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