This is inspired by the same beautiful integral expression for $\zeta(3)$ as this question, but goes in a slightly different direction. Writing the original integral in the form $$\int_0^1\frac{x(1-x)}{\sin\pi x}dx=7\frac{\zeta(3)}{\pi^3} ,$$ it turns out that for $n\in\mathbb N$ there is a unique monic polynomial $p_n$ of degree $n-1$ such that $$\int_0^1x^np_n(x)\frac{1-x}{\sin\pi x}dx=c_{2n+1}\frac{\zeta(2n+1)}{\pi^{2n+1} } $$ with rational $c_k=4(k-1)! \dfrac{2^k-1}{2^k}= (4-2^{2-k})(k-1)! $.

This follows for $\zeta(2n+1)$ from solving the linear system given by the blue lines numbered $n+1,...,2n$ in the other question. As Zurab Silagadze answered it by giving an explicit formula of the coefficients in the blue lines, the $p_n$'s can be calculated. I don't know however if it is possible to give a formula in closed form (meaning here that it should not contain a matrix inversion), but see below.

The first polynomials are $$\begin{align} p_1(x)&=1 \\ p_2(x)&=3-x \\ p_3(x)&=25-20x+x^2 \\ p_4(x)&=455-707x+287x^2-x^3 \\ p_5(x)&=14301-34734x+29046x^2-8304x^3+x^4 \\ p_6(x)&=683067-2289309x+2949276x^2-1721434x^3+382547x^4-x^5 \\ \end{align}$$

The constant terms are supposedly the sequence A272482, thus, correcting the oeis typo $1/(2n)!$, $$[x^0]p_n(x)= {(2n)!}[x^{2n}y^n]\frac{\cos\frac{x(1-y)}{2}} {\cos\frac{x(1+y)}{2}} = \frac 1{4^n} {2n\choose n}\sum_{i=0}^n{n\choose i}E_i,$$ where $E_i$ are the Euler numbers. This seems to suggest something similar for the other coefficients, and thus possibly a closed form.

Are the $p_n$ known? How to find their closed form or generating function?

More generally now, define $$J(m,n,k)=J(n,m,k):= \int_0^1\frac{x^m(1-x)^n}{\sin^k\pi x}dx.$$For this to converge, we need $m,n\geqslant k$.
Experimentally, the situation for $k=2$ is quite similar to the $k=1$ case in that $J(m,n,k)$ is a rational combination of values $\dfrac{\zeta(i)}{\pi^{i+1}}$ with $i$ running over all odd numbers between $\min(m,n)$ and $m+n-1$, e.g. $$J(7,4,2)=\dfrac{105}2\left(-\dfrac{\zeta(5)}{\pi^6}+51\dfrac{\zeta(7)}{\pi^8}-405\dfrac{\zeta(9)}{\pi^{10}}\right).$$

This leads to new possibilities of representing odd zeta values as integrals, this time with $\sin^2\pi x$ in the denominator. Writing as a shortcut $h_m:=J(m,m,2)$, we can for example express $\dfrac{\zeta(2n-1)}{\pi^{2n}}$ as a rational combination of $h_2,\dots,h_n$, i.e. as an integral $$\dfrac{\zeta(2n-1)}{\pi^{2n}}=\int_0^1q_n(x-x^2)\frac{x^2(1-x)^2}{\sin^2\pi x}dx,$$ where $q_n$ is a unique polynomial of degree $n-2$. The first of them are: $$\begin{align} \zeta(3)&=\frac{\pi^4}{6}h_2 \\ \zeta(5)&=\frac{\pi^6}{90}(h_2 +2h_3),\qquad \text{ i. e. } q_2(z)=\frac{1}{90}(1+2z) \quad \text{ etc. }\\ \zeta(7)&=\frac{\pi^8}{1890}(2h_2 +4h_3+3h_4)\\ \zeta(9)&=\frac{\pi^{10}}{28350}(3h_2 +6h_3+5h_4+2h_5)\\ \zeta(11)&=\frac{\pi^{12}}{935550}(10h_2 +20h_3+17h_4+8h_5+2h_6)\\ \zeta(13)&=\frac{\pi^{14}}{638512875}(\color{blue}{691}h_2 +1382h_3+1180h_4+574h_5+175h_6+30h_7)\\ \end{align}$$

