On the series expansion of $frac{operatorname{Li}_3(-x)}{1+x}$ and its usage












5












$begingroup$


Recently I have asked about the evaluation of an integral involving a Trilogarithm $($which can be found here$)$. Pisco provided a quite elegant approach starting with a functional equation of the Trilogarithm. However, his attempt made use of the fact that




$$int_0^{infty}frac{operatorname{Li}_3(-x)}{1+x}x^{s-1}mathrm dx=frac{pi}{sin pi s}[zeta(3)-zeta(3,1-s)]~~~~~text{for }0<s<1$$




Which can be seen as the Mellin Transform of the function $f(x)=frac{operatorname{Li}_3(-x)}{1+x}$. He also supplied a proof of his claim by using the Mellin Inversion Theorem. Anyway, I thought about applying Ramanujan's Master Theorem but for this purpose it is conclusive to show that the given function, lets call it $f(x)$, has a power series of the form



$$f(x)=sum_{n=0}^{infty}(-1)^nfrac{phi(n)}{n!}x^n$$



With the help of WolframAlpha I got that




$$f(x)=frac{operatorname{Li}_3(-x)}{1+x}=frac12sum_{n=1}^{infty}[psi^{(2)}(1+n)-psi^{(2)}(1)](-x)^n~~text{for }|x|<1$$




Using this result combined with Ramanujan's Master Theorem one can easily verify the given relation by setting $phi(n)=frac{Gamma(1+n)}2[psi^{(2)}(1+n)-psi^{(2)}(1)]$ and applying the theorem



$$begin{align}
int_0^{infty}frac{operatorname{Li}_3(-x)}{1+x}x^{s-1}mathrm dx=int_0^{infty}f(x)x^{s-1}mathrm dx&=Gamma(s)phi(-s)\
&=Gamma(s)frac{Gamma(1-s)}2[psi^{(2)}(1-s)-psi^{(2)}(1)]\
&=frac{pi}{sin pi s}frac12[(-2zeta(3,1-s))-(-2zeta(3))]\
&=frac{pi}{sin pi s}[zeta(3)-zeta(3,1-s)]
end{align}$$



Where Euler's Reflection Formula and some basic properties of the Polygamma Functions were used. This attempt affirms Pisco's proof. Up to now this is not a question but more a proof verification hence I am not sure whether all my steps are legitimate like this or not.




Since the power series given by WolframAlpha worked out fairly good I was curious how to derive such a series expansion. First of all I asked for a proof here to see how to approach to $f(x)$. Hence I was not able to identify a pattern within the coefficients of the MacLaurin expansion I thought about using a Cauchy Product here since both functions out of which $f(x)$ is composited have a well-known series expansion therefore this seems to be worth a try. Further the cited proof made use of it in some way aswell. Thus, I tried to compute




$$frac1{1+x}cdotoperatorname{Li}_3(-x)=left(sum_{i=0}^{infty}(-x)^iright)left(sum_{j=1}^{infty}frac{(-x)^j}{j^3}right)=~?$$




But I struggled to get started. I worried about the fact that the first index $i$ starts at $0$ while the second one $j$ at $1$. This would not be a big deal since I could just apply an index shift to one of the two series but then the powers of $(-x)$ would not longer match. Further I am not totally sure how the author of refered proof dogded or accomplished this step.




My question is splitted up into two parts. $(1)$ Is my approach to the equality from above - beside the part where I "cheated" by asking WolframAlpha to expand $f(x)$ as a series - valid i.e. the usage of Ramanujan's Master Theorem here, the simplification of the RHS, etc.? $(2)$ How can one derive the series expansion of $f(x)$? Is it possible - without knowing the explicit power series - to deduce the pattern given by the MacLaurin expansion as values of the Polygamma Function $psi^{(2)}$? Can we apply the Cauchy Prodcut here; if yes how can we take care of the unsuitable indices? I would be interested in a whole derivation of the series expansion for $f(x)$ as well.




