For $\lambda>0$ let

$$I(\lambda)=\int_{0}^{b} f(x) \mathrm{e}^{-\lambda x} d x, \quad \text { with } \quad 0<b<\infty$$

Assume that the function $f(x)$ is continuous on $0<x \leqslant b$, and that

$$f(x) \sim x^{\alpha} \sum_{n=0}^{\infty} a_{n} x^{n \beta}$$

as $x \rightarrow 0_{+}$, where $\alpha>-1$ and $\beta>0$.

(a) Explain briefly why in this case straightforward partial integrations in general cannot be applied for determining the asymptotic behaviour of $I(\lambda)$ as $\lambda \rightarrow \infty$.

(b) Derive with proof an asymptotic expansion for $I(\lambda)$ as $\lambda \rightarrow \infty$.

(c) For the function

$$B(s, t)=\int_{0}^{1} u^{s-1}(1-u)^{t-1} d u, \quad s, t>0$$

obtain, using the substitution $u=e^{-x}$, the first two terms in an asymptotic expansion as $s \rightarrow \infty$. What happens as $t \rightarrow \infty$ ?

[Hint: The following formula may be useful

$$\Gamma(y)=\int_{0}^{\infty} x^{y-1} \mathrm{e}^{-x} d t, \quad \text { for } \quad x>0$$