COMPUTING SOME INFINITE SUMS WITH THE AID OF POLYGAMMA FUNCTION

The initial goal of this post was to proof the following relations seen in this Twitter post.

$\begin{aligned} &\sum_{n=0}^{\infty} \frac{1}{(3 n+1)^{3}}=\frac{117 \zeta(3)+2 \sqrt{3} \pi^{3}}{243} \\ &\sum_{n=0}^{\infty} \frac{1}{(3 n+2)^{3}}=\frac{117 \zeta(3)-2 \sqrt{3} \pi^{3}}{243} \\ &\sum_{n=0}^{\infty} \frac{1}{(4 n+1)^{3}}=\frac{28 \zeta(3)+\pi^{3}}{64} \\ &\sum_{n=0}^{\infty} \frac{1}{(4 n+3)^{3}}=\frac{28 \zeta(3)-\pi^{3}}{64} \\ &\sum_{n=0}^{\infty} \frac{1}{(6 n+1)^{3}}=\frac{91 \zeta(3)+2 \sqrt{3} \pi^{3}}{216} \\ &\sum_{n=0}^{\infty} \frac{1}{(6 n+5)^{3}}=\frac{91 \zeta(3)-2 \sqrt{3} \pi^{3}}{216} \end{aligned}$

However, since I wanted to show all the details of the computation, we will only proof the third and fourth relations, and leave the other four for a future post.

So, in order to proof these two results we will first proof the well known Legendre duplication formula for The Gamma function and then compute some special values of the Quadrigamma(?) function. After these results are established we will be able to proof the two above mentioned relations. As a bonus, in the end of the post we will also compute the value of the Dirichlet beta function of order 3 that comes easily as a consequence of the evaluation of these two infinite sums.

It´s a long post, but I think it´s worth. Hope you enjoy it!

Legendre Duplication Formula for the Gamma Function

Recall the Beta function

$B(x,y)=\int_0^1t^{x-1}(1-t)^{y-1}dt=\frac{\Gamma\left( x\right)\Gamma\left( y\right)}{\Gamma\left(x+y \right)}$

Let $x=y\, \Rightarrow$

$B(x,x)=\int_0^1t^{x-1}(1-t)^{x-1}dt=\frac{\Gamma^2\left( x\right)}{\Gamma\left(2x \right)}$ (1)

$\int_0^1t^{x-1}(1-t)^{x-1}dt=\int_0^{1/2}t^{x-1}(1-t)^{x-1}dt+\int_{1/2}^1t^{x-1}(1-t)^{x-1}dt$

let u=1-t in the second integral, then

$B(x,x)=\int_0^{1/2}t^{x-1}(1-t)^{x-1}dt-\int_{1/2}^0(1-u)^{x-1}u^{x-1}(-du)$

$B(x,x)=2\int_0^{1/2}\left(t(1-t)\right)^{x-1}dt$

Now let $t=\frac{1-\sqrt{w}}{2} \,\,$

Then

$&\qquad w=\sqrt{1-2t}}\\ & \\ &\text{then when \,\,} t=0 \Rightarrow w=1 \\ & \\ &\text{when \,\,} t=\frac{1}{2} \Rightarrow w=o\\ & \\ &\text{and \,\,}dt=-\frac{dw}{4\sqrt{w}}$

Note also that

$t(1-t)=\left(\frac{1}{2}-\frac{w^{1/2}}{2}\right)\left(1-\left(\frac{1}{2}-\frac{w^{1/2}}{2} \right)\right)$

$=\left(\frac{1}{2}-\frac{w^{1/2}}{2}\right)\left(\frac{1}{2}+\frac{w^{1/2}}{2} \right)=\frac{\left(1-w^{1/2}\right)\left(1+w^{1/2}\right)}{4}$

and

$t(1-t)=\frac{(1-w)}{4}$

therefore the integral becomes

$2\int_0^{1/2}\left(t(1-t)\right)^{x-1}dt=\frac{1}{2^{2x-1}}\int_0^1w^{-1/2}(1-w)^{x-1}dw=\frac{1}{2^{2x-1}}B\left(\frac{1}{2},x \right)$ (2)

equating (1) and (2) we get

$\frac{\Gamma^2\left( x\right)}{\Gamma\left(2x \right)}=\frac{2^{1-2x}\Gamma\left( x\right)\Gamma\left( \frac{1}{2}\right)}{\Gamma\left( x+\frac{1}{2}\right)}$

$\boxed{\Gamma\left(2x \right)=\frac{2^{2x-1}\Gamma\left( x\right)\Gamma\left( x+\frac{1}{2}\right)}{\sqrt{\pi}}}$ (3)

Which is Legendre Duplication Formula for the Gamma Function

Duplication formula for Polygamma function

Now take logs in both sides of (3)

