On a Ramanujan Identity and Its Generalizations

Abstract

In the present paper, we propose a new method of derivation of number-theoretic identities which is applied to the proof of the multidimensional analogue of one of the Ramanujan identities. This method allows us to obtain new infinite series representations for the number $\pi$, of the values of the Riemann zeta function, and of the $L$-Dirichlet series at integer points.

1. Introduction

At present, many identities that connect infinite sums of terms of the form $f(n)\cosh^{-1}(cn)$, $f(n)\sinh^{-1}(cn)$, $n=1,2,3,\dots$ ($c>0$ is a constant, $f(n)$ is an arithmetic function) with values of the Riemann zeta function $\zeta(s)$, with Dirichlet $L$-series at integer points, as well as with the Bernoulli and Euler numbers are known ¹. A large number of identities of this type were obtained by Ramanujan, including the equality

$$\sum_{n=1}^{+\infty}\frac{\chi_4(n)}{n} \frac{1}{\cosh(\pi n/2)} =\sum_{\nu=0}^{+\infty}\frac{(-1)^\nu}{(2\nu+1)\cosh\pi(\nu+1/2)} =\frac{\pi}{8}\,, $$

(1.1)

in which $\chi_4(n)$ is a nonprincipal character modulo $4$, i.e., a $4$-periodic arithmetic function such that $\chi_4(1)=1$, $\chi_4(3)=-1$, $\chi_4(0)=\chi_4(2)=0$ (see [7, Chap. 14, Sec. 15]). Ramanujan’s notebooks also contain the following generalization (1.1):

$$\begin{aligned} \, &\alpha^{-k}\sum_{n=1}^{+\infty}\frac{\chi_4(n)}{n^{2k+1}} \frac{1}{\cosh(\alpha n)}+(-\beta)^{-k}\sum_{n=1}^{+\infty} \frac{\chi_4(n)}{n^{2k+1}}\frac{1}{\cosh(\beta n)} \\&\qquad =\frac{\pi}{4}\sum_{m=0}^k(-1)^m \frac{E_{2m}E_{2k-2m}}{(2m)!(2k-2m)!}\,\alpha^{k-m}\beta^m, \end{aligned}$$

which holds for any integer $k$ and for arbitrary positive $\alpha$ and $\beta$, related by the equality $\alpha\beta=\pi^2/4$ (the symbols $E_j$ denote Euler numbers; see [7, Chap. 14, Sec. 21(ii)]). In the present paper, we propose another generalization of identity (1.1). Namely, the following statement holds.

Theorem 1.

For any $r\ge 1$ , the following equality holds:

$$\sum_{\nu_1,\dots,\nu_r=0}^{+\infty} \frac{(-1)^{\nu_1+\dotsb+\nu_r}}{(\nu_1+1/2)\dotsb(\nu_r+1/2)} \frac{1}{\cosh((\pi/2)\sqrt{(\nu_1+1/2)^2+\dotsb+(\nu_r+1/2)^2}\,)} =\frac{1}{r+1}\biggl(\frac{\pi}{2}\biggr)^r. $$

(1.2)

The method on which the generalization was based is, apparently, of interest in itself and may serve as a tool for proving new number-theoretic identities.

2. Proofs of Identity (1.1)

We will give three different proofs of (1.1). The first proof is based on Cauchy’s residue theorem, the second is based on the functional equation for the Dirichlet function $L(s;\chi_4)$, and the third one is based on the “physical” proof, on the solution of the Laplace equation.

2.1. Proof 1: Cauchy’s Residue Theorem

We set

$$g(s)=(s\cos(\pi s)\cosh(\pi s))^{-1}$$

and consider the contour with vertices at the points $s=(1\pm i)N$, $s=0$, where the latter point is traversed along the arc of the circle of radius $\varepsilon$ (here $0<\varepsilon<1/2$ and $N\ge 2$ is an integer). The poles of first order $s=n+1/2$, $n=0,1,2,\dots,N-1$ with residues

$$\frac{(-1)^{n+1}}{\pi(n+1/2)\cosh\pi(n+1/2)}.$$

By Cauchy’s theorem,

$$2\pi i\sum_{n=0}^{N-1}\frac{(-1)^{n+1}}{\pi(n+1/2)\cosh\pi(n+1/2)} =\sum_{r=0}^3I_r,$$

where $I_0$ is the integral along the arc of the circle, $I_1$, $I_2$ are the integrals over the inclined parts of the contour, and $I_3$ is the integral over the vertical segment. Since $g(s)=s^{-1}+O(1)$ as $s\to 0$, we have

$$I_0=\int_{\pi/4}^{-\pi/4} \biggl(\frac{1}{\varepsilon e^{i\varphi}}+O(1)\biggr) \varepsilon e^{i\varphi}i\,d\varphi =-\frac{\pi i}{2}+O(\varepsilon).$$

Further, noting that $\cos(\pi xe^{\pi i/4})=\cosh(\pi xe^{-\pi i/4})$ for real $x$, we see that $I_1+I_2=0$. Finally, setting $s=N+it$, we have $\cos(\pi s)=(-1)^N\cosh(\pi t)$, $|{\cosh(\pi s)}|\ge\sinh(\pi N)$; hence

$$|I_3|\le\int_{-N}^N\frac{dt}{\sqrt{N^2+t^2}\cosh(\pi t)\sinh(\pi N)} \le\frac{2}{N\sinh(\pi N)} \int_0^{+\infty}\frac{dt}{\cosh(\pi t)}=O\biggl(\frac{1}{\sinh(\pi N)}\biggr).$$

Letting $N$ tend to infinity and $\varepsilon$ to zero, we obtain the required equality.

