Given real parameters $\left(\alpha,\beta,\gamma,\delta\right)\in\mathbb{R}^{4}$ such that $0<\alpha<\delta$ and real arguments $\left(x,y\right)\in\left(-\infty,1\right)^{2}$, we can express the Appell $F_{1}$ function via the integral representation
$$\begin{align}
F_{1}{\left(\alpha;\beta,\gamma;\delta;x,y\right)}
&=\frac{1}{\operatorname{B}{\left(\alpha,\delta-\alpha\right)}}\int_{0}^{1}\mathrm{d}t\,\frac{t^{\alpha-1}\left(1-t\right)^{\delta-\alpha-1}}{\left(1-xt\right)^{\beta}\left(1-yt\right)^{\gamma}}.\\
\end{align}$$
Starting from the integral representation of the $F_{1}$ function for the particular set of parameters that we're interested in, we obtain an integral of a simple algebraic function with elementary antiderivative: for any fixed but arbitrary $\left(x,y\right)\in\left(-\infty,1\right)^{2}$,
$$\begin{align}
F_{1}{\left(1;1,\frac12;2;x,y\right)}
&=\frac{1}{\operatorname{B}{\left(1,1\right)}}\int_{0}^{1}\mathrm{d}t\,\frac{1}{\left(1-xt\right)\sqrt{1-yt}}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{1}{\left(1-xt\right)\sqrt{1-yt}}\\
&=\int_{1}^{0}\mathrm{d}u\,\frac{\left(-1\right)\left(1-x\right)}{\left(1-xu\right)^{2}}\cdot\frac{\left(1-xu\right)\sqrt{1-xu}}{\left(1-x\right)\sqrt{\left(1-y\right)-\left(x-y\right)u}};~~~\small{\left[t=\frac{1-u}{1-xu}\right]}\\
&=\frac{1}{\sqrt{1-y}}\int_{0}^{1}\mathrm{d}u\,\frac{1}{\sqrt{\left(1-xu\right)\left[1-\left(\frac{x-y}{1-y}\right)u\right]}}.\\
\end{align}$$
Suppose $a\in\left(0,1\right)$ and $x<1\land x\neq0$. Setting $y=ax$, we then find
$$\begin{align}
F_{1}{\left(1;1,\frac12;2;x,ax\right)}
&=\frac{1}{\sqrt{1-ax}}\int_{0}^{1}\mathrm{d}u\,\frac{1}{\sqrt{\left(1-xu\right)\left[1-\left(\frac{x-ax}{1-ax}\right)u\right]}}\\
&=\frac{1}{\sqrt{1-ax}}\int_{0}^{1}\mathrm{d}u\,\frac{1}{\sqrt{\left(1-xu\right)\left[1-\left(\frac{1-a}{1-ax}\right)xu\right]}}\\
&=\frac{1}{x\sqrt{1-ax}}\int_{0}^{x}\mathrm{d}v\,\frac{1}{\sqrt{\left(1-v\right)\left[1-\left(\frac{1-a}{1-ax}\right)v\right]}};~~~\small{\left[u=\frac{v}{x}\right]}\\
&=\frac{1}{x\sqrt{1-ax}}\int_{1-x}^{1}\mathrm{d}w\,\frac{1}{\sqrt{w\left[1-\left(\frac{1-a}{1-ax}\right)\left(1-w\right)\right]}};~~~\small{\left[v=1-w\right]}\\
&=\frac{1}{x}\int_{1-x}^{1}\mathrm{d}w\,\frac{1}{\sqrt{w\left[\left(1-ax\right)-\left(1-a\right)\left(1-w\right)\right]}}\\
&=\frac{1}{x}\int_{1-x}^{1}\mathrm{d}w\,\frac{1}{\sqrt{w\left[a\left(1-x\right)+\left(1-a\right)w\right]}}\\
&=\frac{1}{x}\int_{0}^{1}\mathrm{d}w\,\frac{1}{\sqrt{w\left[a\left(1-x\right)+\left(1-a\right)w\right]}}\\
&~~~~~-\frac{1}{x}\int_{0}^{1-x}\mathrm{d}w\,\frac{1}{\sqrt{w\left[a\left(1-x\right)+\left(1-a\right)w\right]}}\\
&=\frac{1}{x}\int_{0}^{1}\mathrm{d}t\,\frac{1}{\sqrt{\left(1-t\right)\left[\left(1-ax\right)-\left(1-a\right)t\right]}};~~~\small{\left[w=1-t\right]}\\
&~~~~~-\frac{1}{x}\int_{0}^{1}\mathrm{d}t\,\frac{1}{\sqrt{t\left[a+\left(1-a\right)t\right]}};~~~\small{\left[w=\left(1-x\right)t\right]}\\
&=\frac{1}{x\sqrt{1-ax}}\int_{0}^{1}\mathrm{d}t\,\frac{1}{\sqrt{\left(1-t\right)\left[1-\left(\frac{1-a}{1-ax}\right)t\right]}}\\
&~~~~~-\frac{1}{x}\int_{0}^{1}\mathrm{d}u\,\frac{1}{\sqrt{\left(1-u\right)\left[1-\left(1-a\right)u\right]}};~~~\small{\left[t=1-u\right]}\\
&=\frac{2}{x\sqrt{1-ax}}\,{_2F_1}{\left(\frac12,1;\frac32;\frac{1-a}{1-ax}\right)}-\frac{2}{x}\,{_2F_1}{\left(\frac12,1;\frac32;1-a\right)},\\
\end{align}$$
where in the last line above we've used the Euler integral representation formula to express the remaining integrals in terms of the ${_2F_1}$ function:
$$\int_{0}^{1}\mathrm{d}t\,\frac{t^{\beta-1}\left(1-t\right)^{\gamma-\beta-1}}{\left(1-zt\right)^{\alpha}}=\operatorname{B}{\left(\beta,\gamma-\beta\right)}\,{_2F_1}{\left(\alpha,\beta;\gamma;z\right)};~~~\small{z<1\land0<\beta<\gamma}.$$
$$\tag*{$\blacksquare$}$$
Best Answer
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: $$\boxed{\,\,\int_0^1 t^{-\alpha} \cos (n \pi t)\, dt\sim\frac{\sin\frac{\pi\alpha}{2}}{(\pi n)^{1-\alpha}}\Gamma(1-\alpha)+\frac{(-1)^n}{\Gamma(\alpha)}\sum_{k=1}^\infty (-1)^k\frac{\Gamma(\alpha+2k-1)}{(\pi n)^{2k}}\,\,}$$