Thanks in advance!






proof-verification power-series polylogarithm cauchy-product polygamma







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Regarding the product of two series, just assume $b_0 = 0$: $$[x^n] left( sum_{k = 0}^infty a_k x^k right) sum_{k = 0}^infty b_k x^k = sum_{k = 0}^n a_{n - k} b_k = sum_{k = 1}^n frac {(-1)^n} {k^3} = (-1)^n H_{n, 3}.$$
    $endgroup$
    – Maxim
    Oct 9 '18 at 19:24










  • $begingroup$
    Is this not a little bit of cheating since obviously the second series will not be well defined for $j=0$?
    $endgroup$
    – mrtaurho
    Oct 9 '18 at 19:27










  • $begingroup$
    The second series is $0 x^0 + b_1 x^1 + dots,$, why isn't it well-defined?
    $endgroup$
    – Maxim
    Oct 9 '18 at 19:32












  • $begingroup$
    The second series is $frac{1}{0}-x+frac{x^2}8mpdots$.
    $endgroup$
    – mrtaurho
    Oct 9 '18 at 19:33










  • $begingroup$
    I defined $b_0 = 0$, not $b_0 = (-1)^0/0^3$.
    $endgroup$
    – Maxim
    Oct 9 '18 at 19:35
















5












$begingroup$


Recently I have asked about the evaluation of an integral involving a Trilogarithm $($which can be found here$)$. Pisco provided a quite elegant approach starting with a functional equation of the Trilogarithm. However, his attempt made use of the fact that




$$int_0^{infty}frac{operatorname{Li}_3(-x)}{1+x}x^{s-1}mathrm dx=frac{pi}{sin pi s}[zeta(3)-zeta(3,1-s)]~~~~~text{for }0<s<1$$




Which can be seen as the Mellin Transform of the function $f(x)=frac{operatorname{Li}_3(-x)}{1+x}$. He also supplied a proof of his claim by using the Mellin Inversion Theorem. Anyway, I thought about applying Ramanujan's Master Theorem but for this purpose it is conclusive to show that the given function, lets call it $f(x)$, has a power series of the form



$$f(x)=sum_{n=0}^{infty}(-1)^nfrac{phi(n)}{n!}x^n$$



With the help of WolframAlpha I got that




$$f(x)=frac{operatorname{Li}_3(-x)}{1+x}=frac12sum_{n=1}^{infty}[psi^{(2)}(1+n)-psi^{(2)}(1)](-x)^n~~text{for }|x|<1$$




Using this result combined with Ramanujan's Master Theorem one can easily verify the given relation by setting $phi(n)=frac{Gamma(1+n)}2[psi^{(2)}(1+n)-psi^{(2)}(1)]$ and applying the theorem



$$begin{align}
int_0^{infty}frac{operatorname{Li}_3(-x)}{1+x}x^{s-1}mathrm dx=int_0^{infty}f(x)x^{s-1}mathrm dx&=Gamma(s)phi(-s)\
&=Gamma(s)frac{Gamma(1-s)}2[psi^{(2)}(1-s)-psi^{(2)}(1)]\
&=frac{pi}{sin pi s}frac12[(-2zeta(3,1-s))-(-2zeta(3))]\
&=frac{pi}{sin pi s}[zeta(3)-zeta(3,1-s)]
end{align}$$



Where Euler's Reflection Formula and some basic properties of the Polygamma Functions were used. This attempt affirms Pisco's proof. Up to now this is not a question but more a proof verification hence I am not sure whether all my steps are legitimate like this or not.




Since the power series given by WolframAlpha worked out fairly good I was curious how to derive such a series expansion. First of all I asked for a proof here to see how to approach to $f(x)$. Hence I was not able to identify a pattern within the coefficients of the MacLaurin expansion I thought about using a Cauchy Product here since both functions out of which $f(x)$ is composited have a well-known series expansion therefore this seems to be worth a try. Further the cited proof made use of it in some way aswell. Thus, I tried to compute




$$frac1{1+x}cdotoperatorname{Li}_3(-x)=left(sum_{i=0}^{infty}(-x)^iright)left(sum_{j=1}^{infty}frac{(-x)^j}{j^3}right)=~?$$




But I struggled to get started. I worried about the fact that the first index $i$ starts at $0$ while the second one $j$ at $1$. This would not be a big deal since I could just apply an index shift to one of the two series but then the powers of $(-x)$ would not longer match. Further I am not totally sure how the author of refered proof dogded or accomplished this step.