Experimentally, in $\dfrac{\zeta(2n-1)}{\pi^{2n}}$ the last coefficient (i.e. the one of $h_n$ and the leading term of $q_n$) is $\dfrac{2^{2n-2}}{(2n)!}$ and the one preceding it is $\dfrac{n(n-2)}6\dfrac{2^{2n-2}}{(2n)!}$, while for the first coefficient (equally, the constant term of $q_n$), the occurrence of $\color{blue}{691}$ in the expression for $\zeta(13)$ suggests that it involves the Bernoulli number $B_{2n-2}$.

Any ideas about these polynomials?

Finally, for $k\geqslant 3$ there does not seem to exist any closed form, at least not in terms of zeta values.

What about $J({3,3,3})= \int\limits_0^1\dfrac{x^3(1-x)^3}{\sin^3\pi x}dx$?

  • 6
    $\begingroup$ Maple says $$J(3,3,3) = -{\frac {279\,\zeta \left( 5 \right) }{2\,{\pi}^{5}}}+{\frac {126\, \zeta \left( 3 \right) }{{\pi}^{5}}}+{\frac {5715\,\zeta \left( 7 \right) }{4\,{\pi}^{7}}}-{\frac {1395\,\zeta \left( 5 \right) }{{\pi} ^{7}}} $$ $\endgroup$ May 29, 2017 at 18:54
  • 1
    $\begingroup$ ... and it seems to get $J(n,n,n)$ for all positive integers $n$ as a linear combination of odd $\zeta$ values with coefficients involving $\pi$. $\endgroup$ May 29, 2017 at 19:04
  • $\begingroup$ @RobertIsrael Would you believe it, such an "inhomogeneous" form?! That's just incredible... At first glance, this might even raise some doubts if $\pi^2\zeta(3)$ and $\zeta(5)$ are really linearly independent $-$ but only at first glance, noting that the coefficients here occur over and over again in other $J(m,n,k)$'s, and particularly that $\zeta(n)$ is often accompanied by a factor $2^n-1$. $\endgroup$
    – Wolfgang
    May 29, 2017 at 19:51
  • 3
    $\begingroup$ Not sure if this helps, but maybe the following provides a clue. If we combine: $$\zeta(2n+1)=\frac{\pi^{2n+1}}{4(1-2^{-2n-1})\,(2n)!)}\int_0^1 \frac{(-1)^n\,E_{2n}(u)}{\sin(\pi u)}\,du $$ from Zurab's paper, with your formula above: $$\int_0^1\frac{x^np_n(x)(1-x)}{\sin(\pi x)}dx=c_{2n+1}\frac{\zeta(2n+1)}{\pi^{2n+1} }$$ we get: $$\int_0^1\frac{x^np_n(x)(1-x)}{\sin(\pi x)}dx=\int_0^1 \frac{(-1)^n\,E_{2n}(u)}{\sin(\pi u)}\,du$$ hence: $$\int_0^1x^np_n(x)(1-x)dx=\int_0^1 (-1)^n\,E_{2n}(u)\,du$$ and the LHS-integral has a closed form in $n$. $\endgroup$
    – Agno
    May 30, 2017 at 14:22
  • 2
    $\begingroup$ @RobertIsrael Subsequently to your finding of $J(3,3,3)$, here is what I seem to have found for the general situation: It appears that $$J(m,n,k)= \sum\limits_{i=1}^{[\frac{k+1}2]}\sum\limits_{j=[\frac{\min(m,n)-k+2}2]}^{ [\frac{m+n-k+1}2]}a_{ij}\dfrac{\zeta(2i+2j+1)}{\pi^{k+2j}}$$ with non-zero rationals $a_{ij}$. I have done them all numerically for $k\le8$ and $ m,n\le10$. For even $k$, the lines of the matrix $(a_{ij})$ are proportional: $k=4\Rightarrow$ two identical lines up to sign, $k=6\Rightarrow$ three lines with ratios $1:-5:4$ and $k=8$ four lines with ratios $1:-14:49:-36$. :) $\endgroup$
    – Wolfgang
    May 30, 2017 at 16:25

