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Andrey Rekalo
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The Mittag-Leffler function $E_{\alpha,1}$, $\alpha>0$, is bounded in the sector $$\frac{\alpha\pi}{2}< \arg z<2\pi-\frac{\alpha\pi}{2}.$$

In particular, $e^z=E_{1,1}(z)$ is bounded in $$\frac{\pi}{2}< \arg z<\frac{3\pi}{2}.$$

The Mittag-Leffler function $E_{\alpha,1}$, $\alpha>0$, is bounded in the sector $$\frac{\alpha\pi}{2}< \arg z<2\pi-\frac{\alpha\pi}{2}.$$

The Mittag-Leffler function $E_{\alpha,1}$, $\alpha>0$, is bounded in the sector $$\frac{\alpha\pi}{2}< \arg z<2\pi-\frac{\alpha\pi}{2}.$$

In particular, $e^z=E_{1,1}(z)$ is bounded in $$\frac{\pi}{2}< \arg z<\frac{3\pi}{2}.$$

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Andrey Rekalo
  • 22.3k
  • 12
  • 89
  • 122

The Mittag-Leffler function $E_{\alpha,1}$, $\alpha>0$, is bounded in the sector $$\frac{\alpha\pi}{2}< \arg z<2\pi-\frac{\alpha\pi}{2}.$$