My question is splitted up into two parts. $(1)$ Is my approach to the equality from above - beside the part where I "cheated" by asking WolframAlpha to expand $f(x)$ as a series - valid i.e. the usage of Ramanujan's Master Theorem here, the simplification of the RHS, etc.? $(2)$ How can one derive the series expansion of $f(x)$? Is it possible - without knowing the explicit power series - to deduce the pattern given by the MacLaurin expansion as values of the Polygamma Function $psi^{(2)}$? Can we apply the Cauchy Prodcut here; if yes how can we take care of the unsuitable indices? I would be interested in a whole derivation of the series expansion for $f(x)$ as well.




Thanks in advance!






proof-verification power-series polylogarithm cauchy-product polygamma







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Regarding the product of two series, just assume $b_0 = 0$: $$[x^n] left( sum_{k = 0}^infty a_k x^k right) sum_{k = 0}^infty b_k x^k = sum_{k = 0}^n a_{n - k} b_k = sum_{k = 1}^n frac {(-1)^n} {k^3} = (-1)^n H_{n, 3}.$$
    $endgroup$
    – Maxim
    Oct 9 '18 at 19:24










  • $begingroup$
    Is this not a little bit of cheating since obviously the second series will not be well defined for $j=0$?
    $endgroup$
    – mrtaurho
    Oct 9 '18 at 19:27










  • $begingroup$
    The second series is $0 x^0 + b_1 x^1 + dots,$, why isn't it well-defined?
    $endgroup$
    – Maxim
    Oct 9 '18 at 19:32












  • $begingroup$
    The second series is $frac{1}{0}-x+frac{x^2}8mpdots$.
    $endgroup$
    – mrtaurho
    Oct 9 '18 at 19:33










  • $begingroup$
    I defined $b_0 = 0$, not $b_0 = (-1)^0/0^3$.
    $endgroup$
    – Maxim
    Oct 9 '18 at 19:35














5












5








5


0



$begingroup$


Recently I have asked about the evaluation of an integral involving a Trilogarithm $($which can be found here$)$. Pisco provided a quite elegant approach starting with a functional equation of the Trilogarithm. However, his attempt made use of the fact that




$$int_0^{infty}frac{operatorname{Li}_3(-x)}{1+x}x^{s-1}mathrm dx=frac{pi}{sin pi s}[zeta(3)-zeta(3,1-s)]~~~~~text{for }0<s<1$$




Which can be seen as the Mellin Transform of the function $f(x)=frac{operatorname{Li}_3(-x)}{1+x}$. He also supplied a proof of his claim by using the Mellin Inversion Theorem. Anyway, I thought about applying Ramanujan's Master Theorem but for this purpose it is conclusive to show that the given function, lets call it $f(x)$, has a power series of the form



$$f(x)=sum_{n=0}^{infty}(-1)^nfrac{phi(n)}{n!}x^n$$



With the help of WolframAlpha I got that




$$f(x)=frac{operatorname{Li}_3(-x)}{1+x}=frac12sum_{n=1}^{infty}[psi^{(2)}(1+n)-psi^{(2)}(1)](-x)^n~~text{for }|x|<1$$




Using this result combined with Ramanujan's Master Theorem one can easily verify the given relation by setting $phi(n)=frac{Gamma(1+n)}2[psi^{(2)}(1+n)-psi^{(2)}(1)]$ and applying the theorem



$$begin{align}
int_0^{infty}frac{operatorname{Li}_3(-x)}{1+x}x^{s-1}mathrm dx=int_0^{infty}f(x)x^{s-1}mathrm dx&=Gamma(s)phi(-s)\
&=Gamma(s)frac{Gamma(1-s)}2[psi^{(2)}(1-s)-psi^{(2)}(1)]\
&=frac{pi}{sin pi s}frac12[(-2zeta(3,1-s))-(-2zeta(3))]\
&=frac{pi}{sin pi s}[zeta(3)-zeta(3,1-s)]
end{align}$$



Where Euler's Reflection Formula and some basic properties of the Polygamma Functions were used. This attempt affirms Pisco's proof. Up to now this is not a question but more a proof verification hence I am not sure whether all my steps are legitimate like this or not.