2 Answers 2


Using the reflection formula followed by the recurrence formula and the Beta integral representation (DLMF) \begin{align} \frac{x(1-x)}{\sin \pi x}&=\frac{1}{\pi}x(1-x)\Gamma(x)\Gamma(1-x)\\ &=\frac{1}{\pi}\Gamma(x+1)\Gamma(2-x)\\ &=\frac{2}{\pi}B(1+x,2-x)\\ &=\frac{2}{\pi}\int_0^\infty \frac{t^x}{(1+t)^3}\,dt\\ &=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{e^{u(2x-1)}}{\cosh^3u}\,du \end{align} last expression is obtained with $t=e^{2u}$. Then, for calculating \begin{equation} I_f=\int_0^1\frac{x(1-x)}{\sin \pi x}f(x)\,dx \end{equation} one can express \begin{align} I_f&=\frac{1}{2\pi}\int_0^1\int_{-\infty}^{\infty}\frac{e^{u(2x-1)}}{\cosh^3u}\,duf(x)\,dx\\ &=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{e^{-u}}{\cosh^3u}\,du\int_0^1e^{2ux}f(x)\,dx\\ &=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{e^{-u}}{\cosh^3u}F(u)\,du \label{eq:intf} \end{align} assuming that the change of the integration order is valid and denoting \begin{equation} F(u)=\int_0^1e^{2ux}f(x)\,dx \end{equation} If $F(u)$ is analytic in the half-plane $\Im(u)>0$ and $\left|F(u)\right|=o\left(\frac{e^{4u}}{u} \right)$ for $\left|u\right|\to \infty$,$I_f$ is evaluated by integrating along the real axis closed by the semi-large circle $\Im(u)>0$, using the residue method. Poles are situated at $u_n=i(2n+1)\pi/2$ with $n=0,1,2...$. Residues are $1/2F''(i(2n+1)\pi/2)-F'(i(2n+1)\pi/2)$, where $F''(z)$ and $F'(z)$ are respectively the first and second derivative of $F(z)$. As the half-circle contribution vanishes, it comes \begin{equation} I_f=i\sum_{n=0}^\infty\left[\frac{1}{2}F''(i(2n+1)\frac{\pi}{2})-F'(i(2n+1)\frac{\pi}{2})\right] \end{equation} When $f(x)=1$, to express the original integral, $F(u)=\frac{e^{2u}-1}{2u}$, a simple calculation shows, as expected, that \begin{equation} I=\frac{8}{\pi^3}\sum_{n=0}^\infty \frac{1}{(2n+1)^3}=\frac{7\zeta(3)}{\pi^3} \end{equation}

Another expression for the result is obtained by derivation under the integral \begin{equation} I_f=-2i\sum_{n=0}^\infty \int_0^1x(1-x)f(x)e^{i(2n+1)\pi x}\,dx \end{equation} For a real function $f$, as the summation should be real, it is sufficient to keep the imaginary contribution to the integral: \begin{equation} I_f=2\sum_{n=0}^\infty \int_0^1x(1-x)f(x)\sin\left( (2n+1)\pi x \right)\,dx \end{equation}

