= Solution
The substitution $u=e^{-x}$ changes the <beta function> to
$$
B(s,t)=\int_0^\infty e^{-sx}(1-e^{-x})^{t-1}\,dx.
$$
For fixed $t>0$, its factor near zero expands as
$$
(1-e^{-x})^{t-1}=x^{t-1}\left[1-\frac{t-1}{2}x+O(x^2)\right].
$$
Away from zero the factor is bounded, so the extension from the finite-interval <Watson lemma> to this exponentially damped tail is immediate. Therefore
$$
\boxed{B(s,t)=\frac{\Gamma(t)}{s^t}
\left[1-\frac{t(t-1)}{2s}+O(s^{-2})\right]\quad(s\to\infty).}
$$
The change $u\mapsto1-u$ gives $B(s,t)=B(t,s)$, so, with $s>0$ fixed,
$$
\boxed{B(s,t)=\frac{\Gamma(s)}{t^s}
\left[1-\frac{s(s-1)}{2t}+O(t^{-2})\right]\quad(t\to\infty).}
$$
These expansions hold with the other parameter fixed; the first does not claim uniformity when $t$ grows with $s$. The printed gamma-integral hint has inconsistent variable labels: the integration differential is $dx$ and the positive parameter is $y$.
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