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$\tan (z)=\frac{2}{\pi} \int_0^{\infty} \frac{t^{\frac{2 z}{\pi}}-1}{t^2-1} d t$

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hbghlyj posted 2024-6-10 21:13 |Read mode
functions.wolfram.com/ElementaryFunctions/Tan/07/01/01/0002/
$\displaystyle\tan (z)=\frac{2}{\pi} \int_0^{\infty} \frac{t^{\frac{2 z}{\pi}}-1}{t^2-1} d t \quad 0<\operatorname{Re}(z)<\frac{\pi}{2}$如何證明
  1. Tan[z] == (2/Pi) Integrate[(t^((2 z)/Pi) - 1)/(t^2 - 1), {t, 0, Infinity}] /; 0 < Re[z] < Pi/2
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