Asymptotic of Fourier coefficient and generalised hypergeometric function I would like to study the asymptotic behaviour of the integral
$$
I_n= \int_0^1 t^{-\alpha} \cos (n \pi t)\, dt, 
$$
where  $\alpha\in (0,1/2)$ is a fixed constant. In particular, I am interested in the decay order of $I_n$ (preferably depending on $\alpha$) as $n\to \infty$.
WolframAlpha suggests that
$$
\int_0^1 t^{-\alpha} \cos (n \pi t)\, dt = \frac{_1 F_2(\frac{1-\alpha}{2}; \frac{1}{2},\frac{3-\alpha}{2}; -\frac{1}{4}n^2\pi^2)}{1-\alpha}.
$$
I am not sure how to prove this formula. Moreover, as I am not familiar with the generalized hypergeometric function, I am not sure how to proceed from here.
Is there a reference on the asymptotic behavior of $_1 F_2$ or more directly the quantity $(I_n)_{n\in \mathbb{N}}$?
 A: It is more convenient to consider $\displaystyle I(n,\alpha)=\int_0^1t^{-\alpha}e^{i\pi nt}dt$; then our integral is the areal part of $I(n,\alpha)$. Making the substitution $x=\frac{t}{n}$
$$I(n,\alpha)=\frac{1}{n^{1-\alpha}}\int_0^nx^{-\alpha}e^{i\pi x}dx=\frac{1}{n^{1-\alpha}}\int_0^\infty x^{-\alpha}e^{i\pi x}dx-\frac{1}{n^{1-\alpha}}\int_n^\infty x^{-\alpha}e^{i\pi x}dx=I_1+I_2$$
The first integral is well-known and can be evaluated, for example, by means of integration in the complex plane.
Let's consider the integral along the following contour:

$$\frac{1}{n^{1-\alpha}}\oint z^{-\alpha}e^{i\pi z}dz=\int_r^R z^{-\alpha}e^{i\pi z}dz+I_R+\int_R^r \big(e^\frac{\pi i}{2}z\big)^{-\alpha}e^{-\pi z}idz+I_r\tag{0}$$
There are no singularities inside the contour; therefore, $\displaystyle \oint=0$.
It is not difficult to show that the integrals along the big and small quarter-circles $\,\displaystyle I_r, I_R\to 0\,\,\text{at}\, r\to 0\,\, \text{and}\,\, R\to\infty\,$ correspondingly.
Therefore,
$$I_1=\frac{1}{n^{1-\alpha}}\int_0^\infty x^{-\alpha}e^{i\pi x}dx=\frac{ie^{-\frac{i\pi\alpha}{2}}}{n^{1-\alpha}}\int_0^\infty x^{-\alpha}e^{-\pi x}dx=\frac{e^\frac{i\pi(1-\alpha)}{2}}{n^{1-\alpha}}\int_0^\infty x^{-\alpha}e^{-\pi x}dx$$
$$I_1=\frac{e^\frac{i\pi(1-\alpha)}{2}}{(\pi n)^{1-\alpha}}\Gamma(1-\alpha)\tag{1}$$
$\displaystyle I_2$ we evaluate via integration by part
$$I_2=-\frac{1}{n^{1-\alpha}}\int_n^\infty x^{-\alpha}e^{i\pi x}dx=-\frac{1}{n^{1-\alpha}}\frac{e^{i\pi z}x^{-\alpha}}{i\pi}\bigg|_{x=n}^\infty-\frac{\alpha}{n^{1-\alpha}i\pi}\int_n^\infty x^{-\alpha-1}e^{i\pi x}dx$$
$$=i\frac{(-1)^{n+1}}{\pi n}+\frac{(-1)^{n+1}\alpha}{(\pi n)^2}+\frac{\alpha (1+\alpha)}{\pi^2 n^{1-\alpha}}\int_n^\infty x^{-\alpha-2}e^{i\pi x}dx\tag{2}$$
Taking together (1) and (2)
$$I=\frac{e^\frac{i\pi(1-\alpha)}{2}}{(\pi n)^{1-\alpha}}\Gamma(1-\alpha)+i\frac{(-1)^{n+1}}{\pi n}+\frac{(-1)^{n+1}\alpha}{(\pi n)^2}+\frac{\alpha (1+\alpha)}{\pi^2 n^{1-\alpha}}\int_n^\infty x^{-\alpha-2}e^{i\pi x}dx$$
Taking the real part
$$\int_0^1 t^{-\alpha} \cos (n \pi t)\, dt=\frac{\sin\frac{\pi\alpha}{2}}{(\pi n)^{1-\alpha}}\Gamma(1-\alpha)+\frac{(-1)^{n+1}\alpha}{(\pi n)^2}+O\Big(\frac{1}{n^4}\Big)\,,\,\, n\gg1$$
Integrating several times by part and using $\displaystyle \Gamma(\alpha)\alpha(1+\alpha)...(m+\alpha)=\Gamma(\alpha+m+1)$ we can get a complete asymptotic series:
$$\boxed{\,\,\int_0^1 t^{-\alpha} \cos (n \pi t)\, dt\sim\frac{\sin\frac{\pi\alpha}{2}}{(\pi n)^{1-\alpha}}\Gamma(1-\alpha)+\frac{(-1)^n}{\Gamma(\alpha)}\sum_{k=1}^\infty (-1)^k\frac{\Gamma(\alpha+2k-1)}{(\pi n)^{2k}}\,\,}$$
A: If we go back to the definition of this specific hypergeometric function,
$$\int_0^1 t^{-\alpha} \cos (n \pi t)\, dt =\sum_{k=0}^\infty (-1)^k \frac{(n \pi)^{2k} } {(2k+1-\alpha)\,(2k)! }$$
With
$$a_k=\frac{(n \pi)^{2k} } {(2k+1-\alpha)\,(2k)! }\qquad\implies \qquad \frac{a_{k+1}}{a_k}=\frac{(n\pi)^2\,  (2 k+1-\alpha )}{2 (k+1) (2 k+1) (2 k+3-\alpha )} \sim \left(\frac {n\pi}{2k}\right)^2$$
Edit
If you look at this paper (equation $3.1$) and use it, the asymptotics is immediate.
