Pome

Lemma 18.43.199A4F9F834BF

Lemma 18.43.1.99A4F9F834BF Set $W(a,b)=ab+a^3+b^3$ on $[0,\infty )^2$. As $t\to \infty $,

\[ \begin{gathered} J(t):=\int _0^\infty \int _0^\infty \mathrm e^{-tW(a,b)}\, \mathrm da\, \mathrm db =\frac{\log t}{3t}+O(t^{-1}), \\ \int _0^\infty \int _0^\infty (a+b)\mathrm e^{-tW(a,b)} \, \mathrm da\, \mathrm db=O(t^{-4/3}). \end{gathered} \]

Proof. Put $A=t^{-2/3}$ and $B=t^{-1/3}$; the logarithm comes from the interval $A\leq a\leq B$. For $a>0$, the substitution $u=tab$ gives

\[ H_ t(a):=\int _0^\infty \mathrm e^{-t(ab+b^3)}\, \mathrm db =\frac1{ta}\int _0^\infty \mathrm e^{-u-u^3/(t^2a^3)}\, \mathrm du. \]

Since $0\leq 1-\mathrm e^{-v}\leq v$ for $v\geq 0$,

\[ 0\leq \frac1{ta}-H_ t(a)\leq \frac6{t^3a^4}. \]

Over $A\leq a\leq B$, the integral of this error is $O(t^{-1})$. Replacing $\mathrm e^{-ta^3}$ by one in $\int _ A^ B\mathrm e^{-ta^3}/(ta)\, \mathrm da$ costs at most $\int _ A^ B a^2\, \mathrm da=O(t^{-1})$ as well. Thus the middle interval contributes

\[ \int _ A^ B\mathrm e^{-ta^3}H_ t(a)\, \mathrm da =\frac1t\log \frac BA+O(t^{-1}) =\frac{\log t}{3t}+O(t^{-1}). \]

For $a<A$, the bound $H_ t(a)\leq \int _0^\infty \mathrm e^{-tb^3}\, \mathrm db=O(t^{-1/3})$ gives a contribution $O(t^{-1})$. For $a>B$, use $H_ t(a)\leq 1/(ta)$ and substitute $v=ta^3$ to obtain the bound

\[ \int _ B^\infty \frac{\mathrm e^{-ta^3}}{ta}\, \mathrm da =\frac1{3t}\int _1^\infty \frac{\mathrm e^{-v}}v\, \mathrm dv =O(t^{-1}). \]

Finally, $aH_ t(a)\leq 1/t$ gives

\[ \int _0^\infty \int _0^\infty a\mathrm e^{-tW(a,b)} \, \mathrm da\, \mathrm db \leq \frac1t\int _0^\infty \mathrm e^{-ta^3}\, \mathrm da =O(t^{-4/3}). \]

Symmetry proves the same bound for $b$. $\square$

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