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This question is fantastic! I have not found an integral displaying the symmetry but I wanted to show that it points to a symmetry of the Lerch zeta function, if this is known then perhaps that could explain the symmetry, if it is not known it is very interesting I think.

From $(1)$ in @Sasha: $$\begin{align} f(\alpha,\beta) = \int_0^1 \frac{x^\alpha + x^{-\alpha}}{1+2 x \cos(\pi \beta) + x^2} \mathrm{d} x \tag{1}\\ \end{align} $$ The integrand is comparable to the generating function of the Chebyshev polynomials of the second kind and in fact: $$ \begin{align} {\frac {{x}^{\alpha}+{x}^{-\alpha}}{1+2\,x\cos \left( \pi \,\beta \right) +{x}^{2}}}&=\sum _{n=0}^{\infty }U_n \left( - \cos \left( \pi \,\beta \right) \right) \left( {x}^{n+\alpha}+{x}^{n -\alpha} \right) \tag{2}\\ \end{align} $$ and the Chebyshev polynomial of the second kind satsifies: $$U_n \left( -\cos \left( \pi \,\beta \right)\right)={\frac { \left( -1 \right) ^{n}\sin \left( \left( 1+n \right) \pi \, \beta \right) }{\sin \left( \pi \,\beta \right) }} \tag{3} $$ so after using $(2,3)$ in $(1)$ and switching integration and summation order we obtain a Fourier series: $$ \begin{align} f(\alpha,\beta)&=\frac{1}{\sin \left( \pi \,\beta \right)} \sum _{n=0}^{\infty }\left( -1 \right) ^{n}\sin \left( \left( 1+n \right) \pi \, \beta \right) \left( \dfrac{1}{1+n+\alpha }+ \dfrac{1}{1+n-\alpha } \right)\tag{4}\\ &=-\frac{1}{\sin \left( \pi \,\beta \right)} \sum _{n=-\infty\,(n

e0)}^{\infty } \dfrac{\left( -1 \right) ^{n}\sin \left( n \pi \, \beta \right)}{n+\alpha } \\ &=-\frac{1}{\sin \left( \pi \,\beta \right)} \mathfrak{I}\left[ \sum _{n=1}^{\infty }\left( -1 \right) ^{n}e^{i n\pi \beta } \left( \dfrac{1}{n+\alpha }+ \dfrac{1}{n-\alpha } \right)\right]\\ &=-\frac{1}{\sin \left( \pi \,\beta \right)} \mathfrak{I}\left[\Phi(-e^{i \pi \beta },1,\alpha)+\Phi(-e^{i \pi \beta },1,-\alpha)\right]\\ &=-\frac{1}{\sin \left( \pi \,\beta \right)} \mathfrak{I}\left[\Phi_{+}(-e^{i \pi \beta },1,\alpha)\right]\quad:\quad(n=-\infty..+\infty,n

e 0)\\ &=-\frac{1}{\sin \left( \pi \,\beta \right)} \mathfrak{I}\left[L\left(\frac{\beta+1}{2},\alpha,1\right)+L\left(\frac{\beta+1}{2},-\alpha,1\right)\right]\\ &=-\frac{1}{\sin \left( \pi \,\beta \right)} \mathfrak{I}\left[L_{+}\left(\frac{\beta+1}{2},\alpha,1\right)\right]\\ &=-\frac{1}{2\sin \left( \pi \,\beta \right)}\left[L_{+}\left(\frac{\beta+1}{2},\alpha,1\right)-L_{+}\left(\frac{\beta+1}{2},-\alpha,1\right)\right] \end{align}$$ where $L$ is the Lerch zeta function and $\Phi$ the Lerch transcendant. I can't believe this Fourier series is invariant under $\alpha \leftrightarrow \beta$. I have not as of yet found that symmetry for the Lerch zeta function on line. This series together with the previous demonstration that: $$f(\alpha, \beta) = \pi \frac{ \sin(\pi\alpha \beta)}{\sin(\pi \alpha) \sin(\pi \beta)} \tag{5}$$ show immediately that the following special cases hold which are interesting in their own right: $$\sum _{n=-\infty }^{\infty }{\frac { \left( -1 \right) ^{n}\sin \left( \pi \,xn \right) }{x-n}}={\frac {\pi\sin \left( \pi{x}^{ 2} \right) }{\sin \left( \pi x \right) }}\tag{6}$$ $$\sum _{n=-\infty }^{\infty }{\frac { \left( -1 \right) ^{n}\cos \left( \pi \,xn \right) }{1- \left( n+x \right) ^{2}}}=\pi\sin \left( \pi {x}^{2} \right)\tag{7}$$

