I was reading some articles on the digamma function, and I was wondering if anyone knows how to express the digamma function $\psi^{(0)}(n)$ in terms of a trigonometric function or a logarithmic function. Does anyone know its Fourier Series representation?
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J. M. ain't a mathematician
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Hatchet
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1The poles of the digamma function, as well as its not being periodic, preclude the Fourier series expression, unless you only want to consider some interval that does not contain a pole. At best, all you have is the reflection formula involving the cotangent. – J. M. ain't a mathematician Jul 11 '15 at 04:13
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$\psi_0(n+1)=H_n-\gamma.$ – Lucian Jul 11 '15 at 13:24
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The following formulas may be inserted into the formula given by @Lucian
$$H_n=\sum_{k=1}^n\frac{1}{k} $$
$$ H_n-log\left(n\right)=1-\sum_{k=1}^{\infty}\left(\sum_{i=nk+1}^{n(k+1)}\frac{1}{i}-\frac{1}{k+1}\right) $$ https://math.stackexchange.com/a/1602945/134791
$$ \gamma= \sum_{n=1}^\infty \left(\frac{2}{n}-\sum_{j=n(n-1)+1}^{n(n+1)} \frac{1}{j}\right)=\sum_{k=1}^\infty \left(\frac{2}{k}-\sum_{j=k(k-1)+1}^{k(k+1)} \frac{1}{j}\right) $$
https://math.stackexchange.com/a/1591256/134791
$$\psi_0\left(n+1\right)=H_n-\gamma$$
$$=\sum_{k=1}^n\frac{1}{k}-\sum_{k=1}^\infty \left(\frac{2}{k}-\sum_{j=k(k-1)+1}^{k(k+1)} \frac{1}{j}\right)$$
In the complex plane outside negative integers (http://people.math.sfu.ca/~cbm/aands/page_259.htm) $$\psi(z+1)=-\gamma+\sum_{n=1}^\infty\frac{z}{n(n+z)}=-\gamma+\sum_{n=1}^\infty\left(\frac{1}{n}-\frac{1}{n+z}\right)$$ $$=-\sum_{n=1}^\infty \left(\frac{2}{n}-\sum_{j=n(n-1)+1}^{n(n+1)} \frac{1}{j}\right)+\sum_{n=1}^\infty\left(\frac{1}{n}-\frac{1}{n+z}\right)$$
Finally, $$\psi(z+1)=\sum_{n=1}^\infty \left(-\frac{1}{n}-\frac{1}{n+z} +\sum_{j=n(n-1)+1}^{n(n+1)}\frac{1}{j} \right), z \neq -1,-2,-3,...$$
Jaume Oliver Lafont
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In umbral calculus,
$\psi(x)=\operatorname{eval}\ln(B+x)$
where $B$ is Bernoulli umbra and $\text{eval}$ is index-lowering operator.
Anixx
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