Two dimensional integral of function that depends only on difference of coordinates Question
A book that I'm reading$^{[a]}$ asks to prove that under suitable conditions on the behavior of a function $K:\mathbb{R} \rightarrow \mathbb{R}$ at large arguments, the integral
$$\mathcal{I} = \int_0^t dt_1 \int_0^t dt_2 K(\left \lvert t_1 - t_2 \right \rvert)$$
is proportional to $t$ for sufficiently large $t$.$^{[b]}$
The book claims that a sufficient condition is
$$\lim_{x \rightarrow \infty} K(x) = 0 \, ,$$
but I doubt that because if $K(x)$ decays like $1/x$ then its integral isn't even convergent.
What are the conditions on $K$ under which $\mathcal{I}$ does in fact scale linearly in $t$ in the limit of $t \rightarrow \infty$?
Attempt
I think a more reasonable condition to investigate would be something like this:
Condition: There is a $K^*$ such that for any $\epsilon > 0$ we can find $T$ such that for any $t > T$
$$\left \lvert \int_0^t K(t) \, dt - K^* \right \rvert < \epsilon \, ,$$
i.e. for sufficiently large $t$ the integral of $K$ can be approximated to be $K^*$.
I'm choosing to explore this condition because it's one that makes sense in the context of the book (see footnotes).
Referring to the diagram below, the integral $\mathcal{I}$ is over $A \cup B$, and because of the symmetry of the integrand (i.e. it depends only on $\left \lvert t_1 - t_2 \right \rvert$) the contributions from $A$ and $B$ are equal.
Therefore
$$\mathcal{I} = 2 \int_B K(\left \lvert t_1 - t_2 \right \rvert ) dt_1 \, dt_2 \, .$$
I think we can make progress by attempting to extend the integral to the region $B \cup C$ and argue that the contribution from $C$ doesn't change the large $t$ behavior.
Introduce new coordinates $p$ and $q$ defined by
$$\begin{array}{ll}
q = &(1/\sqrt{2}(t_1 + t_2) \\
p = &(1/\sqrt{2})(t_1 - t_2) \, .
\end{array}$$
If we make the extra assumption that $K$ is positive, then extending the integral to include region $C$ can only increase its value, so
$$\mathcal{I} < 2 \int_{B + C} K(\left \vert t_1 - t_2 \right \rvert) dt_1 dt_2 = \int_0^t dq \int_0^t dp K(p) \stackrel{t\rightarrow \infty}{=}t K^* \, ,$$
which shows that the large $t$ behavior of $\mathcal{I}$ is bounded by $t$.
However, we're supposed to show that $\mathcal{I}$ is proportional to $t$, not bounded by it, and we should be able to do this without the assumption that $K$ is positive.
How do we improve the argument?

[a]: Elements of Nonequilibrium Statistical Mechanics by V. Balakrishnan
[b]: This integral comes up in the study of Brownian motion where $K$ is a so-called "correlation function" of the noise.
 A: I think I figured this out.
Do the integral over region A:
\begin{align}
\mathcal{I}
&= 2 \int_A K(\left \lvert t_1 - t_2 \right \rvert) \\
&= 2 \int_0^t dt_1 \int_0^{t_1} dt_2 \, K(\left \lvert t_1 - t_2 \right \rvert) \\
(t' \equiv t_1 - t_2) \quad &= 2 \int_0^t dt_1 \int_0^{t_1} K(t') \, .
\end{align}
This is an integral over a triangle where here the outer integral scans the horizontal extent of the triangle while the inner integral scans the vertical extent.
Switching the order, we can rewrite the integral as
\begin{align}
\mathcal{I}
&= 2 \int_0^t dt' \, K(t') \int_{t'}^t dt_1 \\
&= 2 \int_0^t dt' \, (t - t') K(t') \\
&= 2 \left( t \int_0^t dt' \, K(t') - \int_0^t dt' \, t' K(t') \right) \\
\frac{d\mathcal{I}}{dt}
&= 2 \left( t K(t) + \int_0^t dt' \, K(t') - t K(t) \right) \\
&= 2 \int_0^t dt' \, K(t') \,.
\end{align}
Now using the assumption on the integral of $K$ given in the question, for any desired accuracy $\epsilon$, there is a value $T$ such that for $t > T$ the integral is within $\epsilon$ of $2 K^*$.
Therefore we've shown that $d\mathcal{I}/dt$ is constant in the limit of large $t$, which is what we wanted to show.
A: I think a good assumption to begin with is $K\in L^1(0,+\infty)$, which is more restrictive than your condition but more comfortable to work with.
I think you don't have $t$ proportionality in general. Choose for instance  $K(x)=\chi_{0<x<1}\sin(2\pi x)$ with support in $[0,1]$, so that $K$ has vanishing integral. Then, in the quantity $\mathcal I$ you have some cancellation. In particular, for $t$ large enough, $\mathcal I$ is a non-zero costant. You might want at least to assume that $K$ has non vanishing integral in order to have actual proportionality.
