Extension of $F_{\alpha}(x)=\int_{0}^{1}\frac{\sin(\pi x)}{t^x (1+\alpha t)}\mathrm{d}t$ for $\alpha>-1$

Context

I have this integral representation$$F_{\alpha}(x)=\int_{0}^{1}\frac{\sin\left(\pi x\right)}{t^{x}\left(1+\alpha t\right)}\mathrm{d}t\qquad \text{for }\alpha>-1$$which converges only for $x<1$ in this form.

But I have found $2$ cases in which it is possible to give an alternative integral representation in such a way as to be able to extend the domain of this function:

$$F_0(x)=\int_{0}^{\pi}\cos\left(q\left(x-1\right)\right)\mathrm{d}q\qquad x\in\mathbb{R}$$and$$F_1(x)=\int_{0}^{\frac{\pi}{2}}\frac{\sin\left(\left(2x-1\right)q\right)}{\sin\left(q\right)}\mathrm{d}q+\frac{\pi}{2}\qquad x\in\mathbb{R}$$

($F_0$ can be expressed in terms of the sinc function and $F_1$ can be expressed in terms of the digamma function).

Question

I would like to know if it is possible to have an alternative integral representation of $F_{\alpha}(x)$ such that its domain can be extended to $x \in \mathbb{R}$ for every $\alpha>-1$.

To be clear: find an $f_{\alpha}(q,x)$ and an $I$ such that

$$\int_I f_{\alpha}(q,x)\mathrm{d}q=F_{\alpha}(x)\qquad\forall x\in\mathbb{R}$$

I already searched on Wolfram, the $F_{\alpha}(x)$ can be expressed in terms of a hypergeometric function, but in that case I found only integrals that converges for $x<1$, because on Wolfram there are no integral representation of$$\sin(\pi x)\dfrac{{}_2F_1(1,1-x,2-x,-\alpha)}{1-x}$$which turns out to be a continuous function due to the vanishing of the sine in correspondence with the asymptotes of the hypergeometric function

Note

Wolfram also returns a possible formulation in terms of the Lerch Phi function, but it is not equivalent to the one with the hypergeometric function

$$\sin(\pi x)\dfrac{{}_2F_1(1,1-x,2-x,-\alpha)}{1-x}$$

is a continuous function

$$\sin(\pi x)\Phi(-\alpha,1,1-x)$$

it is discontinuous.

Example with $\alpha=-0.78$

Graphic example

Graphic example with $\alpha=1$

I would like to find an integral representation that describes the corresponding blue function for every $\alpha>-1$ (near $x=1$ there are numerical problems and it seems to go downwards, but it doesn't).

I underline that $\underline{\text{I don't want to find the resolution of the integral}}$ (I already know that the red function is solved in terms of hypergeometric functions), what I would like to find is an integral representation of the function in blue for every $\alpha$.