Consider the real function $f(t)$ of a real variable $t$ defined by the following contour integral in the complex $s$-plane:

$$f(t)=\frac{1}{2 \pi i} \int_{\Gamma} \frac{e^{s t}}{\left(s^{2}+1\right) s^{1 / 2}} d s,$$

where the contour $\Gamma$ is the line $s=\gamma+i y,-\infty<y<\infty$, for constant $\gamma>0$. By closing the contour appropriately, show that

$$f(t)=\sin (t-\pi / 4)+\frac{1}{\pi} \int_{0}^{\infty} \frac{e^{-r t} d r}{\left(r^{2}+1\right) r^{1 / 2}}$$

when $t>0$ and is zero when $t<0$. You should justify your evaluation of the inversion integral over all parts of the contour.

By expanding $\left(r^{2}+1\right)^{-1} r^{-1 / 2}$ as a power series in $r$, and assuming that you may integrate the series term by term, show that the two leading terms, as $t \rightarrow \infty$, are

$$f(t) \sim \sin (t-\pi / 4)+\frac{1}{(\pi t)^{1 / 2}}+\cdots$$

[You may assume that $\int_{0}^{\infty} x^{-1 / 2} e^{-x} d x=\pi^{1 / 2}$.]