2.2. Proof 2: The Functional Equation

We set

$$f(s)=\biggl(\frac{2}{\pi}\biggr)^s\Gamma(s)L(s;\chi_4)L(s+1;\chi_4)$$

and consider the integral

$$J_1=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}f(s)\,ds,$$

where $1<c<2$ is an arbitrary fixed number. By virtue of the absolute convergence of the following series on the line of integration:

$$L(s;\chi_4)=\sum_{k=1}^{+\infty}\frac{\chi_4(k)}{k^s} =\sum_{k=0}^{+\infty}\frac{(-1)^k}{(2k+1)^s}\,,\qquad L(s+1;\chi_4)=\sum_{k=1}^{+\infty}\frac{\chi_4(k)}{k^{s+1}} =\sum_{k=0}^{+\infty}\frac{(-1)^k}{(2k+1)^{s+1}}$$

we have

$$\begin{aligned} \, J_1 &=\sum_{k,n=0}^{+\infty}\frac{(-1)^{k+n}}{(2n+1)} \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} \biggl(\frac{2}{\pi}\biggr)^s\frac{\Gamma(s)\,ds}{(2k+1)^s(2n+1)^s} \\& =\sum_{k,n=0}^{+\infty}\frac{(-1)^{k+n}}{(2n+1)} \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} \Gamma(s)\biggl(\pi(2k+1)\biggl(n+\frac{1}{2}\biggr)\biggr)^{-s}\,ds. \end{aligned}$$

Hence, using the Mellin transformation formula

$$e^{-x}=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\Gamma(s)x^{-s}\,ds$$

(see [8, Chap. II, Sec. 10.8, Example (b)]), we obtain

$$\begin{aligned} \, J_1 &=\sum_{k,n=0}^{+\infty}\frac{(-1)^{k+n}}{(2n+1)}\,e^{-\pi(2k+1)(n+1/2)} =\sum_{n=0}^{+\infty}\frac{(-1)^n}{2n+1}\,e^{-\pi(n+1/2)} \sum_{k=0}^{+\infty}(-1)^ke^{-2\pi k(n+1/2)} \\& =\sum_{n=0}^{+\infty}\frac{(-1)^n}{(2n+1)} \frac{e^{-\pi(n+1/2)}}{(1+e^{-2\pi(n+1/2)})} =\frac{1}{4}\sum_{n=0}^{+\infty} \frac{(-1)^n}{(n+1/2)\cosh(\pi(n+1/2))}\,. \end{aligned}$$

Denoting the original sum by $S$, we have $S=4J_1$. To evaluate the integral $J_1$, we move the integration to the straight line $\operatorname{Re}s=1-c$. Namely, we set

$$J_2=\frac{1}{2\pi i}\int_{1-c+i\infty}^{1-c-i\infty}f(s)\,ds.$$

Since $s=0$ is a unique singular point of $f(s)$ in the strip $1-c\le\operatorname{Re}s\le c$, in view of Cauchy’s theorem, we have

$$J_1+J_2=\operatorname{res}_{s=0}f(s).$$

But $f(s)=L(1;\chi_4)L(0;\chi_4)s^{-1}+O(1)$ as $s\to 0$, and hence

$$\operatorname{res}_{s=0}f(s)=L(1;\chi_4)L(0;\chi_4).$$

As is well known,

$$L(1;\chi_4)=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\dotsb =\frac{\pi}{4}\,.$$

Setting $s=1$ in the functional equation

$$L(1-s;\chi_4)=\biggl(\frac{\pi}{4}\biggr)^{1/2-s} \frac{\Gamma((s+1)/2)}{\Gamma(1-s/2)}\,L(s;\chi_4) $$

(2.1)

(see, for example, [9, Chap. I, Sec. 4.3, Theorem 1]), we can write

$$L(0;\chi_4)=\biggl(\frac{\pi}{4}\biggr)^{-1/2} \frac{L(1;\chi_4)}{\Gamma(1/2)} =\frac{2}{\sqrt{\pi}}\frac{1}{\sqrt{\pi}}\frac{\pi}{4} =\frac{1}{2}\,.$$

Thus, we have

$$J_1+J_2=\frac{\pi}{8}\,. $$

(2.2)

