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I met the following interesting identity of the Mittag-Leffler function: $$E_\beta (-\lambda t^\beta)=\frac{\lambda}{\pi}\int_0^\infty e^{-t r}r^{\beta-1} \frac{\sin(\beta \pi)}{(r^\beta\cos(\beta \pi)+\lambda)^2+(r^\beta\sin(\beta \pi))^2}dr.$$

I cannot see how to relate this identity to the usual series expansion $$E_\beta (-\lambda t^\beta)=\sum_{k=0}^\infty \frac{(-\lambda t^\beta)^k}{\Gamma(\beta k+1)},$$ or the countour integral form.

Feel free to attempt it through the contour integral way or through the probability interpretation.

Any help or reference is appreciated.

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I am going to assume $\lambda, \beta, t > 0$. Clearly, your equation cannot hold for $\beta \in \mathbb{N}$, since the right-hand side is zero in that case. In fact, it only holds for $\beta \in (0,1)$ (see equation (23) in this paper). However, we can compute the integral for any $\beta \in \mathbb{R}^+ \setminus \mathbb{N}$.

The substitution $r = \lambda^{1/\beta} u$ shows that it is equal to $f_\beta (\lambda^{1/\beta} t)$, where $f_\alpha \colon \mathbb{R}^+ \to \mathbb{R}$ is defined by $$ f_\alpha (x) = \frac{\sin(\pi \alpha)}{\pi \alpha} \int \limits_0^\infty \frac{\mathrm{e}^{-x t^{1/\alpha}}}{1 + 2 \cos(\pi \alpha) t + t^2} \, \mathrm{d} t = \frac{\sin(\pi \alpha)}{\pi} \int \limits_0^\infty \frac{u^{\alpha - 1} \mathrm{e}^{-x u}}{1 + 2 \cos(\pi \alpha) u^{\alpha} + u^{2\alpha}} \, \mathrm{d} u $$ for $\alpha \in \mathbb{R}^+ \setminus \mathbb{N}$. We have (using the principal branch of the complex power function) \begin{align} f_\alpha (x) &= \frac{1}{2 \pi \mathrm{i}} \int \limits_0^\infty \frac{\mathrm{e}^{\mathrm{i} \pi \alpha} - \mathrm{e}^{-\mathrm{i} \pi \alpha}}{(u^\alpha + \mathrm{e}^{-\mathrm{i} \pi \alpha})(u^\alpha + \mathrm{e}^{\mathrm{i} \pi \alpha})} \, u^{\alpha - 1} \mathrm{e}^{-x u} \, \mathrm{d} u \\ &= \frac{1}{2 \pi \mathrm{i}} \int \limits_0^\infty \left(\frac{u^{\alpha - 1} \mathrm{e}^{-x u} }{u^\alpha + \mathrm{e}^{-\mathrm{i} \pi \alpha}} - \frac{u^{\alpha - 1} \mathrm{e}^{-x u} }{u^\alpha + \mathrm{e}^{\mathrm{i} \pi \alpha}} \right) \mathrm{d} u \\ &= \frac{1}{2 \pi \mathrm{i}} \lim_{R \to \infty} \lim_{\varepsilon \to 0^+} \!\! \int \limits_{1/R}^R \! \left(\! \frac{(u \mathrm{e}^{\mathrm{i}(\pi - \varepsilon)})^{\alpha - 1} \mathrm{e}^{-x u} }{(u \mathrm{e}^{\mathrm{i}(\pi - \varepsilon)})^\alpha + 1} \, \mathrm{e}^{\mathrm{i}(\pi - \varepsilon)} - \frac{(u \mathrm{e}^{-\mathrm{i}(\pi - \varepsilon)})^{\alpha - 1} \mathrm{e}^{-x u}}{(u \mathrm{e}^{-\mathrm{i}(\pi - \varepsilon)})^\alpha + 1} \, \mathrm{e}^{-\mathrm{i}(\pi - \varepsilon)} \! \right) \mathrm{d} u \\ &= \frac{1}{2 \pi \mathrm{i}} \lim_{R \to \infty} \lim_{\varepsilon \to 0^+} \left(~\int \limits_{\Gamma_{R, \varepsilon}^+} \frac{z^{\alpha - 1} \mathrm{e}^{x \mathrm{e}^{\mathrm{i} \varepsilon} z} }{z^\alpha + 1} \, \mathrm{d} z + \int \limits_{\Gamma_{R, \varepsilon}^-} \frac{z^{\alpha - 1} \mathrm{e}^{x \mathrm{e}^{-\mathrm{i} \varepsilon} z}}{z^\alpha + 1} \, \mathrm{d} z\right) \\ &= \frac{1}{2 \pi \mathrm{i}} \lim_{R \to \infty} \lim_{\varepsilon \to 0^+} \int \limits_{\Gamma_{R, \varepsilon}} g_\alpha(x, z) \, \mathrm{d} z \, . \end{align} Here $\Gamma_{R, \varepsilon}^+$ is a straight line from $\frac{1}{R} \mathrm{e}^{\mathrm{i}(\pi - \varepsilon)}$ to $R \mathrm{e}^{\mathrm{i}(\pi - \varepsilon)}$, $\Gamma_{R, \varepsilon}^-$ is a straight line from $R \mathrm{e}^{-\mathrm{i}(\pi - \varepsilon)}$ to $\frac{1}{R} \mathrm{e}^{-\mathrm{i}(\pi - \varepsilon)}$ (depicted in the plots below), $\Gamma_{R,\varepsilon} = \Gamma_{R, \varepsilon}^+ \, \cup \, \Gamma_{R, \varepsilon}^-$ and $g_\alpha(x, z) = \frac{z^{\alpha - 1} \mathrm{e}^{x z}}{z^\alpha + 1}$. In the last step we have dropped the $\mathrm{e}^{\pm \mathrm{i} \varepsilon}$ factors in the exponentials. This does not change the value of the limit.

