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HomeZerology of Mittag-Leffler Functions

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Since the Mittag-Leffler function is real on the real, axis, all zeros will either be real or come in complex conjugate pairs. Furthermore, the zeros can be paired in such a way that that a the two zeros in a pair are either complex conjugates or adjacent real zeros. Ther reason this is possible is a combination of contiuity and Rouché’s theorem.

Pick a bounded region symmetric about the real axis which contains two zeros and allow $\alpha$ and $\beta$ to vary within somerange such that, for no value within that range will a zero of the Mittag-Leffler function appear on the boundary of the region. By Rouché’s theorem, the number of zeros contained within tha region will remain 2 for all such values of the parameter.

Pick a one-parameter family of values within this range. Suppose that for one value of the parameter, both zeros are real and that, for another value, they are complex. Since the location of the zeros varies continuously with the parameter and either the real parts or the imaginary parts of both zeros have to be equal, it follows that both zeros must move along the real axis until they coincide, coalesce into a double zero for some value of the parameter, then become a pair of complex zeros.

The zeros of the Mittag-Leffler function $E_{{\alpha_{\beta}}}$ come in pairs — depending on the value of $\alpha$ and $\beta$, either both zeros will be real or they will coincide as a double zero or be complex (in which case, they will be conjugates). Denote the zeros in each pair by $u_{i}$ and $v_{i}$. The numbering of the zeros will be fixed by the requirement that, for values of $\alpha$ and $\beta$ for which all the zeroes are real,

$u_{1}<v_{1}<u_{2}<v_{2}<u_{3}<v_{3}<\cdots$ |

and that, if $u_{i}$ and $v_{i}$ are complex, then the imagiunary part of $u_{i}$ is negative and the imaginary part of $v_{i}$ is positive.

Define $\Delta_{{\alpha\beta}}^{n}$ as $\Delta_{{\alpha\beta}}^{n}=(u_{n}-v_{n})^{2}$. By definition, $\Delta_{{\alpha\beta}}^{n}=0$ if and only if the $n^{{\hbox{th}}}$ pair of zeros coincide.

An integral representation for this quantity $\Delta_{{\alpha\beta}}^{n}$ will now be derived. As a starting point, take the algebraic identity

$(u_{n}-v_{n})^{2}=2(u_{n}^{2}+v_{n}^{2})-(u_{n}+v_{n})^{2}.$ |

The two terms on the right-hand side will be expressed as integrals using residue calculus. Consider the following integrals:

$\displaystyle a_{n}$ | $\displaystyle=$ | $\displaystyle{1\over 2\pi i}\oint_{{C_{n}}}{z{E^{{\prime}}}_{{\alpha\beta}}(z)% \,dz\over E_{{\alpha\beta}}}$ | ||

$\displaystyle b_{n}$ | $\displaystyle=$ | $\displaystyle{1\over 2\pi i}\oint_{{C_{n}}}{z^{2}{E^{{\prime}}}_{{\alpha\beta}% }(z)\,dz\over E_{{\alpha\beta}}}$ |

where $C_{n}$ is a closed countour which surrounds $u_{i}$ and $v_{i}$ but no other zeros of $E_{{\alpha\beta}}$. (By Cauchy’s theorem, the exact choice of $C_{n}$ is irrlevant as long as it encloses the right zeros.) If $u_{i}\neq v_{i}$, then there the integrand has exactly two poles, so the value of the integrand is the sum of the corresponding residues. Since the residue of ${E^{{\prime}}}_{{\alpha\beta}}/E_{{\alpha\beta}}$ is $1$, it follows that $a_{n}=u_{i}+v_{i}$ and $b_{n}=u_{i}^{2}+v_{i}^{2}$. Combining this conclusion with the earlier observation, it follows that $\Delta_{{\alpha\beta}}^{n}=2b_{n}-a_{n}^{2}$; written out fully,

$\Delta_{{\alpha\beta}}^{n}={1\over 2\pi i}\oint_{{C_{n}}}{z^{2}{E^{{\prime}}}_% {{\alpha\beta}}(z)\,dz\over E_{{\alpha\beta}}}+{1\over 2\pi^{2}}\left(\oint_{{% C_{n}}}{z{E^{{\prime}}}_{{\alpha\beta}}(z)\,dz\over E_{{\alpha\beta}}}\right)^% {2}$ |

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