Finally, transforming the integral $J_2$, we obtain

$$\begin{aligned} \, J_2 &=\frac{1}{2\pi}\int_{+\infty}^{-\infty} \biggl(\frac{2}{\pi}\biggr)^{1-c+iu}\Gamma(1-c+iu)L(2-c+iu;\chi_4)L(1-c+iu;\chi_4)\,du \\& =-\frac{1}{2\pi}\int_{-\infty}^{+\infty} \biggl(\frac{2}{\pi}\biggr)^{1-c-it} \Gamma(1-c-it)L(2-c-it;\chi_4)L(1-c-it;\chi_4)\,dt \\& =-\frac{1}{2\pi i}\int_{c-1-i\infty}^{c-1+i\infty} \biggl(\frac{2}{\pi}\biggr)^{-s}\Gamma(-s)L(1-s;\chi_4)L(-s;\chi_4)\,ds \end{aligned}$$

(here we have put $s=c-1+it$). We note that, in the strip $0<c-1\le\operatorname{Re}s\le c$, the integrand has no singularities: the single pole $\Gamma(-s)$ at the point $s=1$ is canceled by the zero $L(-s;\chi_4)$. Therefore, moving the integration to the straight line $\operatorname{Re}s=c$, we find

$$J_2=-\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} \biggl(\frac{2}{\pi}\biggr)^{-s}\Gamma(-s)L(1-s;\chi_4)L(-s;\chi_4)\,ds =-\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}f(-s)\,ds.$$

By (2.1),

$$L(-s;\chi_4)=\biggl(\frac{\pi}{4}\biggr)^{-s-1/2} \frac{\Gamma(1+s/2)}{\Gamma((1-s)/2)}\,L(s+1;\chi_4). $$

(2.3)

Multiplying (2.1) by (2.3), we obtain

$$\begin{aligned} \, f(-s) &=\biggl(\frac{2}{\pi}\biggr)^{-s}\Gamma(-s) \biggl(\frac{\pi}{4}\biggr)^{1/2-s} \frac{\Gamma((s+1)/2)}{\Gamma(1-s/2)} \biggl(\frac{\pi}{4}\biggr)^{-s-1/2} \\&\qquad\qquad\times \frac{\Gamma(1+s/2)}{\Gamma((1-s)/2)}\,L(s;\chi_4)L(s+1;\chi_4) \\& =\frac{2^{3s}}{\pi^s} \frac{\Gamma(-s)\Gamma((s+1)/2)\Gamma(1+s/2)}{\Gamma(1-s/2)\Gamma((1-s)/2)}\,L(s;\chi_4)L(s+1;\chi_4). \end{aligned}$$

By virtue of the duplication formula for the gamma function, we have

$$\Gamma\biggl(\frac{s+1}{2}\biggr) \Gamma\biggl(1+\frac{s}{2}\biggr)=2^{-s}\sqrt{\pi}\,\Gamma(s+1),\qquad \Gamma\biggl(\frac{1-s}{2}\biggr)\Gamma\biggl(1-\frac{s}{2}\biggr) =2^s\sqrt{\pi}\,\Gamma(1-s),$$

and hence

$$\begin{aligned} \, f(-s) &=\frac{2^{3s}}{\pi^s} \frac{\Gamma(-s)2^{-s}\sqrt{\pi}\,\Gamma(s+1)}{2^s\sqrt{\pi}\,\Gamma(1-s)}\,L(s;\chi_4)L(s+1;\chi_4) \\& =\biggl(\frac{2}{\pi}\biggr)^s \frac{\Gamma(-s)\Gamma(s+1)}{(-s)\Gamma(-s)}\,L(s;\chi_4)L(s+1;\chi_4) \\& =-\biggl(\frac{2}{\pi}\biggr)^s\Gamma(s)L(s;\chi_4)L(s+1;\chi_4)=-f(s). \end{aligned}$$

Thus,

$$J_2=-\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}(-f(s))\,ds =\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}f(s)\,ds=J_1.$$

Combining this with (2.2) we obtain

$$2J_1=\frac{\pi}{8}\,,\qquad S=\frac{\pi}{4}\,.$$

2.3. Proof 3: The Laplace Equation

Consider the following two-dimensional boundary-value problem. On a square plate, the temperature of one edge is maintained equal to $T_0$, while the temperature of the other three edges is equal to zero. It is required to find the temperature at the center of the plate in the steady-state mode.

First, we note that the given boundary conditions are a kind of abstraction and never appear in practice. In fact, it can be shown that, for the temperature distribution found below, the heat flow across the edges adjacent to the edge with temperature $T_0$ is infinite. In other words, to maintain the zero temperature on them, a “thermostat” of infinite power is required.

In addition, we intentionally avoid the strict formalization of the solution of this boundary-value problem with discontinuous boundary conditions and restrict ourselves here to a naive understanding of the solution as a function that satisfies the corresponding equation at every point of the plate and is smooth everywhere, except on the boundary.

Let us assume for simplicity that the edges of the plate are of unit length and choose a coordinate system so that its origin coincides with the lower-left corner of the plate and the two sides lie on the positive semi-axes. We will also assume that the side connecting the points $(0,1)$ and $(1,1)$ has temperature $T_0$.

