Digamma Function | Brilliant Math & Science Wiki (original) (raw)

Kishlaya Jaiswal,Aareyan Manzoor,Harsh Shrivastava, and

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\[\psi(z) = \dfrac{\mathrm{d}}{\mathrm{d}z} \ln \Gamma(z) = \dfrac{\Gamma^\prime (z)}{\Gamma(z)}\]

\[\psi(s+1)=\psi(s)+\dfrac{1}{s}\]

Consider

\[\Gamma(s+1)=s\Gamma(s).\]

Taking the natural log, we have

\[\ln\big(\Gamma(s+1)\big)=\ln\big(\Gamma(s)\big)+\ln(s).\]

Differentiating with respect to \( s,\) we get

\[\psi(s+1)=\psi(s)+\dfrac{1}{s}.\ _\square\]

\[\psi(s+1)=-\gamma+\sum_{k=1}^\infty \left(\dfrac{1}{k}-\dfrac{1}{k+s}\right)\]

Consider the Weierstrass representation of the gamma function:

\[\Gamma(s)=\dfrac{e^{-\gamma s}}{s} \prod_{k=1}^\infty e^{s/k} \left(1+\dfrac{s}{k}\right)^{-1}. \]

Taking natural logs,

\[\ln\big(\Gamma(s)\big)=-\gamma s-\ln(s)+\sum_{k=1}^\infty \left(\dfrac{s}{k}-\ln \Big(1+\dfrac{s}{k}\Big)\right).\]

Differentiating with respect to \(s,\) we have

\[\psi(s)=-\gamma-\dfrac{1}{s}+\sum_{k=1}^\infty \left(\dfrac{1}{k}-\dfrac{1}{k+s}\right)=-\gamma+\sum_{k=1}^\infty \left(\dfrac{1}{k}-\dfrac{1}{k+s-1}\right).\]

So, replacing \(s\) with \(s+1\) gives

\[\psi(s+1)=-\gamma+\sum_{k=1}^\infty \left(\dfrac{1}{k}-\dfrac{1}{k+s}\right).\ _\square\]

\[\psi(s+1) = -\gamma + \int_0^1 \dfrac{1-x^s}{1-x} dx\]

Consider

\[\begin{align} \psi(s+1) &=-\gamma+\sum_{n=1}^\infty \left(\dfrac{1}{n}-\dfrac{1}{n+s}\right)\\ &=-\gamma+\sum_{n=1}^\infty \int_0^1 \big(x^{n-1}-x^{n+s-1}\big)dx\\ &=-\gamma+ \int_0^1\sum_{n=1}^\infty \big(x^{n-1}-x^{n+s-1}\big)dx. \end{align}\]

By applying geometric progression sum, we have

\[\psi(s+1) = -\gamma + \int_0^1 \dfrac{1-x^s}{1-x} dx.\ _\square\]

From this, we can find specific values of the digamma function easily; for example, putting \(s=0,\) we get \(\psi(1)=-\gamma.\)

Also, by the integral representation of harmonic numbers,

\[\psi(s+1) = -\gamma + H_s.\]

By Euler's reflection formula, we have the following relation:

\[\psi(1-z) - \psi(z) = \pi \cot \pi z.\]

Euler's reflection formula is as follows:

\[\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin \pi z}.\]

Taking natural logarithm and differentiating the above expression, we observe that

\[\ln\big(\Gamma(z)\big)+\ln\big(\Gamma(1-z)\big) = \log \pi - \log \sin \pi z.\]

On differentiating, we get

\[\begin{align} \dfrac{\Gamma^{\prime}(z)}{\Gamma(z)}-\dfrac{\Gamma^{\prime}(1-z)}{\Gamma(1-z)} &= - \dfrac{\pi \cos \pi z}{\sin \pi z}\\\\ \psi(z) - \psi(1-z) &= - \dfrac{\pi \cos \pi z}{\sin \pi z}\\\\ \psi(1-z) - \psi(z) &= \pi \cot \pi z, \end{align}\]

where \(\psi\) denotes the digamma function, which is the logarithmic derivative of the gamma function. \(_\square\)

\[2\psi(2s)=2\ln(2)+\psi(s)+\psi\left(s+\frac12\right)\]

