Do we have integral test for double series? To determine convergence of
$$\sum_{n=1}^\infty a_n,$$
one can use the integral test if $f(n)=a_n$ satisfies certain properties.
Now, if I would like to determine convergence of double series
$$\sum_{i=1}^\infty \sum_{j=1}^\infty a_{ij},$$
do we have some 'integral test' for it?
I think one can evaluate
$$\int\int f(i,j)\,di\,dj.$$
 A: Let $f : [0,\infty)\times[0,\infty) \to [0,\infty)$ be continuous, and satisfy
$$
x_1 < x_2 \quad \Longrightarrow \quad f(x_1,y) \ge f(x_2,y)
\\
y_1 < y_2 \quad \Longrightarrow \quad f(x,y_1) \ge f(x,y_2).
$$
consider the two series
$$
\sum_{i=0}^\infty \sum_{j=0}^\infty f(i,j)
\tag{1}$$
$$
\sum_{i=1}^\infty \sum_{j=1}^\infty f(i,j)
\tag{$1'$}$$
and the integral
$$
\int_0^\infty \int_0^\infty f(x,y)\;dy\;dx .
\tag{2}$$ 
Note, because $f$ is nonnegative, (1), ($1'$) and (2) exist, but possibly equal $+\infty$.  So to say they "converge" means they have a finite value.  
Now, given $i,j$, note that for all $x,y$ with $i \le x \le i+1, j \le y \le y+1$ we have
$$
f(i,j) \ge f(x,y) \ge f(i+1,j+1).
$$
therefore
$$
f(i,j) \ge \int_i^{i+1} \int_j^{j+1} f(x,y)\;dy\;dx \ge f(i+1,j+1)
$$
Summing these, we get
$$
\sum_{i=0}^\infty\sum_{j=0}^\infty f(i,j) \ge \int_0^\infty \int_0^\infty f(x,y)\;dy\;dx
\tag{3}$$
and
$$
\int_0^\infty \int_0^\infty f(x,y)\;dy\;dx \ge \sum_{i=1}^\infty\sum_{j=1}^\infty f(i,j)
\tag{4}$$
So: we may use (3) to show: if (1) converges, then (2) converges.  And we may use (4) to show: if (2) converges, then ($1'$) converges.
note
Those who like "simple" statments, may be disappointed.  It is possible that (2) converges but (1) diverges because $\sum_i f(i,0)$ and/or $\sum_j f(0,j)$ diverge.
A: Define a function $f(x,y)$ such that $$f(i,j)=a_{ij}$$furthermore define $b_i=\sum_{j=1}^{\infty}f(i,j)$. This means that $$\sum_{i=1}^{\infty}b_i=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}f(i,j)$$is convergent iff $\sum_{j=1}^{\infty}f(x,j)$ is monotone decreasing (a sufficient condition is $f(x_1,y)\ge f(x_2,y)$ when $x_1\le x_2$) and $$\int_{1}^{\infty}\sum_{j=1}^{\infty}f(x,j)dx=\sum_{j=1}^{\infty}\int_{1}^{\infty}f(x,j)dx$$finally by defining $c_n=\int_{1}^{\infty}f(x,n)dx$, the summation $\sum_{1}^{\infty}c_n$ is bounded iff $$\int_{1}^{\infty}f(x,y)dx$$ is monotone decreasing over $y$ (with a similar sufficient condition) and $$\int_{1}^{\infty}\int_{1}^{\infty}f(x,y)dxdy$$is bounded.
A: If $f(t,x)$ is continuous in $[1,\infty)\times[1,\infty)$. Set $D=[1,\infty)$. Also if $f(t,x)$ is non increasing in $D$ meaning that if $t_1,t_2,x_1,x_2\in D$
$$
t_1<t_2\Rightarrow f(t_1,x)\geq f(t_2,x)\textrm{, }\forall x \in D
$$
$$  
x_1<x_2\Rightarrow f(t,x_1)\geq f(t,x_2)\textrm{, }\forall t\in D
$$
If we assume also that
$$
f(n,m)\rightarrow 0\textrm{ as }n,m\rightarrow \infty\tag 1
$$
and
$$
\sum^{n}_{i=1}f(i,1)\textrm{, }\sum^{m}_{j=1}f(1,j)\tag 2
$$
$$
\sum^{n}_{i=1}f(i,m)\textrm{, }\sum^{m}_{j=1}f(n,j) \tag 3
$$
are bounded for all both $n,m>>1$ positive integers, then exists constant$-C$ such that
$$
\sum^{n,m}_{i,j=2}f(i,j)=\int^{n}_{1}\int^{m}_{1}f(t,x)dxdt+C+o(1)\textrm{, as both }n,m\rightarrow\infty.\tag 4
$$
Proof.
Easily using the monotonicity of $f$ we find
$$
\int^{i+1}_{i}\int^{j+1}_{j}(f(t,x)-f(i+1,j+1))dxdt\geq0
$$
and
$$
\int^{i+1}_{i}\int^{j+1}_{j}(f(t,x)-f(i,j))dxdt\leq0
$$
Hence
$$
\sum^{n}_{i=2}\sum^{m}_{j=2}f(i,j)\leq \int^{n}_{1}\int^{m}_{1}f(t,x)dxdt\leq\sum^{n-1}_{i=1}\sum^{m-1}_{j=1}f(i,j)
$$
Hence if we set
$$
k_{n,m}:=\int^{n}_{1}\int^{m}_{1}f(t,x)dxdt-\sum^{n}_{i=2}\sum^{m}_{j=2}f(i,j),
$$
then
$$
0\leq k_{n,m}\leq -f(1,1)+f(n,m)+\sum^{n}_{i=1}f(i,1)+\sum^{m}_{j=1}f(1,j)-\sum^{n}_{i=1}f(i,m)-\sum^{m}_{j=1}f(n,j).
$$
Also
$$
k_{n+1,m}-k_{n,m}=\int^{n+1}_{1}\int^{m}_{1}f(t,x)dxdt-\sum^{n+1}_{i=2}\sum^{m}_{j=2}f(i,j)-
$$
$$
-\left(\int^{n}_{1}\int^{m}_{1}f(t,x)dxdt-\sum^{n}_{i=2}\sum^{m}_{j=2}f(i,j)\right)=
$$
$$
=\int^{n+1}_{n}\int^{m}_{1}f(t,x)dxdt-\sum^{m}_{j=2}f(n+1,j)=
$$
$$
=\int^{n+1}_{n}\left(\sum^{m}_{j=2}\int^{j}_{j-1}\left(f(t,x)-f(n+1,j)\right)dx\right)dt=
$$
$$
=\int^{n+1}_{n}\left(\sum^{m}_{j=2}\int^{j}_{j-1}\left| f(t,x)-f(n+1,j)\right|dx\right)dt\geq 0.
$$
Hence $k_{n+1,m}\geq k_{n,m}$, for all $n,m$ positive integers. In the same way $k_{n,m+1}\geq k_{n,m}$, for all $n,m$ positive integers.
Hence the double sequence $k_{n,m}$ is monotonic and bounded. Hence it must be convergent. From this we get $(4)$.
