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

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  • $\begingroup$ Expression $cos(2πkx)=c\int\limits_0^1K_4(x,u)cos(2πku)du$ is satisfied when $K_4(x,u)= \dfrac1c \delta(u−x).$ Is that part of the task correct? $\endgroup$ – Yuri Negometyanov Feb 16 '18 at 8:16
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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 $n\in\mathbb N.$

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  • $\begingroup$ @PeterSheldrick There is not an error, because the integrals of the both of negative parts of expression were glued to the same expression, which were subtracted from the issue (positive) one. Thank you for comment, I've detalized that part of the answer. $\endgroup$ – Yuri Negometyanov Feb 15 '18 at 22:18
  • $\begingroup$ @PeterSheldrick, I've used 2D intervals method. When $x\ge y$ (factor h(x-y)) took place, we had first splitting set of limits, and otherwise (factor h(y-x)) - the second set. So the negative parts are not the same at all. Then the (discussed) common part were extracted (and subtracted), and differences have left and were included in the final expression. $\endgroup$ – Yuri Negometyanov Feb 15 '18 at 22:50

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