Now, suppose that a function $f_p$ is known such that the integrals \begin{equation} J_p=\int_0^1x(1-x)f_p(x)\sin\left( (2n+1)\pi x \right)\,dx=\frac{A_p}{(2n+1)^{2p+1}} \end{equation} which gives the relation \begin{equation} I_{f_p}=2A_p\sum_{n=0}^\infty\frac{1}{(2n+1)^{2p+1}}=2A_p\left( 1-2^{-2p-1} \right)\zeta(2p+1) \end{equation} Denoting the function $Q^0(x)=x(1-x)f_p(x)$ and $Q^1,Q^2(x)$ its first and second antiderivative. Two successive integrations by parts can be performed: \begin{align} J_p&=-(2n+1)\pi \int_0^1Q^1(x)\cos\left( (2n+1)\pi x \right)\,dx\\ &=(2n+1)\pi\left[Q^2(1)+Q^2(0)\right]-(2n+1)^2\pi^2\int_0^1Q^2(x)\sin\left( (2n+1)\pi x \right)\,dx \label{eq:jp} \end{align} The free parameters in $Q^2(x)$ can be chosen in order that $Q^2(1)=Q^2(0)=0$. With \begin{equation} Q^2(z)=\int_0^zdt\int_0^tQ^0(u)\,du+az+b \end{equation} one may chose $b=0$ and $a=-\int_0^1\,dt\int_0^tQ^0(u)\,du$. Thus \begin{equation} Q^2(z)=\int_0^zdt\int_0^tQ^0(u)\,du-z\int_0^1\,dt\int_0^tQ^0(u)\,du \end{equation} If $f_p(x)$ is a polynomial, then $Q^2(x)$ also. By construction, $x=0$ and $x=1$ are among its roots. It can be written as \begin{equation} Q^2(x)=x(1-x)f_{p+1}(x) \end{equation} or \begin{equation} f_{p+1}(x)=\frac{\int_0^xdt\int_0^tu(1-u)f_p(u)\,du-x\int_0^1\,dt\int_0^tu(1-u)f_p(u)\,du}{x(1-x)} \end{equation} $J_p$ can be written as \begin{equation} J_p=-(2n+1)^2\pi^2\int_0^1x(1-x)f_{p+1}(x)\sin\left( (2n+1)\pi x \right)\,dx \end{equation} One obtain \begin{equation} \int_0^1x(1-x)f_{p+1}(x)\sin\left( (2n+1)\pi x \right)\,dx=-\frac{1}{\pi^2}\frac{A_p}{(2n+1)^{2p+3}} \end{equation} and thus \begin{equation} I_{f_{p+1}}=2A_{p+1}\left( 1-2^{-2p-3} \right)\zeta(2p+3) \end{equation} with \begin{equation} A_{p+1}=-\frac{A_p}{\pi^2} \end{equation} Starting from $f_1(x)=1$ one obtains \begin{align} f_2(x)&=\frac{1}{12}(x^2-x-1)\\ f_3(x)&=\frac{1}{360}(x^4-2x^3-2x^2+3x+3)\\ f_4(x)&=\frac{1}{20160}(x^6-3x^5-3x^4+11x^3+11x^2-17x-17)\\ ... \end{align} which gives \begin{align} \int_0^1\frac{x(1-x)}{\sin\pi x}f_2(x)\,dx&=-\frac{31}{4}\frac{\zeta(5)}{\pi^5}\\ \int_0^1\frac{x(1-x)}{\sin\pi x}f_3(x)\,dx&=\frac{127}{16}\frac{\zeta(7)}{\pi^7}\\ \int_0^1\frac{x(1-x)}{\sin\pi x}f_4(x)\,dx&=-\frac{511}{64}\frac{\zeta(9)}{\pi^9}\\ ... \end{align} Starting from $f_1(x)=x(3-x)$, other series can be obtained. For example \begin{align} &f_2(x)=\frac{1}{60}(2x^4-10x^3+5x^2+5x+5)\\ &f_3(x)=-\frac{1}{5040}(3x^6-21x^2(x^3-x^2-x-1)-49(x+1))\\ &\int_0^1\frac{x(1-x)}{\sin\pi x}f_2(x)\,dx=-\frac{381}{4}\frac{\zeta(7)}{\pi^7}\\ &\int_0^1\frac{x(1-x)}{\sin\pi x}f_3(x)\,dx=-\frac{1533}{16}\frac{\zeta(9)}{\pi^9} \end{align} Other starting points can be obtained by choosing other members of the list proposed in the question above. For example, starting from $f_1(x)=x^2P_3(x)$ above leads to an apparent different expression for $\zeta(9)$: \begin{align} &f_2(x)=-\frac{1}{56}(x^6-27x^5+57x^4-13x^3-13x^2-13x-13)\\ &\int_0^1\frac{x(1-x)}{\sin\pi x}f_2(x)\,dx=-\frac{22995}{8}\frac{\zeta(9)}{\pi^9} \end{align} Obtained polynomials are not of the form $x^pP_p(x)$ as discussed in the question, however the above method may perhaps be adapted in this case.

