Triple sum $\sum\limits_{a=1}^{\infty} \sum\limits_{b=1}^{\infty} \sum\limits_{c=1}^{\infty} \frac{\cos a \cos b \cos c}{a^2 + b^2 + c^2}$

Question

We have poor water heating system in our countryside house (currently it takes 4 hours to warm up the water), and my father has decided to improve it; he bought a water tank and placed it up in the attic, and the last stage is to buy a correct heating element. So he gave me a call and asked to calculate which heating element to buy (more precisely, he asked what electric power does the heating element need to make the descending water hot in 10 minutes; hot was defined as 40 Celsius). The water tank is huge and the heater is placed at one of its dead ends; it wouldn't be correct to say the temperature is uniform and does not vary from one point to another.

The only person for whom I can abandon my work is my father, so I took a standard heat equation: $$\frac{\partial T}{\partial t} = a^2 \left( \frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} + \frac{\partial^2 T}{\partial z^2} \right) + f (x,y,z,t)$$

with initial and boundary conditions $$T(x,y,z,0) = T_0, \hskip 20 pt T(x,y,z,t)\Big|_{\text{boundary}} = T_0$$

and with $$f(x,y,z,t) = \frac{P}{c\rho} \delta(x - x_0) \delta (y - y_0) \delta (z - z_0)$$ for this case (point-like heater located at $(x_0, y_0, z_0)$, $P$ is heater electric power, $c$ is water specific heat capacity, $\rho$ is water density), $a = \sqrt{\kappa/c\rho}$ is water thermal diffusivity, $\kappa$ is water thermal conductivity.

After a tough computation I got the solution:

$$T (x,y,z,t) = T_0 + \frac{P}{lwh \kappa} \sum\limits_{n=1}^{\infty} \sum\limits_{u=1}^{\infty} \sum\limits_{v=1}^{\infty} \frac{\sin \left(\frac{\pi n x}{l} \right) \sin \left(\frac{\pi u y}{w} \right) \sin \left(\frac{\pi v z}{h} \right) \sin \left(\frac{\pi n x_0}{l} \right) \sin \left(\frac{\pi u y_0}{w} \right) \sin \left(\frac{\pi v z_0}{h} \right)}{\left( \frac{\pi n}{l} \right)^2 + \left( \frac{\pi u}{w} \right)^2 + \left( \frac{\pi v}{h} \right)^2 } \cdot $$ $$\cdot \left(1 - \exp\left( - \left( \frac{\pi n a}{l} \right)^2 t - \left( \frac{\pi u a}{w} \right)^2 t - \left( \frac{\pi v a}{h} \right)^2 t \right) \right)$$

So guys, if you will ever install a water tank in your attic, you know where to look. Here $l$, $w$, $h$ are length, width and height of the tank.

My father's question was "which $P$ do I need for achieving $T_{\text{hot}} = 40$ Celsius from $T_0 = 20$ Celsius in $t = 10$ minutes in a point $(x,y,z) = (x_1, y_1, z_1)$?" To resolve it, to find the $P$, I need to numerically compute this triple sum somehow.

Here go my thoughts on the subject:

• the triple sum $\sum\limits_{n=1}^{\infty} \sum\limits_{u=1}^{\infty} \sum\limits_{v=1}^{\infty} \frac{1}{n^2 + u^2 + v^2}$ does not converge; you need to have a power of at least $(3+\varepsilon)$ in denominator

• trigonometric functions in the numerator really matter: $\sum\limits_{n=1}^{\infty} \frac{1}{n}$ does not converge but $\sum\limits_{n=1}^{\infty} \frac{\cos n}{n}$ does

• one can utilise the formula for sine product ($\sin \left(\frac{\pi n x_1}{l}\right) \sin \left(\frac{\pi n x_0}{l}\right) = \dots$) and rewrite the numerator into 8 terms of three cosines

• thus, what I'm interested in is the sum $\sum\limits_{n=1}^{\infty} \sum\limits_{u=1}^{\infty} \sum\limits_{v=1}^{\infty} \frac{\cos n \cos u \cos v}{n^2 + u^2 + v^2}$

• clearly, for numeric computation (e.g. in Mathematica) it would be wonderful to find the precise $N$, $U$, $V$ such that $\sum\limits_{n=1}^{\infty} \sum\limits_{u=1}^{\infty} \sum\limits_{v=1}^{\infty} (\dots) \approx \sum\limits_{n=1}^{N} \sum\limits_{u=1}^{U} \sum\limits_{v=1}^{V} (\dots)$

• There's a decent identity $\int\limits_0^{\infty} \frac{sin at}{a} e^{-bt} dt = \frac{1}{a^2 + b^2}$, which is proved by integration by parts twice. It allows you to split $\frac{1}{a^2 + b^2}$ into factors, into multipliers. Given this split, computing $\sum\limits_{a=1}^{\infty} \sum\limits_{b=1}^{\infty} \frac{1}{a^2 + b^2}$ is relatively easy (well, one can calculate both sums and then prove the resulting integral does not converge). However, I failed to split $\frac{1}{a^2 + b^2 + c^2}$ in this manner.

• With Fourier transform (see wiki, end of the page, row 502)

$$\frac{1}{a^2 + b^2 + c^2} = \iiint\limits_{\mathbb{R}^3} \frac{e^{-2\pi i (a \xi_a + b\xi_a + c\xi_c)}}{\xi_a^2 + \xi_b^2 + \xi_c^2} d\xi_a d\xi_b d\xi_c $$

the triple sum $\sum\limits_{a=1}^{\infty} \sum\limits_{b=1}^{\infty} \sum\limits_{c=1}^{\infty} \frac{\cos a \cos b \cos c}{a^2 + b^2 + c^2}$ could be reduced to ordinary sums such as $\sum\limits_{a=1}^{\infty} \cos a \cdot e^{-2\pi i a \xi_a}$. However, the latter sum does not converge.