In the steady-state mode, the temperature $T$ only depends on the coordinates $(x,y)$ of the point of the plate and does not depend on time. Since there are no heat sources inside the plate, the heat equation degenerates into the Laplace equation

$$\Delta T=\frac{\partial^2T}{\partial x^2}+\frac{\partial^2T}{\partial y^2}=0 $$

(2.4)

(see [10, Chap. I, Sec. 5]), and

$$T(x,1)\equiv T_0,\qquad T(0,y)=T(1,y)=T(x,0)\equiv 0 $$

(2.5)

for $0<x,y<1$. Solving this equation by the method of separation of variables, we finally obtain the expression

$$T(x,y)=\sum_{m=1}^{+\infty}B(m)\sin(\pi mx)\sinh(\pi my).$$

To find the coefficients $B(m)$, we consider the $2$-periodic function

$$g(x)=\frac{4}{\pi}\sum_{\nu=0}^{+\infty} \frac{\sin{\pi(2\nu+1)x}}{2\nu+1}=\begin{cases} 1, &0<x<1, \\ -1, &-1<x<0, \\ 0, &x=0,\pm 1. \end{cases}$$

Therefore, setting

$$B(2\nu)=0,\quad B(2\nu+1)=\frac{4T_0}{\pi}\frac{1}{(2\nu+1)\sinh{\pi(2\nu+1)}}\,,\qquad \nu=0,1,2,\dots,$$

for $0<x<1$, we fulfill condition (3.2):

$$T(x,1)=\frac{4T_0}{\pi}\sum_{\nu=0}^{+\infty} \frac{\sin{\pi(2\nu+1)x}}{2\nu+1}\equiv T_0.$$

Thus,

$$T(x,y)=\frac{4T_0}{\pi}\sum_{\nu=0}^{+\infty} \frac{\sin{\pi(2\nu+1)x}}{(2\nu+1)}\frac{\sinh{\pi(2\nu+1)y}}{\sinh{\pi(2\nu+1)}}\,.$$

In particular, setting $x=y=1/2$, we can write

$$T\biggl(\frac{1}{2}\,,\frac{1}{2}\biggr) =\frac{4T_0}{\pi}\sum_{\nu=0}^{+\infty} \frac{(-1)^\nu}{(2\nu+1)} \frac{\sinh\pi(\nu+1/2)}{\sinh\pi(2\nu+1)} =\frac{T_0}{\pi}\sum_{\nu=0}^{+\infty} \frac{(-1)^\nu}{(\nu+1/2)\cosh(\pi(\nu+1/2))}\,. $$

(2.6)

But the value of $T(1/2,1/2)$ can easily be calculated from symmetry considerations. Indeed, we consider three more boundary-value problems for the initial plate in which the only edge where the nonzero constant temperature $T_0$ is maintained is, respectively, the left ($x=0$, $0<y<1$), right ($x=1$, $0<y<1$), or lower edge ($0<x<1$, $y=0$). Their solutions will obviously be the functions $T (1-y, x)$, $T(y,1-x)$, and $T (x,1-y)$, respectively. But the sum

$$T_1(x,y)=T(x,y)+T(1-y,x)+T(y,1-x)+T(x,1-y)$$

will, obviously, be the solution of the problem corresponding to the case of a constant temperature $T_0$ on the entire boundary of the square (except for its vertices). Assuming the validity of the uniqueness theorem for solving such a problem², we conclude that $T_1(x,y)\equiv T_0$ for $0<x, y<1$. Therefore,

$$T_1\biggl(\frac{1}{2}\,,\frac{1}{2}\biggr) =T_0=4T\biggl(\frac{1}{2}\,,\frac{1}{2}\biggr),\qquad T\biggl(\frac{1}{2}\,,\frac{1}{2}\biggr) =\frac{1}{4}\,T_0. $$

(2.7)

Comparing (2.6) and (2.7), we obtain the required identity.

3. “Physical” Derivation of the Multidimensional Analogue of (1.1)

In order to “guess” the multidimensional analogue of Ramanujan’s identity (1.1), we will consider the following boundary-value problem. Given the unit $(r+1)$-dimensional cube on one of whose $r$-dimensional faces the constant temperature $T_0$ is maintained, while, on the remaining faces of the same dimension, the zero temperature is maintained. It is required to find the temperature at the center of the cube in the steady-state mode.

Let us introduce a rectangular coordinate system $(x_1,\dots,x_{r+1})$ so that the points of the cube satisfy the conditions $0\le x_1,\dots,x_{r+1}\le 1$. The face on which the temperature $T_0$ is maintained is assumed to be defined by the equation $x_{r+1}=1$.

