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#### References to other Issues or PRs Fixes #26523 #### Brief description of what is fixed or changed Added regression tests Implemented the functions _eval_is_finite and _eval_as_leading_term for the lerchphi class. `lerchphi._eval_is_finite` has been implemented with finite complex numbers in mind. A possible extension to handle infinite arguments is left for the future. The divergences considered are the following: $$\Phi(0, s, a) = \frac{1}{a^s}$$ Which diverges when $a=0$ and $s \in \mathbb{R}^+$. When $z \neq 0$, the function diverges when $a \in - \mathbb{N}$ and $s \in \mathbb{R}^+$. $$\Phi(z, s, -k) = \sum\_{n=0}^\infty \frac{z^n}{(n-k)^s}$$ Finally, for $z=1$, the function has the same poles as the zeta function: $s = 1$. No analytic continuation has been considered, except for $\Phi(1,s,a)=\zeta(s,a)$. `lerchphi._eval_as_leading_term` has been implemented. It handles the following limits: $$\lim\_{z \to 1} \Phi(z, 1, a_0) = -\log(\log(z)) - \gamma_E - \psi(a_0)$$ $$\lim\_{z \to 1} \Phi(z, s, a) = \zeta(s, a)$$ Added new functionality to the `zeta._eval_as_leading_term` method. It adds the following cases: $$\lim\_{(s,a) \to (s_0, a_0)} \zeta(s, a) = \zeta(s_0, a_0), \qquad \text{if} \qquad s_0 \neq 1$$ $$\lim\_{(s,a) \to (1, a_0)} \zeta(s, a) = \frac{1}{s-1}$$ Added tests for all the new changes. #### Other comments Road map: Following the discussion at https://github.com/sympy/sympy/issues/26523, in order to solve the bug with the integral $$\int\_0^1 \frac{1-t^z}{1-t} \mathrm{d}t = \psi(z+1) + \gamma_E$$ We need to correctly compute the following limit: $$\lim\_{t \to 1^-} \left\[-t^{z+1}\Phi(t, 1, z+1) - \log(t - 1)\right\] = \psi(z+1) + \gamma_E - i \pi$$ Sympy currently gives the wrong result for this limit, which causes the bug. To avoid returning the wrong value, we need the following: We first need sympy to identify that $\Phi(1, 1, z+1)$ is infinite to avoid simply substituting $t=1$. This requires the implementation of `lerchphi._eval_is_finite`. This is enough to solve the bug in the `gruntz` algorithm. However, Sympy will also try to solve the limit using the `heuristics` algorithm, which also has this bug. The `heuristics` algorithm seems to just do the substitution $t = 1$. So a check to make sure the result is finite is needed. With these, the integral should no longer return the wrong result but rather either raise an error or return the unevaluated result. In order to compute the correct limit, an implementation of `lerchphi._eval_as_leading_term` is needed, which should be able to correctly identify $$\lim\_{\varepsilon \to 0^+} \Phi(1-\varepsilon, 1, z+1) = -\log(\varepsilon) - \gamma_E - \psi(z+1)$$ Finally, a simplification needs to be added somewhere to simplify ```python (-t*t**z*z*lerchphi(t, 1, z + 1)*gamma(z + 1)/gamma(z + 2) - t*t**z*lerchphi(t, 1, z + 1)*gamma(z + 1)/gamma(z + 2) - log(t - 1)) ``` to ```python -t**(z + 1)*lerchphi(t, 1, z + 1) - log(t - 1) ``` #### Release Notes * functions.special * Implemented `lerchphi._eval_is_finite` * Implemented `lerchphi._eval_as_leading_term` * Added functionality to `zeta._eval_as_leading_term`
References to other Issues or PRs
Fixes #26523
Brief description of what is fixed or changed
Added regression tests Implemented the functions _eval_is_finite and _eval_as_leading_term for the lerchphi class.
lerchphi._eval_is_finitehas been implemented with finite complex numbers in mind. A possible extension to handle infinite arguments is left for the future. The divergences considered are the following: $$\Phi(0, s, a) = \frac{1}{a^s}$$ Which diverges when $a=0$ and $s \in \mathbb{R}^+$. When $z \neq 0$, the function diverges when $a \in - \mathbb{N}$ and $s \in \mathbb{R}^+$. $$\Phi(z, s, -k) = \sum_{n=0}^\infty \frac{z^n}{(n-k)^s}$$ Finally, for $z=1$, the function has the same poles as the zeta function: $s = 1$. No analytic continuation has been considered, except for $\Phi(1,s,a)=\zeta(s,a)$.lerchphi._eval_as_leading_termhas been implemented. It handles the following limits: $$\lim_{z \to 1} \Phi(z, 1, a_0) = -\log(\log(z)) - \gamma_E - \psi(a0)$$ $$\lim\{z \to 1} \Phi(z, s, a) = \zeta(s, a)$$ Added new functionality to thezeta._eval_as_leading_termmethod. It adds the following cases: $$\lim_{(s,a) \to (s_0, a_0)} \zeta(s, a) = \zeta(s_0, a_0), \qquad \text{if} \qquad s0 \neq 1$$ $$\lim\{(s,a) \to (1, a_0)} \zeta(s, a) = \frac{1}{s-1}$$ Added tests for all the new changes.Other comments
Road map:
Following the discussion at https://github.com/sympy/sympy/issues/26523, in order to solve the bug with the integral $$\int_0^1 \frac{1-t^z}{1-t} \mathrm{d}t = \psi(z+1) + \gammaE$$ We need to correctly compute the following limit: $$\lim\{t \to 1^-} \left[-t^{z+1}\Phi(t, 1, z+1) - \log(t - 1)\right] = \psi(z+1) + \gamma_E - i \pi$$ Sympy currently gives the wrong result for this limit, which causes the bug. To avoid returning the wrong value, we need the following: We first need sympy to identify that $\Phi(1, 1, z+1)$ is infinite to avoid simply substituting $t=1$. This requires the implementation of
lerchphi._eval_is_finite. This is enough to solve the bug in thegruntzalgorithm. However, Sympy will also try to solve the limit using theheuristicsalgorithm, which also has this bug. Theheuristicsalgorithm seems to just do the substitution $t = 1$. So a check to make sure the result is finite is needed. With these, the integral should no longer return the wrong result but rather either raise an error or return the unevaluated result.In order to compute the correct limit, an implementation of
lerchphi._eval_as_leading_termis needed, which should be able to correctly identify $$\lim_{\varepsilon \to 0^+} \Phi(1-\varepsilon, 1, z+1) = -\log(\varepsilon) - \gamma_E - \psi(z+1)$$ Finally, a simplification needs to be added somewhere to simplifyto
Release Notes
lerchphi._eval_is_finitelerchphi._eval_as_leading_termzeta._eval_as_leading_term