EDIT 04/06/2017 : (sorry for the length of this answer...)

One may characterize more precisely the family of these polynomials. It helps to symmetrize the expressions: \begin{equation} I_f=\int_0^1\frac{x(1-x)}{\sin \pi x}f(x)\,dx=\frac{1}{8}\int_{-1}^1\frac{1-y^2}{\cos\pi y/2}g(y)\,dy \end{equation} with $g(y)=f(x)$ and $x=(1+y)/2$. In this form it is clear that odd contribution of the polynomial $g(y)$ vanishes. The same symmetrization for the proposed decomposition above reads: \begin{equation} I_f=\frac{(-1)^n}{4}\sum_{n=0}^\infty \int_{-1}^1(1-y^2)g(y)\cos\left( (2n+1) \frac{\pi y}{2} \right)\,dy \end{equation} One may adapt the method developed above. If $g_p(y)$ is an even polynomial such as \begin{equation} \sum_{n=0}^\infty \int_{-1}^1(1-y^2)g_p(y)\cos\left( (2n+1) \frac{\pi y}{2} \right)\,dy=\frac{A_p}{(2n+1)^{2p+3}} \end{equation} then, by integrating twice by part, the polynomial \begin{equation} g_{p+1}(y)=\frac{\int_{-1}^ydt\int_{-1}^t(1-u^2)g_p(u)\,du-\frac{y+1}{2}\int_{-1}^1\,dt\int_{-1}^t(1-u^2)g_p(u)\,du}{1-y^2} \end{equation} is such that \begin{equation} \sum_{n=0}^\infty \int_{-1}^1(1-y^2)g_{p+1}(y)\cos\left( (2n+1)\frac{\pi y}{2} \right)\,dy=\frac{A_{p+1}}{(2n+1)^{2p+5}} \end{equation} with $A_{p+1}=-4A_p/\pi^2$. thus \begin{equation} \frac{1}{8}\int_{-1}^1\frac{1-y^2}{\cos\pi y/2}g_{p+1}(y)\,dy=2A_{p+1}\left( 1-2^{-2p-3} \right)\zeta(2p+5) \end{equation} One may show that $g(y)$ is an even polynomial of $y$. For $g_0(y)=1$ one has, as expected \begin{equation} \frac{1}{8}\int_{-1}^1\frac{1-y^2}{\cos\pi y/2}\,dy= \frac{7\zeta(3)}{\pi^3} \end{equation} Then, the recurrence above produces a series of even polynomials $g_p(y)$ of degree $2p$ giving successive integral expressions for $\zeta(2p+3)$. Due to the parity remark, one can conclude that any polynomial $Q(y)$, with its even power coefficient identical to that of $g_p(y)$, is such that \begin{equation} \int_{-1}^1\frac{1-y^2}{\cos\pi y/2}Q(y)\,dy=\left( -1 \right)^p8\left( 2^{2p+3}-1 \right)\frac{\zeta(2p+3)}{\pi^{2p+3}} \end{equation} The condition reads \begin{equation} Q(y)+Q(-y)=2g_p(y) \end{equation} The first polynomials (written with $Y=y^2$) are: \begin{align} g_0(y)&=1\\ g_1(y)&=\frac{1}{12}\left( Y-5 \right)\\ g_2(y)&=\frac{1}{360}\left( Y^2-14Y+61 \right)\\ g_3(y)&=\frac{1}{20160}\left( Y^3-27Y^2+323Y-1385 \right)\\ g_4(y)&=\frac{1}{1814400}\left( Y^4-44Y^3+1006Y^2-11804Y+50521 \right)\\ g_5(y)&=\frac{1}{239500800}\left( Y^{5}-65Y^4+2410Y^3-53954Y^2+631621Y-2702765\right)\\ g_6(y)&=\frac{1}{43589145600}\left(Y^6-90Y^5+4915Y^4-178268Y^3+3980887Y^2-46590634Y+199360981 \right) \end{align} In terms of the non-symmetrized function, any polynomial of the form \begin{equation} f(x)=g_p\left( 2x-1 \right)+P(2x-1) \end{equation} where $P(z)$ is an arbitrary odd polynomial, gives a result proportional to $\zeta(2p+1)$ when integrated as in $I_f$ defined above.

