# Tag Archives: Digamma

## Integral representation of the digamma function using Abel–Plana formula

$$\int^\infty_0 \frac{2x}{(x^2+z^2)e^{2\pi x}-1}\,dx =\log(z)-\psi(z)-\frac{1}{2z}$$ $$\textit{proof}$$ Use Abel–Plana formula $$\sum_{n=0}^\infty f(n) = \int^\infty_0f(x)\,dx+\frac{f(0)}{2} +i\int^\infty_0 \frac{f(ix)-f(-ix)}{e^{2\pi x}-1}\,dx$$ Let $$f(x) = \frac{1}{z+x}$$ Note that $$i(f(ix) -f(-ix))= \frac{i}{z+ix}-\frac{i}{z-ix} = \frac{2x}{z^2+x^2}$$ By integration we have $$\int^\infty_0 \frac{2x}{(x^2+z^2)e^{2\pi x}-1}\,dx =\lim_{N\to \infty}\sum_{n=0}^N \frac{1}{z+n}-\int^N_0 \frac{1}{x+z}\,dx-\frac{1}{2z}$$ The … Continue reading

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## Relation between polygamma and Hurwitz zeta function proof

$\forall \,\, n\geq 1$ $$\psi_{n}(z) \, = \, (-1)^{n+1}n!\,\zeta(n+1,z)$$ $$\textit{proof}$$ Use the series representation of the digamma $$\psi_{0}(z) = -\gamma-\frac{1}{z}+ \sum_{n=1}^\infty\frac{z}{n(n+z)}$$ This can be written as the following $$\psi_{0}(z) = -\gamma + \sum_{k=0}^\infty\frac{1}{k+1}-\frac{1}{k+z}$$ By differentiating with respect to … Continue reading

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## The most ugly looking integral

Prove the following $$I= \log \left\{\frac{\Gamma(b+c+1) \Gamma(c+a+1)\Gamma(a+b+1)}{\Gamma(a+1) \Gamma(b+1) \Gamma(c+1) \Gamma(a+b+c+1)} \right\}$$   where   $$I = \int_0^1 \frac{(1-x^a)(1-x^b)(1-x^c)}{(1-x)(-\log x)}dx$$ $$\textit{proof}$$ First note that since there is a log in the denominator that gives as an idea to use differentiation under … Continue reading

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## Digamma fourth integral representation proof

$$\psi(z) = \log(z) -\frac{1}{2z}-2\int^\infty_0 \frac{t}{(t^2+z^2)(e^{2\pi}-1)}dt\,\,\,\,;\text{ Re}z>0$$ We prove that $$2\int^\infty_0 \frac{t}{(t^2+z^2)(e^{2\pi}-1)}dt= \log(z) -\frac{1}{2z}- \psi(z)$$ First note that $$\frac{2}{e^{2\pi t}-1} =\coth(\pi t)-1$$ Also note that $$\coth(\pi t) = \frac{1}{\pi t}+\frac{2t}{\pi}\sum_{k=1}^\infty\frac{1}{k^2+t^2}$$ Hence we conclude that $$\frac{2t}{e^{2\pi t}-1} =\frac{1}{\pi}-t+\frac{2t^2}{\pi}\sum_{k=1}^\infty\frac{1}{k^2+t^2}$$ Substitute the … Continue reading

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$$\psi \left(a\right)=\int^{\infty}_0 \, \frac{e^{-t}}{t}-\frac{e^{-\left(a t\right)}}{1-e^{-t}}\, dt$$ $$\textit{proof}$$ Let $e^{-t}=x$ $$\int^{1}_0 \, -\frac{1}{\log(x)}-\frac{x^{a-1}}{1-x}\, dx$$ By adding and subtracting 1 $$-\int^{1}_0 \, \frac{1}{\log(x)}+\frac{1}{1-x}\, dx+\int^1_0\frac{1-x^{a-1}}{1-x}\, dx$$ Using the second integral representation $$-\int^{1}_0 \, \frac{1}{\log(x)}+\frac{1}{1-x}\, dx+\gamma+\psi(a)$$ We can prove that $$\int^{1}_0 \, … Continue reading Posted in Digamma function | Tagged , , | Leave a comment ## Second integral representation of digamma proof$$\psi(s+1)\,=\, -\gamma \,+\, \int^{1}_{0}\frac{1-x^s}{1-x}\,dx\textit{proof}$$This can be done by noting that$$\psi(s+1) = -\gamma +\sum_{n=1}^\infty\frac{s}{n(n+s)}$$It is left as an exercise to prove that$$\sum_{n=1}^\infty\frac{s}{n(n+s)} = \int^{1}_{0}\frac{1-x^s}{1-x}\,dx$$Posted in Digamma function | Tagged , , | Leave a comment ## First integral representation of digamma proof$$ \psi(a) = \int^{\infty}_0 \frac{e^{-z}-(1+z)^{-a}}{z}\,dz \textit{proof}$$We begin with the double integral$$\int^{\infty}_0 \int^t_1 \,e^{-xz}\,dx\,dz=\int^{\infty}_{0}\frac{e^{-z}-e^{-tz}}{z}\, dz $$Using fubini theorem we also have$$ \int^t_1 \int^{\infty}_0\,e^{-xz}\,dz \,dx = \int^t_1 \frac{1}{x}\,dx = \log t $$Hence we have the following … Continue reading Posted in Digamma function | Tagged , , | Leave a comment ## Digamma difference formula proof$$\psi(1-x)-\psi(x)=\pi \cot(\pi x) \textit{proof}$$We know by the reflection formula that$$\Gamma(x)\Gamma(1-x)=\pi \csc(\pi x) $$Now differentiate both sides$$\psi(x)\Gamma(x)\Gamma(1-x)-\psi(1-x)\Gamma(x)\Gamma(1-x)=-\pi^2 \csc(\pi x)\,\cot(\pi x) $$Which can be simplified$$\Gamma(x)\Gamma(1-x)\left(\psi(1-x)-\psi(x)\right)=\pi^2 \csc(\pi x)\,\cot(\pi x) $$Further simplifications using ERF results in$$ … Continue reading

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