I am having trouble, trying to compute the integral
\begin{align*}
I(a)= \int_{\mathbb{S}^{d-1}} (1-\cos(a w_1)) \, d \sigma_{d-1}(w)\qquad a>1
\end{align*}
where $w= (w_1,\cdots, w_d)$ and $\mathbb{S}^{d-1}=\{w\in \mathbb{R}^{d}\,:\, w_1^2+\cdots+w_d^2=1\}$.
1- Can the explicit value of $I(a)$ be computed?
2- How can prove that there is $c>0$ such that $I(a)\geq c$ for all $a>1$?
PS: The second question is a conjecture.
Best Answer
This posting is to address question (2) of the OP. Guiseppe already addressed part (1).
Fixing $d$, we have from Fejer's theorem (see theorem in this posting for example) or alternatively, from Riemann-Lebesgue's lemma, that \begin{align} \int^1_{-1}(1-\cos(ax))(1-x^2)^{\frac{d-3}{2}}\,dx&\xrightarrow{a\rightarrow\infty}\Big(\frac{1}{2\pi}\int^{2\pi}_0(1-\cos x)\,dx\big)\Big(\int^1_{-1}(1-x^2)^{\frac{d-3}{2}}\,dx\Big)\\ &=\int^1_0(1-u)^{(d-3)/2}u^{-1/2}\,du=B(\tfrac12,\tfrac{d-1}{2})>0 \end{align} Thus, there is $a_d>0$ such $I(a)>\frac12B(\tfrac12,\tfrac{d-1}{2})$ whenever $a\geq a_d$. Notice that $a\mapsto I(a)$ is positive and continuous on $(0,\infty)$. It follows that there is $C_d>0$ such that $I(a)\geq C_d$ for all $a\geq 1$.
Edit: The integral in the OP can also be written in terms of the Bessel function of the first kind $J_p$:
\begin{align} I(a)&=|\mathbb{S}_{d-1}|-|\mathbb{S}_{d-2}|\int^1_{-1}(1-x^2)^{\frac{d-3}{2}} e^{-iax}\,dx\\ &=\frac{2\pi^{d/2}}{\Gamma(d/2)}-\frac{2\pi^{\frac{d-1}{2}}}{\Gamma(\frac{d-1}{2})}\frac{\Gamma(\frac{d-1}{2})\sqrt{\pi}}{(a/2)^{\frac{d-2}{2}}}J_{\frac{d-2}{2}}(a)\\ &=\frac{2\pi^{d/2}}{\Gamma(d/2)}-\frac{(2\pi)^{\frac{d}{2}}}{a^{\frac{d-1}{2}}}J_{\frac{d-2}{2}}(a) \end{align} where $$ J_p(z)=\sum_{n\geq0}\frac{(-1)^n}{n!\Gamma(n+p+1)}\Big(\frac{z}{2}\Big)^{p+2n} $$