Show that there is a unique polynomial $\int_{x}^{x+1} b_{n}{(t)} dt = x^n$ For each n = 1, 2, 3, . . . show that there is a unique polynomial $ b_{n}{(x)} $ that
satisfies the equation
$\int_{x}^{x+1} b_{n}{(t)} dt = x^n$
n = 1, I find $ b_{1}{(t)} = t + x_{0} $ and find $x_{0} = -\frac{1}{2}$
n = 2, $ b_{2}{(t)} = t^2 - t + \frac{1}{6} $
How  can I prove that the polynomial is unique and always satisfy the equation?
 A: First observe that
$$
\int_x^{x + 1} t^n \,dt = \frac{(x + 1)^{n + 1} - x^n}{n+1} = \frac{1}{n + 1} \sum_{k = 0}^n {n + 1 \choose k} x^k.
$$
Therefore
$$\int_x^{x+1} t^n \,dt = x^n + \text{lower order terms}.$$
Therefore the matrix which writes $\{\int_{x}^{x+1} t^n \,dt : n = 0, 1, 2, 3\dots \}$ in terms of $\{1,x,x^2,x^3,\dots\}$ is triangular with ones down the diagonal. In particular, it is invertible. Because $\{1,x,x^2,x^3,\dots\}$ is a basis for the ring of polynomials, this means that $\{\int_{x}^{x+1} t^n \,dt : n = 0, 1, 2, 3\dots \}$ is also a basis. Therefore every polynomial in $x$ can be written uniquely as a linear combination
$$
\begin{align*}
p(x) &= a_n \int_{x}^{x + 1} t^n \,dt + a_{n - 1}\int_x^{x+1} t^{n - 1} \,dt + \dots + a_1 \int_x^{x+1} t \,dt + a_0 \int_x^{x+1} 1\, dt \\
&= \int_x^{x+1} a_nt^n + a_{n-1}t^{n-1} + \dots + a_1t + a_0 \, dt.
\end{align*}
$$
A: Taking a derivative with respect to $x$ yields $$ -n x^{(-1 + n)} - b_n(x) + b_n(1 + x)=0,$$ which is a simple difference equation with solution $$b_n(t)=-n \zeta_{1 - n}(t),$$ where $\zeta_{1 - n}(t)$ is the Hurwitz zeta function. For $n=1,2...$ it of course gives your solutions.
For a negative integer order, ie $1-n<=0$, the $\zeta$ function is related to the Bernouli polynomials, see here.
A: I think this problem has more to do with linear algebra than real analysis.
Let $\phi: \mathbb R_n[X]\to\mathbb R_n[X], P\mapsto (x\mapsto \int_x^{x+1}P(t)dt) $.


*

*$\phi$ is well defined. Let $Q:x\mapsto \int_x^{x+1}P(t)dt$. $Q$ is indeed a polynomial, and since $$\int_x^{x+1}t^n dt = \frac{(x+1)^{n+1}-x^n}{n+1}=\frac{1}{n+1}x^{n}+\ldots$$ $Q$ has degree less than $n$.

*$\phi$ is linear (I leave the details to you).

*$\phi$ is injective: let $P\in \mathbb R_n[X]$ such that $\phi(P)=0$. Take the derivative with respect to $x$ to get $\forall x \in \mathbb R,\; P(x+1)-P(x)=0$. Since $P$ is a polynomial, this holds true for complex $x$, thus, if $x_0$ is a complex root of $P$, then for all $n$, $x_0+n$ is a root of $P$, hence $P=0$.
$P$ is linear and injective between linear spaces that have same dimensions, it must therefore be one-to-one.
