Does a certain property of Tornheim's Double Series generalize to the positive reals?

Question

On the first page of this paper, a number of properties of Tornheim's Double Series are listed. These series are defined as follows:

$$T(r,s,t) = \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{m^{r}n^{s}(m+n)^{t}}, \qquad (r, s, t \text{ nonnegative integers}).$$ This series is convergent when the following inequalities hold: $$r+t >1, \quad s+t>1, \quad r+s+t > 2 \qquad (*). $$

I am particularly interested in property (1.4) of Tornheim's Double Series, as described in the aforementioned paper. It is as follows: $$T(r,0,t) + T(t,0,r) = \zeta(r)\zeta(t) - \zeta(r+t), \quad r \geq 2. $$ Here, $\zeta(\cdot)$ is the Riemann Zeta function.

Questions

1. Does property (1.4) also hold when $r, s$ and $t$ are not nonnegative integers? For instance, does it also hold when $r, t \in \mathbb{R}_{>0}$ ? If not, what about $r, t$ of the form $r = \frac{2a+1}{2}$ and $t = \frac{2b+1}{2}$, in which $a, b \in \mathbb{Z}_{>0}$ ? (Assuming the triple $(r,s,t)$ still satisfies the inequalities described in $(*)$ .)
2. How can property (1.4) be proved? (Preferably in the generalized case, if it indeed can be generalized.)

Answer 1

$\begin{array}\\ T(r,0,t) &= \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \dfrac{1}{m^{r}(m+n)^{t}}\\ &= \sum_{m=1}^{\infty}\dfrac1{m^r}\sum_{n=1}^{\infty} \dfrac{1}{(m+n)^{t}}\\ &= \sum_{m=1}^{\infty}\dfrac1{m^r}\sum_{n=m+1}^{\infty} \dfrac{1}{n^{t}} \qquad\text{(see below})\\ &= \sum_{m=1}^{\infty}\dfrac1{m^r}(\zeta(t)-\sum_{n=1}^{m} \dfrac{1}{n^{t}})\\ &= \zeta(t)\sum_{m=1}^{\infty}\dfrac1{m^r}-\sum_{m=1}^{\infty}\dfrac1{m^r}\sum_{n=1}^{m} \dfrac{1}{n^{t}}\\ &= \zeta(t)\zeta(r)-\sum_{m=1}^{\infty}\sum_{n=1}^{m}\dfrac1{m^r} \dfrac{1}{n^{t}}\\ &= \zeta(t)\zeta(r)-\sum_{n=1}^{\infty}\sum_{m=n}^{\infty}\dfrac1{m^r} \dfrac{1}{n^{t}}\\ &= \zeta(t)\zeta(r)-\sum_{n=1}^{\infty}\dfrac{1}{n^{t}}\sum_{m=n}^{\infty}\dfrac1{m^r} \\ &= \zeta(t)\zeta(r)-\sum_{n=1}^{\infty}\dfrac{1}{n^{t}}(\dfrac1{n^r}+\sum_{m=n+1}^{\infty}\dfrac1{m^r} )\\ &= \zeta(t)\zeta(r)-(\sum_{n=1}^{\infty}\dfrac{1}{n^{t}}\dfrac1{n^r}+\sum_{n=1}^{\infty}\dfrac{1}{n^{t}}\sum_{m=n+1}^{\infty}\dfrac1{m^r} )\\ &= \zeta(t)\zeta(r)-\sum_{n=1}^{\infty}\dfrac{1}{n^{r+t}}-\sum_{n=1}^{\infty}\dfrac{1}{n^{t}}\sum_{m=n+1}^{\infty}\dfrac1{m^r}\\ &= \zeta(t)\zeta(r)-\zeta(r+t)-\sum_{n=1}^{\infty}\dfrac{1}{n^{t}}\sum_{m=n+1}^{\infty}\dfrac1{m^r}\\ &= \zeta(t)\zeta(r)-\zeta(r+t)-T(t, 0, r) \qquad\text{(see above - swap m and n)}\\ \end{array} $

(I think I have done this before.)