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.stack.imgur.com/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.



What is the optimal asymptotic behavior of this integral over the sphere?
$\newcommand\la\lambda\renewcommand{\S}{\mathbb S}\newcommand{\si}{\sigma}$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.