$\ln \Gamma\left(2x \right)=(2x-1)\ln 2 + \ln \Gamma\left( x\right) + \ln \Gamma\left( x+\frac{1}{2}\right) - \frac{1}{2} \ln \pi$ (4)

differentiating (4) with respect to x we get

$2 \psi \left(2x \right)=2 \ln 2+ \psi(x)+\psi \left( x+\frac{1}{2}\right)$ (5)

differentiating (5) with respect to x we get

$4\psi^{\prime} \left(2x \right)= \psi^{\prime}(x)+\psi^{\prime} \left( x+\frac{1}{2}\right)$ (6)

differentiating (6) with respect to x we get

$8\psi^{\prime \prime} \left(2x \right)= \psi^{\prime \prime}(x)+\psi^{\prime \prime} \left( x+\frac{1}{2}\right)$ (7)

Recurrence equation for Polygamma

Recall the recurrence equation of the Gamma Function

$\Gamma(x+1)=x\Gamma(x)$ (8)

Taking log of (8) and differentiating with respect to we get

$\psi(x+1)=\frac{1}{x}+\psi(x)$ (9)

differentiating (9) with respect to x we obtain

$\psi^{\prime}(x+1)=-\frac{1}{x^2}+\psi^{\prime}(x)$ (10)

differentiating (10) with respect to x we obtain

$\psi^{\prime \prime}(x+1)=\frac{2}{x^3}+\psi^{\prime \prime}(x)$ (11)

Infinite series representation of Polygamma

Recall also the infinite series representation of the Digamma function

$\psi(x+1)=-\gamma+\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+x} \right)$ (12)

If we enforce $n \longrightarrow n+1$ in (12) we get

$\psi(x+1)=-\gamma+\sum_{n=0}^{\infty}\frac{1}{n+1}-\sum_{n=0}^{\infty}\frac{1}{n+1+x}$

Now substitute the L.H.S of the above equation by (9)

$\psi(x)+\frac{1}{x}=-\gamma+\sum_{n=0}^{\infty}\frac{1}{n+1}-\sum_{n=0}^{\infty}\frac{1}{n+1+x}$

$\psi(x)=-\gamma+\sum_{n=0}^{\infty}\frac{1}{n+1}-\frac{1}{x}-\sum_{n=0}^{\infty}\frac{1}{n+1+x}$

$\psi(x)=-\gamma+\sum_{n=0}^{\infty}\frac{1}{n+1}-\sum_{n=-1}^{\infty}\frac{1}{n+1+x}$

Shiftting the index of the second sum on the R.H.S

$\psi(x)=-\gamma+\sum_{n=0}^{\infty}\frac{1}{n+1}-\sum_{n=0}^{\infty}\frac{1}{n+x}$

$\psi(x)=-\gamma+\sum_{n=0}^{\infty}\left(\frac{1}{n+1}-\frac{1}{n+x} \right)$ (12´)

differentiating (12) and (12´) with respect to x we obtain

$\psi^{\prime}(x+1)=\sum_{n=1}^{\infty}\frac{1}{(n+x)^2}$ (13)

$\psi^{\prime}(x)=\sum_{n=0}^{\infty}\frac{1}{(n+x)^2}$ (13´)

differentiating (13) and (13´) with respect to x we obtain

$\psi^{\prime \prime}(x+1)=-2\sum_{n=1}^{\infty}\frac{1}{(n+x)^3}$ (14)

$\psi^{\prime \prime}(x)=-2\sum_{n=0}^{\infty}\frac{1}{(n+x)^3}$ (14´)

Reflection formula of Polygamma

Recall the reflection formula of the Gamma function

$\Gamma\left( x\right)\Gamma\left( 1-x\right)=\frac{\pi}{\sin (\pi x)}$ (15)

Taking logs of both sides of (15) and then differentiating with respect to x we obtain

$\psi(x)-\psi(1-x)=-\pi \cot(\pi x)$ (16)

Differentiating (16) with respect to x we get

$\psi^{\prime}(x)+\psi^{\prime}(1-x)=\frac{\pi^2} {\sin^2(\pi x)}$ (17)

Differentiating (17) with respect to x we get

$\psi^{\prime \prime}(x)-\psi^{\prime \prime}(1-x)=-2\pi^3\frac{\cot(\pi x)} {\sin^2(\pi x)}$ (18)

Special values of the $\psi^{\prime \prime}(x)$

If we let x=0 in (14) we obtain

$\psi^{\prime \prime}(1)=-2\sum_{n=1}^{\infty}\frac{1}{n^3}$

$\psi^{\prime \prime}(1)=-2 \zeta(3)$ (19)

Now, if we let $x=\frac{1}{2}$ in(7), we get

$8\psi^{\prime \prime} \left(1 \right)= \psi^{\prime \prime}\left(\frac{1}{2} \right)+\psi^{\prime \prime} \left( \frac{1}{2}+\frac{1}{2}\right)$