  • 1
    $\begingroup$ Very nice! When I first looked at it, I obtained experimentally the same polynomials that start with $f_1\equiv1$. By noticing that the coefficients come in pairs (i.e. that $f_{n+1}(x)$ minus the leading $x^{2n}$ term has a factor $(x+1)$), I then figured that there are some degrees of freedom, and if we want to obtain unique polynomials, we must add constraints, e.g. require the last $n$ terms to vanish. So I came up with the factor $x^n$ under the integral. BTW I happened to do all that shortly before Zurab Silagadze posted his answer to the other question, so what was timely. :) $\endgroup$
    – Wolfgang
    Jun 3, 2017 at 8:32
  • 1
    $\begingroup$ Thank you! I had a lot of fun. I understand the constraint you imposed to the polynomial factor. There can exist many different class of solutions. I just added a starting point at $f_1=x^2p_3(x)$ in your list to show that different expressions can be found. As you noticed in your comment, at least with this method, the structure of the polynomials seems to be very characteristic. I also corrected some typos. $\endgroup$
    – Paul Enta
    Jun 3, 2017 at 9:33
  • 1
    $\begingroup$ @Paul Enta. I believe the closed form for your polynomials is: $$g_n(y)=-\frac{1}{(n+1)\,(2n+1)!}\,\sum_{m=0}^{n+1}\,y^{2m}\sum_{k=0}^{m}\,E \big( 2\,(n+1-k) \big)\binom{2(n+1)}{2k}$$ Where $E(x)$ is an Euler number. If my formula is indeed correct, I like to conjecture that you made a typo in $g_5$ (that is also indexed wrongly by naming it $g_2$), where the coefficient $95$ should actually be $65$. Grateful if you could confirm. Thanks! $\endgroup$
    – Agno
    Jun 5, 2017 at 22:52
  • $\begingroup$ @Agno You are perfectly right. I corrected the typos in $g_5$ and add the expression for $g_6$ which fits your formula too. Congratulations. How did you get this result? $\endgroup$
    – Paul Enta
    Jun 6, 2017 at 20:51
  • $\begingroup$ Thanks Paul. I found the formula through "reverse engineering" with the help of this particular Sloane's integer sequence oeis.org/A086646. I discovered that the integers in this triangle correspond to your coefficients as finite sums of Euler numbers weighted by a binomial. One wishes a more direct mathematical proof of course, but I don't have that yet. Note that the closed form becomes even more elegant when you let the index start at 1. It then becomes: $$g_n(y)=-\frac{1}{n\,(2n-1)!}\,\sum_{m=0}^{n}\,y^{2m}\sum_{k=0}^{m}\,E \big(2n-2k \big)\binom{2n}{2k}$$. $\endgroup$
    – Agno
    Jun 6, 2017 at 21:06