If $T=T(x_1,\dots,x_{r+1})$ is the temperature at the point $(x_1,\dots,x_{r+1})$ of the cube, then

$$\Delta T=\frac{\partial^2T}{\partial x_1^2}+\dotsb +\frac{\partial^2T}{\partial x_{r+1}^2}=0.$$

Setting $T=f_1(x_1)\dotsb f_{r+1}(x_{r+1})$ and solving this equation by the method of separation of variables, we obtain

$$\frac{\Delta T}{T}=\frac{f_1''(x_1)}{f_1(x_1)}+\dotsb +\frac{f_{r+1}''(x_{r+1})}{f_{r+1}(x_{r+1})}=0,$$

whence

$$\frac{f''_j(x_{j})}{f_{j}(x_{j})} =-m_j^2,\quad j=1,\dots,r,\qquad \frac{f_{r+1}''(x_{r+1})}{f_{r+1}(x_{r+1})} =m_{r+1}^2=m_1^2+\dotsb+m_r^2$$

and, therefore,

$$\begin{aligned} \, f_j(x_{j}) &=A_j\cos{(m_jx_j)}+B_j\sin{(m_jx_j)},\quad j=1,\dots,r, \\ f_{r+1}(x_{r+1}) &=A_{r+1}\cosh(m_{r+1}x_{r+1})+B_{r+1}\sinh(m_{r+1}x_{r+1}). \end{aligned}$$

The conditions

$$T(0,x_2,\dots,x_{r+1}) =T(x_1,0,x_{3},\dots,x_{r+1})=\dotsb =T(x_1,x_2,\dots,x_r,0)\equiv 0$$

lead to the equalities $A_j=0$, $j=1,\dots,r+1$. The conditions

$$T(1,x_2,\dots,x_{r+1})=T(x_1,1,x_3,\dots,x_{r+1}) =\dotsb=T(x_1,x_2,\dots,x_{r-1},1,x_{r+1})\equiv 0$$

imply that $\sin m_j=0$, $m_j=\pi k_j$, $k_j$ are integers, $j=1,\dots,r$. Therefore, the required solution is of the form

$$T(x_1,\dots,x_{r+1}) =\sum_{k_1,\dots,k_r=1}^{+\infty} B(k_1,\dots,k_r)\sin(\pi k_1x_1)\dotsb\sin(\pi k_rx_r) \sinh\bigl(\pi x_{r+1}\sqrt{k_1^2+\dotsb+k_r^2}\,\bigr). $$

(3.1)

Further, it is easy to see that the product

$$T_0g(x_1)\dotsb g(x_r) =T_0\biggl(\frac{4}{\pi}\biggr)^r\sum_{\nu_1,\dots,\nu_r=0}^{+\infty} \frac{\sin\pi(2\nu_1+1)x_1}{(2\nu_1+1)}\dotsb \frac{\sin\pi(2\nu_r+1)x_r}{(2\nu_r+1)} $$

(3.2)

is identically equal to $T_0$ in the domain $0<x_1,\dots,x_r<1$. Comparing the expansions (3.1) and (3.2), we set $B(k_1,\dots,k_r)=0$ if at least one of the numbers $k_j$ is even, and we set

$$B(2\nu_1+1,\dots,2\nu_r+1) =\frac{T_0(4/\pi)^r}{(2\nu_1+1)\dotsb(2\nu_r+1)} \frac{1}{\sinh(\pi\sqrt{(2\nu_1+1)^2+\dotsb+(2\nu_r+1)^2}\,)}\,.$$

Then the function

$$\begin{aligned} \, T(x_1,\dots,x_{r+1}) &=T_0\biggl(\frac{4}{\pi}\biggr)^r\sum_{\nu_1,\dots,\nu_r=0}^{+\infty} \frac{\sin\pi(2\nu_1+1)x_1}{(2\nu_1+1)}\dotsb \frac{\sin\pi(2\nu_r+1)x_1}{(2\nu_r+1)} \\&\qquad\qquad{}\times \frac{\sinh(\pi x_{r+1}\sqrt{(2\nu_1+1)^2+\dotsb+(2\nu_r+1)^2}\,)} {\sinh(\pi\sqrt{(2\nu_1+1)^2+\dotsb+(2\nu_r+1)^2}\,)} \end{aligned}$$

is the solution of the original problem. Taking $x_1=\dotsb=x_{r+1}=1/2$, we obtain

$$\begin{aligned} \, T\biggl(\frac{1}{2}\,,\dots,\frac{1}{2}\biggr) &=\frac{T_0}{2}\biggl(\frac{2}{\pi}\biggr)^r \sum_{\nu_1,\dots,\nu_r=0}^{+\infty} \frac{(-1)^{\nu_1+\dotsb+\nu_r}}{(\nu_1+1/2)\dotsb(\nu_r+1/2)} \\&\qquad\qquad{}\times \frac{1}{\cosh((\pi/2)\sqrt{(\nu_1+1/2)^2+\dotsb+(\nu_r+1/2)^2}\,)}\,. \end{aligned}$$

Just as was done in the case $r=1$, the value of the temperature at the center of the hypercube is easily obtained from symmetry considerations. Obviously, the constant function $T(x_1,\dots,x_{r+1})\equiv T_0$ is the solution of the problem corresponding to the case in which the temperature $T_0$ is maintained on all the $r$-dimensional faces. This function is the sum of the solutions of $2(r+1)$ problems coinciding with the original problem, but differing only by the number of the single $r$-dimensional face where the nonzero constant temperature $T_0$ is maintained, i.e., is the sum of the following $2(r+1)$ terms:

$$T(x_1,\dots,x_{j-1},x_{r+1},x_{j+1},\dots,x_j),\qquad T(x_1,\dots,x_{j-1},x_{r+1},x_{j+1},\dots,1-x_j),$$

where $j=1,\dots,r+1$. Setting $x_1=\dotsb=x_{r+1}=1/2$, we obtain

$$2(r+1)T\biggl(\frac{1}{2}\,,\dots,\frac{1}{2}\biggr)=T_0,\qquad T\biggl(\frac{1}{2}\,,\dots,\frac{1}{2}\biggr) =\frac{T_0}{2(r+1)}\,.$$

This yields the identity in the theorem.