Consider

\[\sqrt{\pi} \ \Gamma(2s)=2^{2s-1} \Gamma(s)\Gamma\left(s+\frac12\right).\]

Taking logs,

\[\ln\big(\sqrt{\pi}\big)+\ln\big(\Gamma(2s)\big)=(2s-1)\ln(2)+\ln\big(\Gamma(s)\big)+\ln\left(\Gamma\Big(s+{\small\frac12}\Big)\right).\]

Differentiating with respect to \(s,\) we have

\[2\psi(2s)=2\ln(2)+\psi(s)+\psi\left(s+\frac12\right).\ _\square\]

Prove

\[\sum_{n=0}^\infty \dfrac{1}{n^2+1} =\dfrac{\pi+1}{2} + \dfrac{\pi}{e^{2\pi}-1}.\]

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We have

\[ S = \sum_{n=0}^{\infty} \frac{1}{n^2+1} = \frac{1}{2i} \sum_{n=0}^{\infty} \left( \frac{1}{n-i} - \frac{1}{n+i} \right). \]

Rewriting this summation gives

\[\begin{align} 2iS &= \sum_{n=1}^{\infty} \left( \frac{1}{n-1-i} - \frac{1}{n-1+i} \right)\\ &=\sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n-1+i} \right)-\sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n-1-i} \right). \end{align}\]

By the series representation of the digamma function, this is just

\[\begin{align} 2iS &=\psi(i)-\psi(-i)\\ &=\psi(i)-\psi(1-i)-\dfrac{1}{i}\\ &=-\pi \cot(i\pi)+i\\ &=\pi i \coth(\pi)+i. \end{align}\]

Simplifying further, we get

\[S= \dfrac{1+\pi\coth(\pi)}{2}\implies S=\dfrac{\pi+1}{2} + \dfrac{\pi}{e^{2\pi}-1}.\ _\square\]

The \(n^\text{th}\) polygamma function is given by

\[\psi_n (s)= \dfrac{d^n}{ds^n} \psi(s)=\psi^{(n)}(s).\]

We can get many properties from this; for example, by differentiating the series representation \(n\) times, we have

\[\psi_n(s)=(-1)^{n+1}n! \sum_{k=1}^\infty \dfrac{1}{(k+s-1)^{n+1}}=(-1)^{n+1}n! \sum_{k=0}^\infty \dfrac{1}{(k+s)^{n+1}}=(-1)^{n+1} n!\zeta(n+1,s),\]

where \(\zeta(n+1,s)\) is the Hurwitz zeta function. Putting \(s=1,\) we can get \(\psi_n(1)=(-1)^{n+1} n!\zeta(n+1).\)

From this, we can also get the Taylor series of the digamma function:

\[\begin{align} \psi(s)&=\sum_{n=0}^\infty \dfrac{\psi^{(n)}(1)(s-1)^n}{n!}\\ &=-\gamma+\sum_{n=1}^\infty \dfrac{\psi_n(1)(s-1)^n}{n!}\\ &=-\gamma-\sum_{n=1}^\infty (-1)^n\zeta(n+1)(s-1)^n\\ \psi(s+1)&=-\gamma-\sum_{n=1}^\infty \zeta(n+1)(-s)^n. \end{align}\]

We can differentiate the integral representation \(n\) times to get

\[\psi_n(s+1)=\int_0^1 \dfrac{\ln^n(x) x^s}{x-1}dx.\]

We can also do this to the functional equation to get

\[\psi_n(s+1)=\psi_n(s)+(-1)^nn! z^{-n-1}.\]

\[\sum _{ k=1 }^{ \infty }{ \dfrac { { \psi }^{ 1 }\left( k \right) }{ k } } =\sum _{ n=1 }^{ \infty }{ \dfrac { A }{ { n }^{ B } } } \]

If the equation above holds true for positive integers \(A\) and \(B\), find \(A+B\).

Inspired by Ishan.

The correct answer is: 5

The digamma functions can be implemented in Mathematica as follows:

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