(a) Consider the integral

$$I(k)=\int_{0}^{\infty} f(t) e^{-k t} d t, \quad k>0$$

Suppose that $f(t)$ possesses an asymptotic expansion for $t \rightarrow 0^{+}$of the form

$$f(t) \sim t^{\alpha} \sum_{n=0}^{\infty} a_{n} t^{\beta n}, \quad \alpha>-1, \quad \beta>0$$

where $a_{n}$ are constants. Derive an asymptotic expansion for $I(k)$ as $k \rightarrow \infty$ in the form

$$I(k) \sim \sum_{n=0}^{\infty} \frac{A_{n}}{k^{\gamma+\beta n}}$$

giving expressions for $A_{n}$ and $\gamma$ in terms of $\alpha, \beta, n$ and the gamma function. Hence establish the asymptotic approximation as $k \rightarrow \infty$

$$I_{1}(k)=\int_{0}^{1} e^{k t} t^{-a}\left(1-t^{2}\right)^{-b} d t \sim 2^{-b} \Gamma(1-b) e^{k} k^{b-1}\left(1+\frac{(a+b / 2)(1-b)}{k}\right)$$

where $a<1, b<1$.

(b) Using Laplace's method, or otherwise, find the leading-order asymptotic approximation as $k \rightarrow \infty$ for

$$I_{2}(k)=\int_{0}^{\infty} e^{-\left(2 k^{2} / t+t^{2} / k\right)} d t$$

[You may assume that $\Gamma(z)=\int_{0}^{\infty} t^{z-1} e^{-t} d t$ for $\operatorname{Re} z>0$,

$$\text { and that } \left.\int_{-\infty}^{\infty} e^{-q t^{2}} d t=\sqrt{\pi / q} \text { for } q>0 .\right]$$