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?



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