Using the other three contours shown below we can rewrite the remaining integral as $$ f_\alpha (x) = \frac{1}{2 \pi \mathrm{i}} \lim_{R \to \infty} \lim_{\varepsilon \to 0^+} \left(~\int \limits_{C_{R, \varepsilon}} g_\alpha(x, z) \, \mathrm{d} z - \int \limits_{c_{R, \varepsilon}} g_\alpha(x, z) \, \mathrm{d} z - \int \limits_{\gamma_{R, \varepsilon}} g_\alpha(x, z) \, \mathrm{d} z\right) . $$ The limit of the third integral vanishes, as $$ \left \lvert ~ \int \limits_{\gamma_{R, \varepsilon}} g_\alpha(x, z) \, \mathrm{d} z \, \right \rvert \leq \frac{2 (\pi - \varepsilon) \mathrm{e}^{x/R}}{R^\alpha - 1} \stackrel{\varepsilon \to 0^+}{\longrightarrow} \frac{2 \pi \mathrm{e}^{x/R}}{R^\alpha - 1} \stackrel{R \to \infty}{\longrightarrow} 0 \, . $$ For small values of $\varepsilon$ the contour $c_{R,\varepsilon}$ encloses all poles of $g_\alpha(x,\cdot)$, so the second integral can be evaluated using the residue theorem. The limit of the first integral is just the usual contour integral representation of the Mittag-Leffler function. Therefore, \begin{align} f_\alpha (x) &= \operatorname{E}_\alpha (- x^\alpha) - \!\!\!\!\!\!\!\!\!\!\!\!\! \sum \limits_{z \in S^1 \setminus \{-1\} : \, z^\alpha = - 1} \!\!\!\!\!\!\!\!\!\!\!\!\! \operatorname{Res} (g_\alpha(x,\cdot), z) \\ &= \operatorname{E}_\alpha (- x^\alpha) - \frac{2}{\alpha} \sum \limits_{k=0}^{\left \lfloor \frac{\alpha-1}{2}\right \rfloor} \exp\left[x \cos \left(\frac{(2k+1) \pi}{\alpha}\right)\right] \cos\left[x \sin \left(\frac{(2k+1) \pi}{\alpha}\right)\right] \, . \end{align} For $\alpha \in (0,1)$ the additional sum vanishes ($g_\alpha(x,\cdot)$ does not have any poles enclosed by $c_{R,\varepsilon}$), so $f_\alpha (x) = \operatorname{E}_\alpha (-x^\alpha)$ holds.


Contours The five contours used above for $R > 2$. The radii of the larger and smaller circles are $2$ and $\frac{1}{R}$, respectively.

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  • $\begingroup$ Beautiful proof. Thank you. $\endgroup$ – gouwangzhangdong Feb 17 at 12:53

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