Formula for a sequence defined on $K_1(x,y) := y+0$ if $x \geq y$ and $y-1$ otherwise Define $K_1:[0,1]^2\rightarrow\mathbb{R}$ as
$$K_1(x,y) := x - \frac{1}{2} - \begin{cases}
\ +(x - y - \frac{1}{2}) & \text{if $x \geq y$},\\
\ -(y - x - \frac{1}{2}) & \text{otherwise}
\end{cases}$$
then with
$$K_n(x,y) := \int_0^1K_1(x,u)K_{n-1}(u,y)\textrm{d}u$$
cf. equation $(35)$ part $3$
show that for $n\geq 1$
$$
n!K_n(x,y) = B_n(x) - \begin{cases} B_n(x-y) ~~\textrm{ if } x\geq y\\
~\\
(-1)^nB_n(y-x)~~\textrm{ otherwise }\end{cases}
$$
Here $B_n$ are the Bernoulli Polynomials.
Verify that
\begin{align}
+\sin(2\pi k x)=(2\pi k)^1\int_0^1K_1(x,u)\cos(2\pi k u)\textrm{d}u\\
-\cos(2\pi k x)=(2\pi k)^2\int_0^1K_2(x,u)\cos(2\pi k u)\textrm{d}u\\
-\sin(2\pi k x)=(2\pi k)^3\int_0^1K_3(x,u)\cos(2\pi k u)\textrm{d}u\\
+\cos(2\pi k x)=(2\pi k)^4\int_0^1K_4(x,u)\cos(2\pi k u)\textrm{d}u
\end{align}
for all $x\in [0,1]$ and $k\in \mathbb{Z}$, $k\neq 0$,
as well as
$$
B_{n+m}(x)=\frac{(n+m)!}{m!}\int_0^1K_n(x,u)B_m(u)\textrm{d}u\\
$$
 A: Functions $K_n(x, y), n\in\mathbb N,$ are considered in the area 
$$\mathbb S = \{(x, y) \in [0,1]^2\}.\tag1$$
Besides this, can be used step function 
$$h(x) = 
\begin{cases}
1, \text{ if }x \in (0, 1],\\
0, \text{ otherwize}
\end{cases}\tag2$$
for brief notation of 2D intervals method.
In particular, we can present the issue condition in the forms of
$$K_1(x, y) = 
\begin{cases}
y\text{ if } x \ge y\\
y-1\text{ if } x < y
\end{cases}
= yh(x - y) + (y - 1)h(y - x). \tag3$$
In this way, taking in account identity
$$(-1)^nB(x) = B(1-x),\quad n\in\mathbb N\tag4$$
easy to see the replacement
$$B_n(x) - 
\begin{cases}
B_n(x - y)\text{ if } x\ge y\\
(-1)^nB_n(y - x)\text{ if } x < y
\end{cases}=
\begin{cases}
B_n(x) - B_n(x - y)\text{ if } x\ge y\\
B_n(x) - B_n(x + 1 - y)\text{ if } x < y.
\end{cases}.
$$
So we can prove the identity
$$n!K_n(x, y) = B_n(x) - h(x - y)B_n(x - y) - h(y - x)B_n(x + 1 - y).\quad k \ge 1\tag5$$
Let us prove it by induction.
At first, for $B_1(x) = x - \frac12$
$$B_1(x) - B_1(y) = x - y\tag6,$$
and one can write
$$B_1(x)  - h(x - y)B_1(x - y) - h(y - x)B_1(x + 1 - y) = h(x - y) (B_1(x) - B_1(x - y) ) + h(y - x)(B_1(x) - B_1(x + 1 - y)) = yh(x - y) + (y - 1)h(y - x) = K_1(x)),$$
so identity $(5)$ is satisfied for $k = 1$.
