What is the optimal asymptotic behavior of this integral over the sphere?

Question

Let $k_{1},\dots, k_{d}>1$ be integers and consider the integral $$J_{\lambda }=\int_{\mathbb{S}^{d-1}}e^{-\lambda \left(x^{2k_{1}}_{1}+\dots+ x^{2k_{d}}_{d}\right)} d\sigma(x)$$ where $d\sigma$ denotes the standard surface measure on $\mathbb{S}^{d-1}$, the unit sphere in $\mathbb{R}^{d}$, $d\geq 2$.

I can not figure out the asymptotic behaviour of $I_{\lambda}$ ad $\lambda\rightarrow \infty$.

Obviously, by the dominated convergence theorem, $I_{\lambda}\rightarrow 0$. We can also write $$J_{\lambda }=2\int_{\substack{ (x_{1},\dots,x_{d-1})\in\mathbb{R}^{d-1}\\ x_{1}^{2}+\dots+x_{d-1}^{2}<1}} e^{-\lambda \left(x^{2k_{1}}_{1}+\dots+ x^{2k_{d-1}}_{d-1}\right)-\lambda\left(1-x_{1}^{2}-\dots-x_{d-1}^{2}\right)^{k_{d}}} \\\frac{1}{\sqrt{1-x_{1}^{2}-\dots-x_{d-1}^{2}}} dx_{1}\dots dx_{d-1}.$$ For this formula and the transformation behind it, see e.g. the attached extract from Appendix D in Grafakos's Classical Fourier Analysis: [1]: https://i.sstatic.net/hNwed.png

Now, when $x_{1}^{2}+\dots+x_{d-1}^{2}<1$ there exist two positive constants $c_{1}, c_{2}$ such that $$c_{2} (x^{2}_{1}+\dots+ x^{2}_{d-1})^{k_{max}}\leq x^{2k_{1}}_{1}+\dots+ x^{2k_{d-1}}_{d-1}\leq c_{1} (x^{2}_{1}+\dots+ x^{2}_{d-1})^{k_{min}},$$ where $k_{min}=\min_{1\leq i \leq d-1}{k_{i}}$ and $k_{max}=\max_{1\leq i \leq d-1}{k_{i}}$.

Therefore, using spherical coordinates, we have $$\int_{0}^{1} \frac{e^{-\lambda r^{2k_{max}}-\lambda\left(1-r^{2}\right)^{k_{d}}}}{\sqrt{1-r^{2}}} dr\gtrsim J_{\lambda }\gtrsim\int_{0}^{1} \frac{e^{-\lambda r^{2k_{min}}-\lambda\left(1-r^{2}\right)^{k_{d}}}}{\sqrt{1-r^{2}}} dr.$$

And since $1-r^2\leq r^2$ iff $1/\sqrt{2}\leq r$, we deduce that $J_{\lambda}$ decays faster than $$\int_{0}^{1/\sqrt{2}} \frac{e^{-2\lambda r^{2k_{MM}}}}{\sqrt{1-r^{2}}} dr,$$ where $k_{MM}=\max\{k_{max},k_{d}\}$ and slower than $$\int_{1/\sqrt{2}}^{1} \frac{e^{-2\lambda r^{2k_{M}}}}{\sqrt{1-r^{2}}} dr,$$ where $k_{M}=\min\{k_{min},k_{d}\}$.

It would be very helpful to find the asymptotic behaviour of either one of the last two integrals.

Answer 1

$\newcommand\la\lambda\renewcommand{\S}{\mathbb S}\newcommand{\si}{\sigma}$Let us show that \begin{equation*} J_\la=e^{-\la(m+o(1))} \tag{1}\label{1} \end{equation*} (as $\la\to\infty$), where \begin{equation*} m:=\min_{x\in\S^{d-1}}s(x),\quad s(x):=\sum_1^d x_j^{2k_j} \end{equation*} for $x=(x_1,\dots,x_d)$.

Indeed, take any real $h>0$. Note that $m=s(y)$ for some $y\in\S^{d-1}$. Since the function $s$ is continuous, there is a neighborhood $N_h$ of $y$ on $\S^{d-1}$ such that $s\le m+h$ on $N_h$. Also, $c_h:=\si(N_h)>0$. So, \begin{equation*} J_\la\ge\int_{N_h}e^{-\la s(x)}\si(dx) \ge e^{-\la(m+h)}\,c_h=e^{-\la(m+h+o(1))}. \end{equation*} On the other hand, \begin{equation*} J_\la\le e^{-\la m}\,\si(\S^{d-1})=e^{-\la(m+o(1))}. \end{equation*} Since $h>0$ is arbitrary, \eqref{1} follows.