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 $$ B_{n+m}(x)=\frac{(n+m)!}{m!}\int_0^1K_n(x,u)B_m(u)\textrm{d}u\\ $$

Solution 1:

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 $n\in\mathbb N.$