Prove that $\int_{0}^\pi x^{2k} \cos(h x) dx\geq 0$.

Question

Prove that $$\int_{0}^\pi x^{2k} \cos(h x) dx\geq 0$$ for all $k\in\mathbb N$ and $h$ even integer. I have tried with Induction Principle (for $h$) but without success.

Answer 1

Let $h=2n$, and $f$ be convex on $[0,\pi]$. Then $g(x)=\frac1{2n}\,f\!\left(\frac{x}{2n}\right)$ is convex on $[0,2\pi n]$. $$ \begin{align} \int_0^\pi f(x)\cos(hx)\,\mathrm{d}x &=\int_0^\pi f(x)\cos(2nx)\,\mathrm{d}x\tag{1}\\ &=\int_0^{2\pi n}\frac1{2n}\,f\!\left(\frac{x}{2n}\right)\cos(x)\,\mathrm{d}x\tag{2}\\ &=\int_0^{2\pi n}g(x)\cos(x)\,\mathrm{d}x\tag{3}\\ &=\sum_{j=0}^{n-1}\int_0^{2\pi}g(x+2\pi j)\cos(x)\,\mathrm{d}x\tag{4}\\ \end{align} $$ Explanation
$(1)$: substitute $h=2n$
$(2)$: substitute $x\mapsto\frac{x}{2n}$
$(3)$: $g(x)=\frac1{2n}\,f\!\left(\frac{x}{2n}\right)$
$(4)$: $\cos(x+2\pi j)=\cos(x)$

If $h$ is convex, then for $z\ge y$, we have $h(z+\pi)-h(y+\pi)\ge h(z)-h(y)$. Therefore, $$ \begin{align} \int_0^{2\pi}h(x)\cos(x)\,\mathrm{d}x &=\int_0^{\pi/2}\left(h(x)-h(\pi-x)-h(\pi+x)+h(2\pi-x)\right)\cos(x)\,\mathrm{d}x\\ &=\int_0^{\pi/2}\left(\left[h(2\pi-x)-h(\pi+x)\right]-\left[h(\pi-x)-h(x)\right]\right)\cos(x)\,\mathrm{d}x\\[6pt] &\ge0\tag{5} \end{align} $$ Combining $(4)$ and $(5)$ and noting that $f(x)=x^{2k}$ is convex on $[0,\pi]$ gives the desired result.

Answer 2

Do induction on even k and odd k separately. First note that this is true for k=1 and k=2 (This will be the base case). Use integration by parts twice to calculate,

$\int_{0}^{\pi}x^{2k}\cos(hx)dx = \frac{2k\pi^{2k-2}}{h^2}+\frac{2k(2k-1)}{h^2}\int_{0}^{\pi}x^{2(k-1)}\cos(hx)dx$

The result then follows by induction on k odd, k even separately.