Then, let identity $(5)$ is satisfied for the case $n-1,$
$$(n-1)!K_{n - 1}(x, y) = B_{n - 1}(x) - h(x - y)B_{n - 1}(x - y) - h(y - x)B_{n - 1}(x + 1 - y).$$
Taking in account $(1), (6)$ and using different splittings of the integrals in the cases $x \ge y$ and $x < y,$ one can get
\begin{aligned}
(n-1)!\int_0^1\,K_1(x, u)K_{n - 1}(u,y)\,du = \int_0^1(uh(x - u) + (u - 1)h(u - x))\\
\times(B_{n - 1}(u) - h(u - y)B_{n - 1}(u - y) - h(y - u)B_{n - 1}(u + 1 - y))du\\
\end{aligned}
\begin{aligned}
 = \int_0^xuB_{n - 1}(u)\,du + \int_x^1(u - 1)B_{n - 1}(u)\,du\\
\end{aligned}
\begin{aligned}
-  h(x - y)\left(\int_0^yuB_{n - 1}(1 + u - y)\,du + \int_y^xuB_{n - 1}(u - y)\,du\\
+ \int_x^1(u - 1)B_{n - 1}(u - y)\,du\right)\\
\end{aligned}
\begin{aligned}
 - h(y - x)\left(\int_0^xuB_{n - 1}(1 + u - y)\,du + \int_x^y(u - 1)B_{n - 1}(1 + u - y)\,du\\
 + \int_y^1(u - 1)B_{n - 1}(u - y)\,du\right).
\end{aligned}
Standartization of $B_{n-1}$ arguments can be achieved by linear substitutions, and that allows to reconstruct the issue integral to the forms of
\begin{aligned}
(n-1)!\int_0^1\,K_1(x, u)K_{n - 1}(u,y)\,du
 = \int_0^xB_{n - 1}(u)\,du + \int_0^1(u-1)B_{n - 1}(u)\,du \\
\end{aligned}
\begin{aligned}
-  h(x - y)\left(\int_{1 - y}^1(u + y - 1)B_{n - 1}(u)\,du + \int_0^{x - y}(u + y)B_{n - 1}(u)\,du\\
+ \int_{x-y}^{1-y}(u + y - 1)B_{n - 1}(u)\,du\right)\\
\end{aligned}
\begin{aligned}
 - h(y - x)\left(\int_{1-y}^{x + 1 - y}(u + y - 1)B_{n - 1}(u)\,du + \int_{x + 1 - y}^1(u + y - 2)B_{n - 1}(u)\,du\\
 + \int_0^{1-y}(u + y - 1)B_{n - 1}(u)\,du\right).
\end{aligned}
\begin{aligned}
& = \int_0^xB_{n - 1}(u)\,du  + \int_0^1(u-1)B_{n - 1}(u)\,du - \int_0^1(u + y - 1)B_{n - 1}(u)\,du \\
& -  h(x - y)\int_0^{x - y}B_{n - 1}(u)\,du + h(y - x)\int_{1 - y + x}^1B_{n - 1}(u)\,du
\end{aligned}
\begin{aligned}
& = \int_0^xB_{n - 1}(u)\,du  - y\int_0^1B_{n - 1}(u)\,du \\
& -  h(x - y)\int_0^{x - y}B_{n - 1}(u)\,du + h(y - x)\int_{1 - y + x}^1B_{n - 1}(u)\,du
\end{aligned}
And now it has become possible to use the identities
$$\int_a^x\,B_{n - 1}(t)\,dt = \frac1n(B_n(x) - B_n(a)),\tag7$$
$$B_n(1) - B_n(0) = 0,\quad n > 1,\tag8$$
with the result
\begin{aligned}
&n!\int_0^1\,K_1(x, u)K_{n - 1}(u,y)\,du = B_n(x) - B_n(0 ) - y(B_n(1) - B_n(0))\\
& -  h(x - y)(B_n(x - y) - B_n(0)) + h(y - x)(B_n(1) - B_n(x + 1 - y))\\
& = B_n(x) - h(x - y)B_n(x - y) - h(y - x)B_n(x + 1 - y) - (y - h(y - x))(B_n(1) - B_n(0))\\
& = B_n(x) - h(x - y)B_n(x - y) - h(y - x)B_n(x + 1 - y),
\end{aligned}
$$\boxed{n!K_n(x, y) = B_n(x) - h(x - y)B_n(x - y) - h(y - x)B_n(x + 1 - y).}$$
So identity $(5)$ is satisfied for arbitrary $n\in\mathbb N.$
