Show that for the Fourier coefficients of the function $f(x)=\frac{1}{\sqrt{x}},\; 0<x\leq 2\pi$, as $n\to \infty$ the following asymptotic relations hold:
$a_n \sim \frac{1}{\sqrt{2\pi n}}$ and $b_n \sim \frac{1}{\sqrt{2\pi n}}$.
My attempt: Since I suppose in a similar way we can prove both of the relations, I tried solving the problem for $a_n$ coefficients:
$$a_n=\frac{1}{\pi}\int_0^{2\pi}f(x)\cos{nx}dx=\frac{1}{\pi}\int_0^{2\pi}\frac{\cos{nx}}{\sqrt{x}}dx=\frac{1}{\pi\sqrt{n}}\int_0^{\sqrt{2n\pi}}\cos{t^2}dt$$
and here I'm stuck!
We may write $$ a_n = \frac{1}{\pi }\operatorname{Re} \int_0^{2\pi } {\frac{{\mathrm{e}^{\mathrm{i}nx} }}{{\sqrt x }}\mathrm{d}x} = \frac{1}{\pi }\operatorname{Re} \int_0^{ + \infty } {\frac{{\mathrm{e}^{\mathrm{i}nx} }}{{\sqrt x }}\mathrm{d}x} - \frac{1}{\pi }\operatorname{Re} \int_{2\pi }^{ + \infty } {\frac{{\mathrm{e}^{\mathrm{i}nx} }}{{\sqrt x }}\mathrm{d}x} . $$ The second part, via integration by parts, is $$ \frac{1}{\pi }\operatorname{Re} \int_{2\pi }^{ + \infty } {\frac{{\mathrm{e}^{\mathrm{i}nx} }}{{\sqrt x }}\mathrm{d}x} = \frac{1}{{2\pi n}}\int_{2\pi }^{ + \infty } {\frac{{\sin (nx)}}{{x^{3/2} }}\mathrm{d}x} = \mathcal{O}\!\left( {\frac{1}{n}} \right). $$ The first one, by deforming the contour of integration, is \begin{align*} \frac{1}{\pi }\operatorname{Re} \int_0^{ + \infty } {\frac{{\mathrm{e}^{\mathrm{i}nx} }}{{\sqrt x }}\mathrm{d}x} & = \frac{1}{\pi }\operatorname{Re} \left[ {\mathrm{e}^{\frac{\pi }{4}\mathrm{i}} \int_0^{ - \mathrm{i}\infty } {\frac{{\mathrm{e}^{ - nt} }}{{\sqrt t }}\mathrm{d}t} } \right] = \frac{1}{\pi }\operatorname{Re} \left[ {\mathrm{e}^{\frac{\pi }{4}\mathrm{i}} \int_0^{ + \infty } {\frac{{\mathrm{e}^{ - nt} }}{{\sqrt t }}\mathrm{d}t} } \right] \\ & = \frac{1}{\pi }\operatorname{Re} \left[ {\mathrm{e}^{\frac{\pi }{4}\mathrm{i}} \frac{1}{{\sqrt n }}\Gamma\! \left( {\frac{1}{2}} \right)} \right] = \frac{1}{\pi }\frac{1}{{\sqrt 2 }}\frac{1}{{\sqrt n }}\sqrt \pi = \frac{1}{{\sqrt {2\pi n} }}. \end{align*} Hence, $$ a_n = \frac{1}{{\sqrt {2\pi n} }} + \mathcal{O}\left( {\frac{1}{n}} \right) $$ as required. The treatment of $b_n$ is similar.