$$\psi(1-x)-\psi(x)=\pi \cot(\pi x) $$

$$\textit{proof}$$

We know by the reflection formula that

$$\Gamma(x)\Gamma(1-x)=\pi \csc(\pi x) $$

Now differentiate both sides

$$\psi(x)\Gamma(x)\Gamma(1-x)-\psi(1-x)\Gamma(x)\Gamma(1-x)=-\pi^2 \csc(\pi x)\,\cot(\pi x) $$

Which can be simplified

$$\Gamma(x)\Gamma(1-x)\left(\psi(1-x)-\psi(x)\right)=\pi^2 \csc(\pi x)\,\cot(\pi x) $$

Further simplifications using ERF results in

$$ \psi(1-x)-\psi(x)=\pi \cot(\pi x) \,$$

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