这是一份imperial帝国理工大学 MATH19872作业代写的成功案例
(i) $\phi^{(k)}(0)=0$ for $k2 n$;
(ii) $\phi^{(k)}(0)$ and $\phi^{(k)}(1)$ are integers for all $k \in \mathbf{N}$.
Suppose that $\pi^{2}=p / q$ for some positive integers $p, q$, and define
$$
F=q^{n} \sum_{k=0}^{n}(-1)^{2 n-k} \pi^{2(n-k)} \phi^{(2 k)}
$$
Show that $F(0)$ and $F(1)$ are integers, and that
$$
F^{\prime \prime}(x)+\pi^{2} F(x)=\pi^{2} p^{n} \phi(x) .
$$
Setting
$$
G(x)=F^{\prime}(x) \sin \pi x-\pi F(x) \cos \pi x
$$
show that
$$
G^{\prime}(x)=\pi^{2} p^{n} \phi(x) \sin \pi x
$$
and hence that
$$
\pi \int_{0}^{1} p^{n} \phi(x) \sin \pi x \mathrm{~d} x
$$
MATH19872 COURSE NOTES ：
Proof. Let $F$ be a Lebesgue primitive of $f$. Then $F \in \mathcal{P}{f}$, , $$ F{}(\eta)-F_{}(\xi) \leq F(\eta)-F(\xi) \quad(\xi \leq \eta) .
$$
Since a finite union of sets of measure zero has measure zero, it follows from this inequality and Theorem (2.1.10) that
$$
f \leq F_{}^{\prime} \leq F^{\prime}=f $$ almost everywhere. Hence $F_{}^{\prime}=f$ almost everywhere, and $F_{*}$ is a Lebesgue primitive of $f$.