4. Proof of the Theorem

Before giving a rigorous proof of identity (1.2), we will present several auxiliary statements.

Lemma 1.

Let $a_n$ , $b_n$ , $n=1,2,\dots,N$ , be arbitrary sequences, where $b_n>0$ for any $n$ , and let $r\ge 2$ be an integer. Then the following identity holds:

$$\sum_{n_1,\dots,n_r=1}^Na_{n_1}\dotsb a_{n_r} \frac{b_{n_r}}{b_{n_1}+\dotsb+b_{n_r}} =\frac{1}{r}\biggl(\,\sum_{n=1}^Na_n\biggr)^r.$$

Proof.

We have the following chain of obvious equalities:

$$\begin{aligned} \, \biggl(\,\sum_{n=1}^Na_n\biggr)^r &=\sum_{n_1,\dots,n_r=1}^Na_{n_1}\dotsb a_{n_r} =\sum_{n_1,\dots,n_r=1}^Na_{n_1}\dotsb a_{n_r} \frac{b_{n_1}+\dotsb+b_{n_r}}{b_{n_1}+\dotsb+b_{n_r}} \\& =\sum_{n_1,\dots,n_r=1}^Na_{n_1}\dotsb a_{n_r}\,\frac{b_{n_1}}{b_{n_1}+\dotsb+b_{n_r}}+\dotsb \\&\qquad +\sum_{n_1,\dots,n_r=1}^Na_{n_1}\dotsb a_{n_r}\,\frac{b_{n_r}}{b_{n_1}+\dots+b_{n_r}}\,. \end{aligned}$$

The last $r$ sums coincide so that

$$\sum_{n_1,\dots,n_r=1}^Na_{n_1}\dotsb a_{n_r}\,\frac{b_{n_r}}{b_{n_1}+\dotsb+b_{n_r}} =\frac{1}{r}\biggl(\,\sum_{n=1}^Na_n\biggr)^r,$$

as was required.

Lemma 2.

Let $x$ be a real number. Then

$$\frac{1}{\cosh(\pi x)}=\frac{4}{\pi}\sum_{n=1}^{+\infty} \frac{\chi_4(n)n}{n^2+4x^2} =\frac{4}{\pi}\sum_{k=0}^{+\infty} \frac{(-1)^k(2k+1)}{(2k+1)^2+4x^2}\,.$$

Proof.

This identity was given as a problem in [11, Chap. 7, Sec. 7.4, Example 6]. One of the possible methods of proof consists in multiplying both sides of the identity

$$\frac{1}{\cosh(\pi y)}=2\sum_{k=0}^{+\infty}(-1)^ke^{-\pi(2k+1)y}$$

by $\cos(2\pi xy)$ and integrating over $y$, $0\le y<+\infty$. Then the required result is obtained from the formulas

$$\begin{gathered} \, \int_0^{+\infty}\frac{\cos(2\pi xy)}{\cosh(\pi y)}\,dy =\frac{1}{2\cosh(\pi x)}\,, \\ \int_0^{+\infty}e^{-\pi(2k+1)y}\cos(2\pi xy)\,dy =\frac{1}{\pi}\frac{2k+1}{(2k+1)^2+4x^2}\,, \end{gathered}$$

the first of which is established by using the Cauchy residue theorem and the second, by using the Euler formulas.

Proof of the theorem.

Replace $r$ by $r+1$ in the identity from Lemma 1 and set $a_n=\chi_4(n)/n$, $b_n=n^2$. Then, for any $N\ge 1$, we have

$$\sum_{n_1,\dots,n_{r+1}=1}^N \frac{\chi_4(n_1)\dotsb\chi_4(n_r)}{n_1\dotsb n_r} \frac{\chi_4(n_{r+1})}{n_{r+1}} \frac{n_{r+1}^2}{(n_1^2+\dotsb+n_r^2+n_{r+1}^2)} =\frac{1}{r+1}\biggl(\,\sum_{n=1}^N\frac{\chi_4(n)}{n}\biggr)^{r+1}.$$

Letting $N$ tend to infinity, we obtain

$$\sum_{n_1,\dots,n_{r+1}=1}^{+\infty} \frac{\chi_4(n_1)\dotsb\chi_4(n_r)}{n_1\dotsb n_r} \frac{\chi_4(n_{r+1})n_{r+1}}{(n_1^2+\dotsb+n_r^2+n_{r+1}^2)} =\frac{1}{r+1}\biggl(\frac{\pi}{4}\biggr)^{r+1}.$$

