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.
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 $$ J_p(z)=\sum_{n\geq0}\frac{(-1)^n}{n!\Gamma(n+p+1)}\Big(\frac{z}{2}\Big)^{p+2n} $$
INCOMPLETE.
Certainly you can compute that when $d$ is odd.
For all $d\ge 2$, the surface measure reads $$ d\sigma_{d-1}(w)=(1-w_1^2)^{\frac{d-3}{2}}dw_1 d\sigma_{d-2}, $$ so our integral reads \begin{equation*} \begin{split} I(a)&=\int_{\mathbb{S}^{d-1}} (1-\cos(a w_1)) \, (1-w_1^2)^{\frac{d-3}{2}}dw_1 d\sigma_{d-2} \\ &=\int_{-1}^1 (1-\cos(aw_1))(1-w_1^2)^\frac{d-3} {2}\, dw_1 \int_{\mathbb S^{d-2}} d\sigma_{d-2}\\ &=\lvert\mathbb S^{d-2}\rvert \int_{-1}^1 (1-\cos(aw_1))(1-w_1^2)^\frac{d-3} {2}\, dw_1. \end{split} \end{equation*} When $d$ is odd, the term $\frac{d-3}{2}$ is integer and the right-hand side integral is of the form $\int_{-1}^1 (1-\cos(ax))P(x)\, dx$ for a polynomial $P$, which can be explicitly computed by repeated integration by parts.
For $d=3$ this is very simple, as you do not need any integration by parts. Recalling $\lvert \mathbb S^{d-2}\rvert=\lvert\mathbb S^1\rvert=2\pi$, $$ I(a)=4\pi\left(1-\frac{\sin a}{a}\right).$$
Proving your conjecture in this $d=3$ case is also immediate, as $\frac{\sin a}{a}$ has a strict global maximum at $a=0$, so $I(a)\ge C_\epsilon >0$ for $\lvert a \rvert \ge \epsilon$.
It may be the case that you can compute $I(a)$ also for even $d$, but that requires some trick to get rid of the square root in $(1-w_1^2)^\frac{d-3} {2}$.
We want to calculate $$F(s)=\int\limits_{\partial \mathbb B(0,1)}(1-\cos(sx_1)) \mathrm d^{n-1}\boldsymbol x$$
Of course you should switch to hyperspherical coordinates: $$F(s)=\int\limits_{\theta_{n-1}=0}^{2\pi}~\underbrace{\int\limits_{\theta_{n-2}=0}^\pi\cdots\int\limits_{\theta_{1}=0}^\pi}_{n-2~\text{of these}}\big(1-\cos(s\cos\theta_1)\big)~\prod_{k=1}^{n-1}\sin(\theta_i)^{n-1-k}\mathrm d\theta_k$$ Now the integrals are separable. $$F(s)=\int_{0}^{2\pi}\int_0^\pi\cdots \int_0^\pi\mathrm d\theta_{n-1} \prod_{k=1}^{n-3}\sin(\theta_k)^{n-1-k}\mathrm d\theta_k~\cdot~\int_0^\pi \big(1-\cos(s\cos\theta_1)\big) \sin(\theta_1)^{n-2} \mathrm d\theta_1 \\ =2\pi \prod_{k=1}^{n-3}\int_0^\pi \sin(\theta_k)^{n-k-1}\mathrm d\theta_k~\cdot~\int_0^\pi \big(1-\cos(s\cos\theta_1)\big) \sin(\theta_1)^{n-2} \mathrm d\theta_1\tag{*}$$
All the integrals of the form $\int_0^\pi \sin(t)^k\mathrm dt=2\int_0^{\pi/2}\sin(t)^k\mathrm dt$ are evaluated easily with the Beta function, so all that is left is the integral $$\int_0^\pi \cos\big(s\cos t\big)\sin(t)^{n-2}\mathrm dt $$ There is a well known integral representation of the Bessel function $$J_{\nu}(z)=\frac{(z/2)^\nu}{\sqrt{\pi} ~\Gamma(\nu+1/2)}\int_0^\pi\cos(z\cos \theta)\sin(\theta)^{2\nu}\mathrm d\theta \\ \Re(\nu)>-1/2$$ Which in our case means $$\int_0^\pi \cos\big(s\cos t\big)\sin(t)^{n-2}\mathrm dt \\ =\frac{\sqrt{\pi}~\Gamma\left(\frac{n-1}{2}\right)}{(s/2)^{\frac{n-2}{2}}}J_{\frac{n-2}{2}}(s)$$
So now just go back to $(\ast )$ and fill in the gaps.
