Monthly Archives: December 2016

Zeta for Even integers proof (Bernoulli numbers)

$$\zeta(2k) \, = \, (-1)^{k-1} B_{2k} \frac{2^{2k-1}}{(2k)!}{\pi}^{2k}$$ $$\textit{proof}$$ We start by the product formula of the sine function $$\frac{\sin(z)}{z} = \prod_{n=1}^\infty \left(1-\frac{z^2}{n^2 \, \pi^2} \right)$$ Take the logarithm to both sides $$\log(\sin(z)) – \log(z) = \sum_{n=1}^\infty \log \left(1-\frac{z^2}{n^2 \, \pi^2} … 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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Digamma third integral representation proof

$$\psi \left(a\right)=\int^{\infty}_0 \, \frac{e^{-t}}{t}-\frac{e^{-\left(a t\right)}}{1-e^{-t}}\, dt$$ $$\textit{proof}$$ 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

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

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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

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Cosine sine integral

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

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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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Gamma Euler Reflection formula proof

$$\Gamma \left({z}\right) \Gamma \left({1 – z}\right) = \frac \pi {\sin \left({\pi z}\right)}\,\,\, \, , \forall z \notin \mathbb{Z}$$ $$\textit{proof}$$ 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

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Relation between binomial sum and harmonic numbers

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

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