Asymptotic expansion on the following integral of exponential function

Question

I wish to obtain the asymptotic for the following integral $$ \int_{r: \|r\|\le 1} \exp(M\cdot a^Tr) \, dr, $$ where $a$ is a given vector in $\mathbb{R}^d$, $\|\cdot\|$ is a general norm function and $M\to\infty$. I know the maximizer of $\max\{a^Tr: \|r\|\le 1\}$ would be $\|a\|_*$. However, still confused how to simplify such kind of integral.

Answer 1

$\newcommand\de\delta\newcommand\ep\varepsilon$The logarithmic asymptotic is given by the formula $$I(M):=\int_{r\colon\|r\|\le1}\exp(M\,a^Tr)\,dr=\exp(M(1+o(1))\,\|a\|_*) \tag{1}\label{1}$$ (as $M\to\infty$), where $\|a\|_*:=\sup\{a^T r\colon r\in B_1\}\ne0$ and $B_R:=\{r\colon\|r\|<R\}$.

Indeed, the unit ball $B_1$ is a nonempty bounded open set. So, $m:=|B_1|\in(0,\infty)$, where $|\cdot|$ is the Lebesgue measure. Similarly, for any $\de\in(0,1)$, the set $A_\de:=B_1\setminus B_{1-\de}$ contains a nonempty open set, and hence $\ep_\de:=|A_\de|\in(0,m]\subset(0,\infty)$. Now write $$I(M)\le\int_{B_1}\exp(M\,\|a\|_*)\,dr=\exp(M\,\|a\|_*)\,m$$ and $$I(M)\ge\int_{A_\de}\exp(M\,a^Tr)\,dr \ge\exp((1-\de)M\,\|a\|_*)\ep_\de.$$ Letting now $M\to\infty$, we get $$I(M)\le\exp((1+o(1))M\,\|a\|_*)$$ and $$I(M)\ge\exp((1-\de+o(1))M\,\|a\|_*),$$ for each $\de\in(0,1)$. Thus, \eqref{1} follows. $\quad\Box$