Transforming the sum over $n_{r+1}$ by using the identity from Lemma 2 (in which we must set $x=\sqrt{(n_1/2)^2+\dotsb+(n_r/2)^2}\,$), we find

$$\frac{\pi}{4}\sum_{n_1,\dots,n_{r+1}=1}^{+\infty} \frac{\chi_4(n_1)\dotsb\chi_4(n_r)}{n_1\dots n_r} \frac{1}{\cosh(\pi\sqrt{(n_1/2)^2+\dotsb+(n_r/2)^2}\,)} =\frac{1}{r+1}\biggl(\frac{\pi}{4}\biggr)^{r+1}.$$

Hence we easily obtain the required relation.

5. Concluding Remarks

Choosing the sequences $a_n$, $b_n$ in Lemma 1 in a suitable way, we can obtain some other interesting identities.

So, setting $a_n=1/2^n$, $b_n=2^n$, $r=2$, for any $N\ge 1$, we see that

$$\sum_{m,n=1}^N\frac{1}{2^m(2^m+2^n)} =\frac{1}{2}\biggl(\,\sum_{n=1}^N\frac{1}{2^n}\biggr)^2.$$

Letting $N$ tend to infinity, we can write

$$\sum_{m,n=1}^{+\infty}\frac{1}{2^m(2^m+2^n)}=\frac{1}{2}\,.$$

The choice $a_n=(-1)^n/n$, $b_n=n^2$, $r=2$ yields

$$\sum_{m,n=1}^N\frac{(-1)^{m+n}m}{n(m^2+n^2)} =\frac{1}{2}\biggl(\,\sum_{m=1}^N\frac{(-1)^m}{m}\biggr)^2.$$

Hence, as $N\to+\infty$, we see that

$$\sum_{m,n=1}^{+\infty}\frac{(-1)^{m+n}m}{n(m^2+n^2)} =\frac{1}{2}(\ln 2)^2.$$

Taking $a_n=1/n^k$, $b_n=n^k$, where $k\ge 2$ is an integer, we obtain

$$\sum_{n_1,\dots,n_{r+1}=1}^{+\infty} \frac{1}{n_1^k\dotsb n_r^k(n_1^k+\dotsb+n_r^k+n_{r+1}^k)} =\frac{1}{r+1}\biggl(\,\sum_{n=1}^{+\infty} \frac{1}{n^k}\biggr)^{r+1} =\frac{\zeta^{r+1}(k)}{r+1}\,.$$

In particular, for $k=2,3$ and $r=1$, the above equalities imply the following identities:

$$\sum_{m,n=1}^{+\infty}\frac{1}{m^2(m^2+n^2)} =\frac{\pi^4}{72}\,,\qquad \sum_{m,n=1}^{+\infty} \frac{1}{m^3(m^3+n^3)}=\frac{1}{2}\,\zeta^2(3).$$

The use of Lemma 1 combined with formulas involving the Fourier transform opens up almost unlimited possibilities for deriving new identities for the number $\pi$ and the values of $\zeta(s)$ and $L(s;\chi)$ (here $\chi$ is a nonprincipal Dirichlet character modulo an arbitrary integer $q\ge 3$) at integer points, etc. We will restrict ourselves to just one example.

First, we note that, for any real (not necessarily integer) positive $m$ and $n$, the following identity holds:

$$\frac{m^3}{m^4+n^4}=\int_0^{+\infty}e^{-my/\sqrt{2}} \sin\biggl(\frac{\pi}{4}+\frac{my}{\sqrt{2}}\biggr)\cos(ny)\,dy$$

(see, for example, formulas (19), (29) from [12, Chap. I, Sec. 1.2]). Further, assuming $m$ to be an integer, we multiply both sides by $\chi_4(m)$ and sum over all $m\ge 1$. Denoting the left-hand side by $S(n)$, we obtain

$$\begin{aligned} \, S(n) &=\sum_{m=1}^{+\infty}\frac{m^3\chi_4(m)}{m^4+n^4} =\operatorname{Im}\biggl\{\sum_{m=1}^{+\infty} \int_0^{+\infty}e^{-my/\sqrt{2}}\,e^{i(\pi/4+my/\sqrt{2})}\cos{(ny)}\,dy\biggr\} \\& =\operatorname{Im}\biggl\{e^{\pi i/4}\int_0^{+\infty} \sum_{m=1}^{+\infty}e^{-(my/\sqrt{2})(1-i)}\chi_4(m)\cos{(ny)}\,dy\biggr\} \end{aligned}$$