In this recent paper on log tangent integrals, Theorem 1 (2.2, in the published version) expresses that, if $n$ is a positive integer, \begin{equation} \int_0^{\tfrac{\pi}{2}}E_{2n-1}\left( \frac{2}{\pi}x \right)\log(\tan x)\,dx= \frac{(-1)^{n-1}(2n-1)!}{\pi^{2n-1}}\left( 2-2^{-2n} \right)\zeta(2n+1) \end{equation} where $E_n(x)$ are the Euler polynomials. This expression can be written as \begin{equation} \int_0^{1}E_{2n-1}\left( x \right)\log(\tan \frac{\pi}{2}x)\,dx= \frac{(-1)^{n-1}2(2n-1)!}{\pi^{2n}}\left( 2-2^{-2n} \right)\zeta(2n+1) \end{equation} Defining the antiderivatives \begin{equation} F_{2n-1}(x)=\int_0^xE_{2n-1}\left( t \right)\,dt \end{equation} one may notice that $F_{2n-1}(0)=F_{2n-1}(1)=0$, as $E_{2n-1}( 1-x )=-E_{2n-1}( x )$. Then, integrating by parts, it comes \begin{equation} \int_0^1\frac{ F_{2n-1}(x)}{\sin\pi x}\,dx= \frac{(-1)^{n}4(2n-1)!}{\pi^{2n+1}}\left( 1-2^{-2n-1} \right)\zeta(2n+1) \end{equation} As $x=0,1$ are two roots of $F_{2n-1}(x)$, we conclude that the polynomials \begin{equation} f_{2n-1}(x)=\frac{1}{x(1-x)}\int_0^xE_{2n-1}\left( t \right)\,dt \end{equation} verify \begin{equation} \int_0^1 f_{2n-1}(x)\frac{x(1-x) }{\sin\pi x}\,dx= \frac{(-1)^{n}4(2n-1)!}{\pi^{2n+1}}\left( 1-2^{-2n-1} \right)\zeta(2n+1) \end{equation} More generally, for symmetry reasons, any polynomial \begin{equation} g_{2n-1}(x)=f_{2n-1}(x)+P(2x-1) \end{equation} where $P(x)$ is an arbitrary odd polynomial, gives the same result. This result gives an explicit representation of the polynomials derived in my previous answer.

Edit: Using the derivative property for the Euler polynomials, $E_{2n-1}(x)=(2n)^{-1}dE_{2n}(x)/dx$, one can express \begin{align} f_{2n-1}(x)&=\frac{1}{2n}\frac{1}{x(1-x)}\left[E_{2n}(x)-E_{2n}(0)\right]\\ &=\frac{1}{x(1-x)}\left[\frac{1}{2n}E_{2n}(x)+\frac{1}{n(2n+1)}\left( 2^{2n+1}-1 \right)B_{2n+1}\right] \end{align} where $B_{2n+1}$ is a Bernoulli number.

  • $\begingroup$ Very nice! I just see an inconsistency between the $f_n$ you define in your first answer and the $f_{2n-1}$ you define here. Apart from $n\leftrightarrow 2n-1$, you also lost a factor $-\frac2{(2n-1)!}$ somewhere, e.g. the $f_4$ of your first answer is wolframalpha.com/input/… $\endgroup$
    – Wolfgang
    Sep 27, 2017 at 13:32
  • $\begingroup$ Thanks! You are right, besides the change of the indices I realize that normalization of the polynomials in the previous answer was somewhat imprecise. I edited the present answer to add an explicit expression of the polynomials. By the way, this method may provide additional expressions, for instance: $$ \int_0^1 \frac{G_{2n-1}(x)}{\sin^2 \tfrac{\pi}{2}x}\,dx$$ is proportional to $\zeta(2n+1)$, with \begin{equation} G_{2n-1}(x)=\int_0^x \,du\int_0^uE_{2n-1}(t)\,dt \end{equation} $\endgroup$
    – Paul Enta
    Sep 27, 2017 at 22:09

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