Since the power series given by WolframAlpha worked out fairly good I was curious how to derive such a series expansion. First of all I asked for a proof here to see how to approach to $f(x)$. Hence I was not able to identify a pattern within the coefficients of the MacLaurin expansion I thought about using a Cauchy Product here since both functions out of which $f(x)$ is composited have a well-known series expansion therefore this seems to be worth a try. Further the cited proof made use of it in some way aswell. Thus, I tried to compute




$$frac1{1+x}cdotoperatorname{Li}_3(-x)=left(sum_{i=0}^{infty}(-x)^iright)left(sum_{j=1}^{infty}frac{(-x)^j}{j^3}right)=~?$$




But I struggled to get started. I worried about the fact that the first index $i$ starts at $0$ while the second one $j$ at $1$. This would not be a big deal since I could just apply an index shift to one of the two series but then the powers of $(-x)$ would not longer match. Further I am not totally sure how the author of refered proof dogded or accomplished this step.




My question is splitted up into two parts. $(1)$ Is my approach to the equality from above - beside the part where I "cheated" by asking WolframAlpha to expand $f(x)$ as a series - valid i.e. the usage of Ramanujan's Master Theorem here, the simplification of the RHS, etc.? $(2)$ How can one derive the series expansion of $f(x)$? Is it possible - without knowing the explicit power series - to deduce the pattern given by the MacLaurin expansion as values of the Polygamma Function $psi^{(2)}$? Can we apply the Cauchy Prodcut here; if yes how can we take care of the unsuitable indices? I would be interested in a whole derivation of the series expansion for $f(x)$ as well.




Thanks in advance!






proof-verification power-series polylogarithm cauchy-product polygamma







share|cite|improve this question











$endgroup$




Recently I have asked about the evaluation of an integral involving a Trilogarithm $($which can be found here$)$. Pisco provided a quite elegant approach starting with a functional equation of the Trilogarithm. However, his attempt made use of the fact that




$$int_0^{infty}frac{operatorname{Li}_3(-x)}{1+x}x^{s-1}mathrm dx=frac{pi}{sin pi s}[zeta(3)-zeta(3,1-s)]~~~~~text{for }0<s<1$$




Which can be seen as the Mellin Transform of the function $f(x)=frac{operatorname{Li}_3(-x)}{1+x}$. He also supplied a proof of his claim by using the Mellin Inversion Theorem. Anyway, I thought about applying Ramanujan's Master Theorem but for this purpose it is conclusive to show that the given function, lets call it $f(x)$, has a power series of the form



$$f(x)=sum_{n=0}^{infty}(-1)^nfrac{phi(n)}{n!}x^n$$



With the help of WolframAlpha I got that




$$f(x)=frac{operatorname{Li}_3(-x)}{1+x}=frac12sum_{n=1}^{infty}[psi^{(2)}(1+n)-psi^{(2)}(1)](-x)^n~~text{for }|x|<1$$




Using this result combined with Ramanujan's Master Theorem one can easily verify the given relation by setting $phi(n)=frac{Gamma(1+n)}2[psi^{(2)}(1+n)-psi^{(2)}(1)]$ and applying the theorem



$$begin{align}
int_0^{infty}frac{operatorname{Li}_3(-x)}{1+x}x^{s-1}mathrm dx=int_0^{infty}f(x)x^{s-1}mathrm dx&=Gamma(s)phi(-s)\
&=Gamma(s)frac{Gamma(1-s)}2[psi^{(2)}(1-s)-psi^{(2)}(1)]\
&=frac{pi}{sin pi s}frac12[(-2zeta(3,1-s))-(-2zeta(3))]\
&=frac{pi}{sin pi s}[zeta(3)-zeta(3,1-s)]
end{align}$$



Where Euler's Reflection Formula and some basic properties of the Polygamma Functions were used. This attempt affirms Pisco's proof. Up to now this is not a question but more a proof verification hence I am not sure whether all my steps are legitimate like this or not.




Since the power series given by WolframAlpha worked out fairly good I was curious how to derive such a series expansion. First of all I asked for a proof here to see how to approach to $f(x)$. Hence I was not able to identify a pattern within the coefficients of the MacLaurin expansion I thought about using a Cauchy Product here since both functions out of which $f(x)$ is composited have a well-known series expansion therefore this seems to be worth a try. Further the cited proof made use of it in some way aswell. Thus, I tried to compute




$$frac1{1+x}cdotoperatorname{Li}_3(-x)=left(sum_{i=0}^{infty}(-x)^iright)left(sum_{j=1}^{infty}frac{(-x)^j}{j^3}right)=~?$$




But I struggled to get started. I worried about the fact that the first index $i$ starts at $0$ while the second one $j$ at $1$. This would not be a big deal since I could just apply an index shift to one of the two series but then the powers of $(-x)$ would not longer match. Further I am not totally sure how the author of refered proof dogded or accomplished this step.




