kumar · 10 months, 1 week ago

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@Aditya Kumar 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 Aman Rajput · 10 months, 1 week ago

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@Aman Rajput Can u provide a formula to evaluate it? Aditya Kumar · 10 months, 1 week ago

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@Aditya Kumar 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 \(\zeta(2,4)\) Aman Rajput · 10 months, 1 week ago

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@Aman Rajput Ahahahaha. I'm a fool. Since this problem involves challenging ideas I went through polygamma. Aditya Kumar · 10 months, 1 week ago

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@Aditya Kumar I have edited it .. Check again Aman Rajput · 10 months, 1 week 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,

\(\text{A} = \boxed{\left[\zeta(3)\right]^2 - \dfrac{25}{12} \zeta(6)} \) Ishan Singh · 10 months, 1 week ago

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