$\psi^{\prime \prime}\left(\frac{1}{2} \right)=7 \psi^{\prime \prime}(1)$

$\psi^{\prime \prime}\left(\frac{1}{2} \right)=-14 \zeta(3)$ (20)

Now let $x=\frac{1}{4}$ in (7)

$8\psi^{\prime \prime} \left(\frac{1}{2} \right)= \psi^{\prime \prime}\left(\frac{1}{4} \right)+\psi^{\prime \prime} \left( \frac{3}{4}\right)$

$\psi^{\prime \prime}\left(\frac{1}{4} \right)+\psi^{\prime \prime} \left( \frac{3}{4}\right)=-112 \zeta(3)$ (21)

On the other hand setting $x=\frac{1}{4}$ in (18) we obtain

$\psi^{\prime \prime}\left(\frac{1}{4}\right)-\psi^{\prime \prime}\left(\frac{3}{4}\right)=-2\pi^3\frac{\cot\left(\frac{\pi}{4}\right)} {\sin^2\left(\frac{\pi}{4}\right)}$

$\psi^{\prime \prime}\left(\frac{1}{4}\right)-\psi^{\prime \prime}\left(\frac{3}{4}\right)=-4\pi^3$ (22)

Adding (21) and (22)

$\psi^{\prime \prime}\left(\frac{1}{4}\right)=-2 \pi^3 -56 \zeta(3)$ (23)

Subtracting (22) from (21) we get

$\psi^{\prime \prime}\left(\frac{3}{4}\right)=2 \pi^3 -56 \zeta(3)$ (24)

We have the beautiful results

$\psi^{\prime \prime}(1)=-2 \zeta(3)$

$\psi^{\prime \prime}\left(\frac{1}{2} \right)=-14 \zeta(3)$

$\psi^{\prime \prime}\left(\frac{1}{4}\right)=-2 \pi^3 -56 \zeta(3)$

$\psi^{\prime \prime}\left(\frac{3}{4}\right)=2 \pi^3 -56 \zeta(3)$

We can now proof the following results

$\sum_{n=0}^{\infty} \frac{1}{(4 n+1)^{3}}=\frac{28 \zeta(3)+\pi^{3}}{64}$ (25)

$\sum_{n=0}^{\infty} \frac{1}{(4 n+3)^{3}}=\frac{28 \zeta(3)-\pi^{3}}{64}$ (26)

lets start with (25)

$\sum_{n=0}^{\infty} \frac{1}{(4 n+1)^{3}}=\frac{1}{64}\sum_{n=0}^{\infty} \frac{1}{( n+\frac{1}{4})^{3}}$

From (14´) we know that

$\sum_{n=0}^{\infty}\frac{1}{(n+x)^3}=-\frac{\psi^{\prime \prime}\left(x\right)}{2}$

$\sum_{n=0}^{\infty} \frac{1}{(4 n+1)^{3}}=-\frac{\psi^{\prime \prime}\left(\frac{1}{4}\right)}{128}$

$\sum_{n=0}^{\infty} \frac{1}{(4 n+1)^{3}}=\frac{2 \pi^3 +56 \zeta(3)}{128}$

And (25) is proved

$\boxed{\sum_{n=0}^{\infty} \frac{1}{(4 n+1)^{3}}=\frac{28 \zeta(3)+\pi^{3}}{64}}$

Similarly (26)

$\sum_{n=0}^{\infty} \frac{1}{(4 n+3)^{3}}=\frac{1}{64}\sum_{n=0}^{\infty} \frac{1}{( n+\frac{3}{4})^{3}}$

$\sum_{n=0}^{\infty} \frac{1}{(4 n+3)^{3}}=-\frac{\psi^{\prime \prime}\left(\frac{3}{4}\right)}{128}$

Plugging (24) we proof (26)

$\boxed{\sum_{n=0}^{\infty} \frac{1}{(4 n+3)^{3}}=\frac{28 \zeta(3)-\pi^{3}}{64}}$

Once we established (25) and (26) we can proof another well known infinite sum, namely

$\beta(3)=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2 n+1)^{3}}$ (27)

Dirichlet beta function of order 3. First lets split (27) between odd and even terms

$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2 n+1)^{3}}=\sum_{n=0}^{\infty} \frac{(-1)^{2n}}{(2( 2n)+1)^{3}}+\sum_{n=0}^{\infty} \frac{(-1)^{2n+1}}{(2 (2n+1)+1)^{3}}$

$=\sum_{n=0}^{\infty} \frac{1}{(4n+1)^{3}}-\sum_{n=0}^{\infty} \frac{1}{( 4n+3)^{3}}$

Which is exactly the difference between

$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2 n+1)^{3}}=\left(\frac{28 \zeta(3)+\pi^{3}}{64}\right)-\left(\frac{28 \zeta(3)-\pi^{3}}{64}\right)$

and finally

$\boxed{\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)^3}=\frac{\pi^3}{32}}$
(28)

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