My question is splitted up into two parts. $(1)$ Is my approach to the equality from above - beside the part where I "cheated" by asking WolframAlpha to expand $f(x)$ as a series - valid i.e. the usage of Ramanujan's Master Theorem here, the simplification of the RHS, etc.? $(2)$ How can one derive the series expansion of $f(x)$? Is it possible - without knowing the explicit power series - to deduce the pattern given by the MacLaurin expansion as values of the Polygamma Function $psi^{(2)}$? Can we apply the Cauchy Prodcut here; if yes how can we take care of the unsuitable indices? I would be interested in a whole derivation of the series expansion for $f(x)$ as well.




Thanks in advance!






proof-verification power-series polylogarithm cauchy-product polygamma




proof-verification power-series polylogarithm cauchy-product polygamma






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 27 at 6:24







mrtaurho

















asked Oct 7 '18 at 14:05









mrtaurhomrtaurho

5,98551641




5,98551641








  • 1




    $begingroup$
    Regarding the product of two series, just assume $b_0 = 0$: $$[x^n] left( sum_{k = 0}^infty a_k x^k right) sum_{k = 0}^infty b_k x^k = sum_{k = 0}^n a_{n - k} b_k = sum_{k = 1}^n frac {(-1)^n} {k^3} = (-1)^n H_{n, 3}.$$
    $endgroup$
    – Maxim
    Oct 9 '18 at 19:24










  • $begingroup$
    Is this not a little bit of cheating since obviously the second series will not be well defined for $j=0$?
    $endgroup$
    – mrtaurho
    Oct 9 '18 at 19:27










  • $begingroup$
    The second series is $0 x^0 + b_1 x^1 + dots,$, why isn't it well-defined?
    $endgroup$
    – Maxim
    Oct 9 '18 at 19:32












  • $begingroup$
    The second series is $frac{1}{0}-x+frac{x^2}8mpdots$.
    $endgroup$
    – mrtaurho
    Oct 9 '18 at 19:33










  • $begingroup$
    I defined $b_0 = 0$, not $b_0 = (-1)^0/0^3$.
    $endgroup$
    – Maxim
    Oct 9 '18 at 19:35














  • 1




    $begingroup$
    Regarding the product of two series, just assume $b_0 = 0$: $$[x^n] left( sum_{k = 0}^infty a_k x^k right) sum_{k = 0}^infty b_k x^k = sum_{k = 0}^n a_{n - k} b_k = sum_{k = 1}^n frac {(-1)^n} {k^3} = (-1)^n H_{n, 3}.$$
    $endgroup$
    – Maxim
    Oct 9 '18 at 19:24










  • $begingroup$
    Is this not a little bit of cheating since obviously the second series will not be well defined for $j=0$?
    $endgroup$
    – mrtaurho
    Oct 9 '18 at 19:27










  • $begingroup$
    The second series is $0 x^0 + b_1 x^1 + dots,$, why isn't it well-defined?
    $endgroup$
    – Maxim
    Oct 9 '18 at 19:32












  • $begingroup$
    The second series is $frac{1}{0}-x+frac{x^2}8mpdots$.
    $endgroup$
    – mrtaurho
    Oct 9 '18 at 19:33










  • $begingroup$
    I defined $b_0 = 0$, not $b_0 = (-1)^0/0^3$.
    $endgroup$
    – Maxim
    Oct 9 '18 at 19:35








1




1




$begingroup$
Regarding the product of two series, just assume $b_0 = 0$: $$[x^n] left( sum_{k = 0}^infty a_k x^k right) sum_{k = 0}^infty b_k x^k = sum_{k = 0}^n a_{n - k} b_k = sum_{k = 1}^n frac {(-1)^n} {k^3} = (-1)^n H_{n, 3}.$$
$endgroup$
– Maxim
Oct 9 '18 at 19:24