Now, from $(4)$ and $(5)$ and the variable change $\alpha=x,\, \beta=2y-1$, with $-1<x<1,\,0\le y<1$, we have: $$L_{+}(y,x,1)=-L_{+}(y,-x,1)-2\pi i\dfrac{\sin(2\pi x(y-\frac{1}{2})}{\sin(x)} \tag{8}$$ where we recognise the trigonometric term as the Dirichlet Kernel (with $x\rightarrow2x$ and for $y$ generalised to non-integer). If we then use differentiation with respect to $x$ as a raising operator we obtain the reflection formula in the $x$ variable: $$L_{+}(y,x,k)=\left( -1 \right) ^{k-1}L_{+}(y,-x,k) -2\pi i \dfrac{ \left( -1 \right) ^{k-1}}{(k-1)!}{\frac {\partial ^{k-1}}{\partial {x}^{k-1}}} {\frac {\sin \left( 2\pi x \left( y-\frac{1}{2} \right) \right) } {\sin \left( \pi x \right) }} \tag{9}$$ where the order becomes $k$ from simple differentiation of the function definition, and it also follows from reversing summation order in the function definition that: $$L_{+}(y,x,k)=\sum_{n=-\infty (n

e0)}^{\infty}\dfrac{e^{2\pi in y}}{(n+x)^k}=(-1)^k L_{+}(-y,-x,k)\tag{10}$$ and so $(9)$ can also be viewed as a reflection formula in $y$: $$L_{+}(y,x,k)=L_{+}(-y,x,k) -2\pi i \dfrac{ \left( -1 \right) ^{k-1}}{(k-1)!}{\frac {\partial ^{k-1}}{\partial {x}^{k-1}}} {\frac {\sin \left( 2\pi x \left( y-\frac{1}{2} \right) \right) } {\sin \left( \pi x \right) }} \tag{11}$$ or as an explicit formula for the imaginary part: $$\mathfrak{I}\left(L_{+}(y,x,k)\right)= \pi\dfrac{ \left( -1 \right) ^{k-1}}{(k-1)!}{\frac {\partial ^{k-1}}{\partial {x}^{k-1}}} {\frac {\sin \left( 2\pi x \left( y-\frac{1}{2} \right) \right) } {\sin \left( \pi x \right) }} \tag{11}$$ As a final application, evaluating $(9)$ and $(10)$ at $x=0$ and recognising that $(10)$ is then proportional to the Fourier series of the Bernoulli polynomials $B(m,y)$, we obtain the Taylor series for the Dirichlet kernel in the $x$ variable: $$2\sum _{m=0}^{\infty }\left( -1 \right) ^{m}{\frac { B \left( 2m+1,y \right) }{ \left( 2m+1 \right) !}}{x}^{2m}={ \frac {\sin \left( x \left( y-\frac{1}{2} \right) \right) }{\sin \left( \frac{x}{2} \right) }} \tag{12}$$

Still no closer to displaying the symmetry as an integral but I just wanted to show some interesting consequences of the symmetry and the relation itself. Also, if we wanted to preserve the symmetry and generalise equation $(1)$ then we could do so by applying any symmetric differential operator as a raising operator e.g. ${\partial_{\alpha}}{\partial_{\beta}}$.