Difference between revisions of "Integral representation of polygamma 2"

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(Created page with "<div class="toccolours mw-collapsible mw-collapsed"> <strong>Theorem:</strong> The following formula holds: $$\psi^{(m)}(z)=-\displa...")
 
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<strong>[[Integral representation of polygamma 2|Theorem]]:</strong> The following formula holds:
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<strong>[[Integral representation of polygamma 2|Theorem]]:</strong> The following formula holds for $\mathrm{Re}(z)>0$ and $m>0$:
 
$$\psi^{(m)}(z)=-\displaystyle\int_0^1 \dfrac{t^{z-1}}{1-t} \log^m(t) \mathrm{d}t,$$
 
$$\psi^{(m)}(z)=-\displaystyle\int_0^1 \dfrac{t^{z-1}}{1-t} \log^m(t) \mathrm{d}t,$$
 
where $\psi^{(m)}$ denotes the [[polygamma]] and $\log$ denotes the [[logarithm]].
 
where $\psi^{(m)}$ denotes the [[polygamma]] and $\log$ denotes the [[logarithm]].

Revision as of 19:23, 3 June 2016

Theorem: The following formula holds for $\mathrm{Re}(z)>0$ and $m>0$: $$\psi^{(m)}(z)=-\displaystyle\int_0^1 \dfrac{t^{z-1}}{1-t} \log^m(t) \mathrm{d}t,$$ where $\psi^{(m)}$ denotes the polygamma and $\log$ denotes the logarithm.

Proof: