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How do u derive this identity below?

$$\Phi(-1,m+1,z)=\frac{1}{(-2)^{m+1}m!}\left(\psi^{(m)}\left(\frac{z}{2}\right)-\psi^{(m)}\left(\frac{z+1}{2}\right)\right)$$

If Lerch transcendent is defined as this $$\Phi(-1,m+1,z):=\sum_{k=0}^\infty\frac{(-1)^k}{(z+k)^{m+1}}$$

If also the polygamma function is defined as this $$\psi^{(n)}(z):=\frac{\mathrm d^{n+1}}{\mathrm dz^{n+1}}\log\Gamma(z)=-\gamma \delta_{n0}-\frac{(-1)^nn!}{z^{n+1}}+\sum_{k=1}^\infty\left(\frac{1}{k}\delta_{n0}-\frac{(-1)^nn!}{(k+z)^{n+1}}\right)$$

K.defaoite
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Richie
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1 Answers1

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It is well known (and easy to derive starting with $\;\displaystyle\psi(z)=\psi(z+1)-\frac{1}{z}$) that for $m>0$ :

$$\tag{1}\psi^{(m)}(z) = (-1)^{m+1}\, m! \sum_{k=0}^\infty \frac{1}{(z+k)^{m+1}}$$ so that $$\tag{2}\psi^{(m)}\left(\frac z2\right) = (-1)^{m+1}\, m! \sum_{k=0}^\infty \frac{2^{m+1}}{(z+2k)^{m+1}}$$ $$\tag{3}\psi^{(m)}\left(\frac {z+1}2\right) = (-1)^{m+1}\, m! \sum_{k=0}^\infty \frac{2^{m+1}}{(z+1+2k)^{m+1}}$$

subtract $(2)$ and $(3)$, divide by $(-2)^{m+1}m!$ and conclude.

(the case $m=0$ is similar with the additional terms canceling in the subtraction)

Raymond Manzoni
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