State the Riemann-Lebesgue lemma as applied to the integral

$$\int_{a}^{b} g(u) e^{i x u} d u$$

where $g^{\prime}(u)$ is continuous and $a, b \in \mathbb{R}$.

Use this lemma to show that, as $x \rightarrow+\infty$,

$$\int_{a}^{b}(u-a)^{\lambda-1} f(u) e^{i x u} d u \sim f(a) e^{i x a} e^{i \pi \lambda / 2} \Gamma(\lambda) x^{-\lambda}$$

where $f(u)$ is holomorphic, $f(a) \neq 0$ and $0<\lambda<1$. You should explain each step of your argument, but detailed analysis is not required.

Hence find the leading order asymptotic behaviour as $x \rightarrow+\infty$ of

$$\int_{0}^{1} \frac{e^{i x t^{2}}}{\left(1-t^{2}\right)^{\frac{1}{2}}} d t$$