# Monthly Archives: December 2016

## 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 $$\sum_{n=1}^\infty\frac{s}{n(n+s)} = \int^{1}_{0}\frac{1-x^s}{1-x}\,dx$$

## 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

Prove that $$\int_0^{2\pi} \cos(\sin{x})e^{\cos{x}} dx = 2\pi$$ $$\textit{proof}$$ \begin{align} \int_0^{2\pi} \cos(\sin{x})e^{\cos{x}} dx &=Re\left( \int_0^{2\pi} e^{i\sin(x)}e^{\cos{x}} dx\right) \\ &= Re \left(\int_0^{2\pi} e^{e^{i x}} dx\right)\\ &= Re \left(\sum_{n=0}^\infty \frac{1}{n!}\int_0^{2\pi} e^{inx} dx\right) \\ &= \sum_{n=0}^\infty \frac{1}{n!}\int_0^{2\pi} \cos(nx) dx \\ &= \int_0^{2\pi} dx = … Continue reading Posted in Uncategorized | 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

$$\Gamma \left({z}\right) \Gamma \left({1 – z}\right) = \frac \pi {\sin \left({\pi z}\right)}\,\,\, \, , \forall z \notin \mathbb{Z}$$ $$\textit{proof}$$ We can use the sine product formula $$\frac{\pi}{\sin(\pi z)} = \frac{1}{z}\prod_{n=1}^\infty\left(1-\frac{z^2}{n^2}\right)^{-1}$$ Now we start by noting that $$\Gamma(z)\Gamma(1-z) = -z\Gamma(z)\Gamma(-z)$$ … Continue reading
Prove that $$\sum_{r=1}^n {n\choose r}(-1)^{r+1}\dfrac{1}{r}=\sum_{r=1}^n \dfrac{1}{r}$$ $$proof$$ Start by $$\sum_{r=0}^n {n\choose r}x^r=(1+x)^n$$ Which can be converted to integration $$\sum_{r=1}^n {n\choose r}\frac{(-1)^{r}}{r}=\int^{-1}_0 \frac{(x+1)^n-1}{x} dx$$ By substitution we have $$\sum_{r=1}^n {n\choose r}\frac{(-1)^{r+1}}{r}=\int^{1}_0 \frac{t^n-1}{t-1} dt = H_n$$ Note the last step by expanding … Continue reading