Suppose I have a sum of the form:
$$ \sum_{j=0}^n \sum_{i=0}^m c_{i,j}$$
And I sub :
$$ j+ i \to x$$
And,
$$ i-j \to y$$
Then:
$$ \frac{i+j}{2} = x$$
And,
$$ \frac{i-j}{2} = y$$
If I sub:
$$ \sum_{j=0}^n \sum_{i=0}^m c_{ \frac{x+y}{2} , \frac{x-y}{2} }$$
How do I figure out the new bounds of summation..?
I will give an answer for the infinite case:
$$\sum_{j=0}^{\infty} \sum_{i=0}^{\infty} c_{i,j}=\sum_{x=0}^{\infty} \sum_{\begin{matrix}y=-x\\ \color{red}{\text{step} \ 2} \end{matrix}}^{y=x} c_{x,y} $$
The "$\color{red}{\text{step} \ 2}$" meaning that the values taken by $y$ are:
$$y=-x, \ -x+2, \ -x+4 \ \cdots x. \tag{1}$$
Why that ? let us write the $i,j$ tables under this form:
$$\text{Abscissas i : } \ \ \begin{matrix} ..&..&..&..&..&\\ 0&1&2&3&4&\cdots\\ 0&1&2&3&4&\cdots\\ 0&1&2&3&4&\cdots\\ 0&1&2&3&4&\cdots\\ 0&1&2&3&4&\cdots\\ \end{matrix} \ \ \ \ \ \text{Ordinates j : } \ \ \begin{matrix} ..&..&..&..&..&\\ 4&4&4&4&4&\cdots\\ 3&3&3&3&3&\cdots\\ 2&2&2&2&2&\cdots\\ 1&1&1&1&1&\cdots\\ 0&0&0&0&0&\cdots\\ \end{matrix}$$
From which we deduce the tables of $x$ and $y$:
$$\text{Values of x : } \ \ \begin{matrix} ..&..&..&..&..&\\ 4&5&6&7&8&\cdots\\ \color{blue}{3}&4&5&6&7&\cdots\\ 2&\color{blue}{3}&4&5&6&\cdots\\ 1&2&\color{blue}{3}&4&5&\cdots\\ 0&1&2&\color{blue}{3}&4&\cdots\\ \end{matrix} \ \ \ \ \ \ \ \text{Values of y : } \ \ \begin{array}{rrrrrc} ..&..&..&..&..&\\ -4&-3&-2&-1&0&\cdots\\ \color{red}{-3}&-2&-1&0&1&\cdots\\ -2&\color{red}{-1}&0&1&2&\cdots\\ -1&0&\color{red}{1}&2&3&\cdots\\ 0&1&2&\color{red}{3}&4&\cdots\\ \end{array}$$
(check that $i+j=x$ and $i-j=y$).
In the table of $x$ values, one sees that $x$ can take (along diagonals with "equations" $i+j=x$, one of them has been highlighted in $\color{blue}{\text{blue}}$) all positive integer values, indicating that it has to be taken as the "driving" variable. $x$ being fixed, the corresponding values taken by $y$ (in $\color{red}{\text{red}}$) are indeed described by (1).