$begingroup$
Regarding the product of two series, just assume $b_0 = 0$: $$[x^n] left( sum_{k = 0}^infty a_k x^k right) sum_{k = 0}^infty b_k x^k = sum_{k = 0}^n a_{n - k} b_k = sum_{k = 1}^n frac {(-1)^n} {k^3} = (-1)^n H_{n, 3}.$$
$endgroup$
– Maxim
Oct 9 '18 at 19:24












$begingroup$
Is this not a little bit of cheating since obviously the second series will not be well defined for $j=0$?
$endgroup$
– mrtaurho
Oct 9 '18 at 19:27




$begingroup$
Is this not a little bit of cheating since obviously the second series will not be well defined for $j=0$?
$endgroup$
– mrtaurho
Oct 9 '18 at 19:27












$begingroup$
The second series is $0 x^0 + b_1 x^1 + dots,$, why isn't it well-defined?
$endgroup$
– Maxim
Oct 9 '18 at 19:32






$begingroup$
The second series is $0 x^0 + b_1 x^1 + dots,$, why isn't it well-defined?
$endgroup$
– Maxim
Oct 9 '18 at 19:32














$begingroup$
The second series is $frac{1}{0}-x+frac{x^2}8mpdots$.
$endgroup$
– mrtaurho
Oct 9 '18 at 19:33




$begingroup$
The second series is $frac{1}{0}-x+frac{x^2}8mpdots$.
$endgroup$
– mrtaurho
Oct 9 '18 at 19:33












$begingroup$
I defined $b_0 = 0$, not $b_0 = (-1)^0/0^3$.
$endgroup$
– Maxim
Oct 9 '18 at 19:35




$begingroup$
I defined $b_0 = 0$, not $b_0 = (-1)^0/0^3$.
$endgroup$
– Maxim
Oct 9 '18 at 19:35










1 Answer
1






active

oldest

votes


















1












$begingroup$

Combining Maxim's suggestion to define the $0^{th}$ coefficient of the Trilogarithm series to be zero and ysharifi's given proof here on AoPS it is rather simple to obtain the given expansion. Using the Cauchy Product leads to



$$begin{align}
operatorname{Li}_3(-x)cdotfrac1{1+x}&=left(sum_{n=1}^{infty}frac{(-x)^n}{n^3}right)cdotleft(sum_{n=0}^{infty}(-x)^nright)\
&=sum_{n=1}^{infty}sum_{k=1}^nfrac1{k^3}(-x)^n\
&=sum_{n=1}^{infty}left[sum_{k=1}^{infty}frac1{k^3}-sum_{k=n+1}^{infty}frac1{k^3}right](-x)^n\
&=sum_{n=1}^{infty}left[sum_{k=0}^{infty}frac1{(1+k)^3}-sum_{k=0}^{infty}frac1{(1+n+k)^3}right](-x)^n\
&=sum_{n=1}^{infty}left[-frac12psi^{(2)}(1)+frac12psi^{(2)}(1+n)right](-x)^n\
end{align}$$




$$frac{operatorname{Li}_3(-x)}{1+x}=frac12sum_{n=1}^{infty}[psi^{(2)}(1+n)-psi^{(2)}(1)](-x)^n$$







share|cite|improve this answer









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    1 Answer
    1






    active

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    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Combining Maxim's suggestion to define the $0^{th}$ coefficient of the Trilogarithm series to be zero and ysharifi's given proof here on AoPS it is rather simple to obtain the given expansion. Using the Cauchy Product leads to



    $$begin{align}
    operatorname{Li}_3(-x)cdotfrac1{1+x}&=left(sum_{n=1}^{infty}frac{(-x)^n}{n^3}right)cdotleft(sum_{n=0}^{infty}(-x)^nright)\
    &=sum_{n=1}^{infty}sum_{k=1}^nfrac1{k^3}(-x)^n\
    &=sum_{n=1}^{infty}left[sum_{k=1}^{infty}frac1{k^3}-sum_{k=n+1}^{infty}frac1{k^3}right](-x)^n\
    &=sum_{n=1}^{infty}left[sum_{k=0}^{infty}frac1{(1+k)^3}-sum_{k=0}^{infty}frac1{(1+n+k)^3}right](-x)^n\
    &=sum_{n=1}^{infty}left[-frac12psi^{(2)}(1)+frac12psi^{(2)}(1+n)right](-x)^n\
    end{align}$$




    $$frac{operatorname{Li}_3(-x)}{1+x}=frac12sum_{n=1}^{infty}[psi^{(2)}(1+n)-psi^{(2)}(1)](-x)^n$$