(the validity of the change of order of summation and integration can easily be justified). Further, setting $s=y(1-i)/\sqrt{2}=ye^{-\pi i/4}$ for brevity, we can write

$$\sum_{m=1}^{+\infty}\chi_4(m)e^{-ms} =\sum_{\mu=0}^{+\infty}e^{-(4\mu+1)s}-\sum_{\mu=0}^{+\infty}e^{-(4\mu+3)s} =\frac{e^{-s}-e^{-3s}}{1-e^{-4s}} =\frac{e^{-s}}{1+e^{-2s}}=\frac{2}{\cosh(\pi s)}\,,$$

so that

$$S(n)=2\operatorname{Im}\biggl\{e^{\pi i/4} \int_0^{+\infty}\frac{\cos(ny)}{\cosh(ye^{-\pi i/4})}\,dy\biggr\}.$$

Using the identity

$$\int_0^{+\infty}\frac{\cos(\alpha x)}{\cosh(\gamma x)}\,dx =\frac{\pi}{2\gamma}\frac{1}{\cosh(\pi\alpha/(2\gamma))}$$

for $\alpha>0$, $\operatorname{Re}\gamma>0$ (see [13, Chaps. 3, 4, formula 3.981.10]), we obtain

$$S(n)=\frac{1}{2}\operatorname{Im}\biggl\{e^{\pi i/4} \frac{\pi}{2}\,e^{\pi i/4}\cosh^{-1}\biggl(\frac{\pi n}{2}\,e^{\pi i/4}\biggr)\biggr\} =\frac{\pi}{4}\operatorname{Re}\cosh^{-1}\biggl(\frac{\pi n}{2}\,e^{\pi i/4}\biggr).$$

But

$$\cosh\biggl(\frac{\pi n}{2}\,e^{\pi i/4}\biggr) =\cosh\frac{\pi n}{2\sqrt{2}}\cos\frac{\pi n}{2\sqrt{2}} +i\sinh\frac{\pi n}{2\sqrt{2}}\sin\frac{\pi n}{2\sqrt{2}}\,,$$

whence

$$\begin{aligned} \, S(n) &=\frac{\pi}{4}\cosh\frac{\pi n}{2\sqrt{2}}\cos\frac{\pi n}{2\sqrt{2}}\Bigm/ \biggl(\cosh^2\frac{\pi n}{2\sqrt{2}}\cos^2\frac{\pi n}{2\sqrt{2}} +\sinh^2\frac{\pi n}{2\sqrt{2}}\sin^2\frac{\pi n}{2\sqrt{2}}\biggr) \\& =\frac{\pi}{4}\cosh\frac{\pi n}{2\sqrt{2}}\cos\frac{\pi n}{2\sqrt{2}}\Bigm/ \biggl(\sinh^2\frac{\pi n}{2\sqrt{2}}+\cos^2\frac{\pi n}{2\sqrt{2}}\biggr) \\& =\frac{\pi}{2}\cosh\frac{\pi n}{2\sqrt{2}}\cos\frac{\pi n}{2\sqrt{2}}\Bigm/ \biggl(\cosh\frac{\pi n}{\sqrt{2}}+\cos\frac{\pi n}{\sqrt{2}}\biggr). \end{aligned}$$

Replacing $n$ by $n\sqrt{2}$, we obtain

$$\sum_{m=1}^{+\infty}\frac{m^3\chi_4(m)}{m^4+4n^4} =\frac{\pi}{2}\cosh\frac{\pi n}{2}\cos\frac{\pi n}{2}\Bigm/ (\cosh(\pi n)+\cos(\pi n)). $$

(5.1)

In particular, for an integer $n\equiv 1\pmod{2}$, we see that

$$\sum_{m=1}^{+\infty}\frac{m^3\chi_4(m)}{m^4+4n^4}=0.$$

Since

$$\frac{m^3}{m^4+4n^4}=\frac{1}{m}-\frac{4n^4}{m(m^4+4n^4)}\,,$$

we have

$$4n^4\sum_{m=1}^{+\infty}\frac{\chi_4(m)}{m(m^4+4n^4)} =\sum_{m=1}^{+\infty}\frac{\chi_4(m)}{m}=\frac{\pi}{4}\,,$$

whence

$$\sum_{m=1}^{+\infty}\frac{\chi_4(m)}{m^5+4mn^4}=\frac{\pi}{(2n)^4}\,.$$

Setting $n=1$, we can write

$$\sum_{m=1}^{+\infty}\frac{\chi_4(m)}{m^5+4m}=\frac{\pi}{16}\,.$$

Substituting $n=2k$ ($k$ is an integer) into (5.1), after obvious transformations, we obtain the equality

$$\sum_{m=1}^{+\infty}\frac{m^3\chi_4(m)}{m^4+2^6k^4} =\frac{\pi}{4}\frac{(-1)^k}{\cosh(\pi k)}\,,$$

which, in turn, yields the equality

$$\sum_{m=1}^{+\infty}\frac{\chi_4(m)}{m^5+64k^4m} =\frac{\pi}{(4k)^4}\biggl(1-\frac{(-1)^k}{\cosh(\pi k)}\biggr).$$

It is of interest to compare equality (5.1) with the Ramanujan identities contained in [14, Chap. 30, Secs. 13, 14].

Funding

The research of the second author was carried out at the Steklov Mathematical Institute of Russian Academy of Sciences and was supported by the Grant of the Russian Science Foundation (project no. 19-11-00001).