# Fubini's theorem with unbounded but summable integrand

I am studying the function \begin{align*} %\int_{-\infty}^\infty x \rightarrow \int_{-\infty}^\infty \mathrm{arcsch} (|x-y|) \frac{2y^2+by-1}{\sqrt{1-y^2}} \mathbf{1}_{(-1,1)} \mathrm{d}y %\mathrm{d}x \end{align*} where $$\mathbf{1}_{(-1,1)}$$ is the indicator function. $$\mathrm{arcsch} (|x-y|)$$ has log-singularity in $$x=y$$, $$1/\sqrt{1-y^2}$$ has $$1/\sqrt{.}$$ singularity in $$y=\pm 1$$, so even for $$x=\pm 1$$ this integral is finite (in fact, I observe that it is even a continuous function, although I haven't managed to prove it should be). The integral of the absolute value of the integrand above will also be finite.

I have thus confidently applied Fubini's theorem to the double integral: \begin{align*} I :&= -\int_{-\infty}^\infty \int_{-\infty}^\infty \mathrm{arcsch} (|x-y|) \frac{2y^2+by-1}{\sqrt{1-y^2}} \mathbf{1}_{(-1,1)} \mathrm{d}y \mathrm{d}x \\ &=\int_{-\infty}^\infty \int_{-\infty}^\infty \mathrm{arcsch} (|z|) \frac{2(x-z)^2+b(x-z)-1}{\sqrt{1-(x-z)^2}} \mathbf{1}_{(-1,1)} \mathrm{d}z \mathrm{d}x \\ &=\int_{-\infty}^\infty \int_{-\infty}^\infty \mathrm{arcsch} (|z|) \frac{2(x-z)^2+b(x-z)-1}{\sqrt{1-(x-z)^2}} \mathbf{1}_{(-1,1)} \mathrm{d}x \mathrm{d}z \\ &=\int_{-\infty}^\infty \mathrm{arcsch} (|z|) \int_{-\infty}^\infty \frac{2(x-z)^2+b(x-z)-1}{\sqrt{1-(x-z)^2}} \mathbf{1}_{(-1,1)} \mathrm{d}x\, \mathrm{d}z \end{align*}

Now, note that $$\int_{-\infty}^\infty \frac{2y^2+by-1}{\sqrt{1-y^2}} \mathbf{1}_{(-1,1)} \mathrm{d}y = 0$$, so I conclude that $$I=0$$. However, this is not what numerical integration of the above definition of $$I$$ tells me (performing quadrature first on $$y$$, then on $$x$$). Which is wrong: applying Fubini's theorem, or numerical integration?