    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Combining Maxim's suggestion to define the $0^{th}$ coefficient of the Trilogarithm series to be zero and ysharifi's given proof here on AoPS it is rather simple to obtain the given expansion. Using the Cauchy Product leads to



      $$begin{align}
      operatorname{Li}_3(-x)cdotfrac1{1+x}&=left(sum_{n=1}^{infty}frac{(-x)^n}{n^3}right)cdotleft(sum_{n=0}^{infty}(-x)^nright)\
      &=sum_{n=1}^{infty}sum_{k=1}^nfrac1{k^3}(-x)^n\
      &=sum_{n=1}^{infty}left[sum_{k=1}^{infty}frac1{k^3}-sum_{k=n+1}^{infty}frac1{k^3}right](-x)^n\
      &=sum_{n=1}^{infty}left[sum_{k=0}^{infty}frac1{(1+k)^3}-sum_{k=0}^{infty}frac1{(1+n+k)^3}right](-x)^n\
      &=sum_{n=1}^{infty}left[-frac12psi^{(2)}(1)+frac12psi^{(2)}(1+n)right](-x)^n\
      end{align}$$




      $$frac{operatorname{Li}_3(-x)}{1+x}=frac12sum_{n=1}^{infty}[psi^{(2)}(1+n)-psi^{(2)}(1)](-x)^n$$







      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Combining Maxim's suggestion to define the $0^{th}$ coefficient of the Trilogarithm series to be zero and ysharifi's given proof here on AoPS it is rather simple to obtain the given expansion. Using the Cauchy Product leads to



        $$begin{align}
        operatorname{Li}_3(-x)cdotfrac1{1+x}&=left(sum_{n=1}^{infty}frac{(-x)^n}{n^3}right)cdotleft(sum_{n=0}^{infty}(-x)^nright)\
        &=sum_{n=1}^{infty}sum_{k=1}^nfrac1{k^3}(-x)^n\
        &=sum_{n=1}^{infty}left[sum_{k=1}^{infty}frac1{k^3}-sum_{k=n+1}^{infty}frac1{k^3}right](-x)^n\
        &=sum_{n=1}^{infty}left[sum_{k=0}^{infty}frac1{(1+k)^3}-sum_{k=0}^{infty}frac1{(1+n+k)^3}right](-x)^n\
        &=sum_{n=1}^{infty}left[-frac12psi^{(2)}(1)+frac12psi^{(2)}(1+n)right](-x)^n\
        end{align}$$




        $$frac{operatorname{Li}_3(-x)}{1+x}=frac12sum_{n=1}^{infty}[psi^{(2)}(1+n)-psi^{(2)}(1)](-x)^n$$







        share|cite|improve this answer









        $endgroup$



        Combining Maxim's suggestion to define the $0^{th}$ coefficient of the Trilogarithm series to be zero and ysharifi's given proof here on AoPS it is rather simple to obtain the given expansion. Using the Cauchy Product leads to



        $$begin{align}
        operatorname{Li}_3(-x)cdotfrac1{1+x}&=left(sum_{n=1}^{infty}frac{(-x)^n}{n^3}right)cdotleft(sum_{n=0}^{infty}(-x)^nright)\
        &=sum_{n=1}^{infty}sum_{k=1}^nfrac1{k^3}(-x)^n\
        &=sum_{n=1}^{infty}left[sum_{k=1}^{infty}frac1{k^3}-sum_{k=n+1}^{infty}frac1{k^3}right](-x)^n\
        &=sum_{n=1}^{infty}left[sum_{k=0}^{infty}frac1{(1+k)^3}-sum_{k=0}^{infty}frac1{(1+n+k)^3}right](-x)^n\
        &=sum_{n=1}^{infty}left[-frac12psi^{(2)}(1)+frac12psi^{(2)}(1+n)right](-x)^n\
        end{align}$$




        $$frac{operatorname{Li}_3(-x)}{1+x}=frac12sum_{n=1}^{infty}[psi^{(2)}(1+n)-psi^{(2)}(1)](-x)^n$$








        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 16 '18 at 18:58









        mrtaurhomrtaurho

        5,98551641




        5,98551641






























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