# Tag Archives: proof

## Euler reflection formula proof using contour integration

$$\int_{0}^{\infty}\frac{x^{\alpha}}{x+1}\,dx=- \pi \csc(\pi \alpha)$$ $$\textit{proof}$$ Consider the following function $$f(z) = \frac{z^{\alpha}}{1+z} = \frac{e^{\alpha \log(z)}}{1+z}$$ As we know the function $\log(z)$ is multi-valued defined as $$\log(z) = \ln|z|+i\theta +2k\pi i$$ This maps the complex plain more than once … Continue reading

## Contour integraion of a rational function of logarithm and exponential

$$\int_{0}^\infty \frac{\log(x)\cos(x)}{(x^2+1)^2}\,dx = – \frac{\pi \mathrm{Ei}(1)}{4e}-\frac{\pi}{4e}$$   $$\textit{proof}$$ Consider the following function $$f(z) = \frac{\log(z) }{(z^2+1)^2}e^{iz}$$ Now consider the principle logarithm where $$\log(z) = \log|r|+i \theta \,\,\, , \theta \,\in (-\pi , \pi]$$ Consider the following contour   Then by … Continue reading

## Euler Hypergeometric transformation proof

$${}_2F_1 \left(a,b;c;z\right)=(1-z)^{c-a-b} {}_2F_1 \left(c-a,c-b;c;z\right)$$ $$\textit{proof}$$ In the Pfaff transformations let $z \to \frac{z}{z-1}$ , proved here  $${}_2F_1 \left(a,b;c;\frac{z}{z-1}\right)=(1-z)^{-a} {}_2F_1 \left(a,c-b;c;z\right)$$ and $${}_2F_1 \left(a,b;c;\frac{z}{z-1}\right)=(1-z)^{-b} {}_2F_1 \left(c-a,b;c;z\right)$$ By equating the two transformations $$(1-z)^{-a} {}_2F_1 \left(a,c-b;c;z\right)=(1-z)^{-b} {}_2F_1 \left(c-a,b;c;z\right)$$ Now use the transformation … Continue reading

$${}_2F_1 \left(a,b;c;z\right)=(1-z)^{-a} {}_2F_1 \left(a,c-b;c;\frac{z}{z-1}\right)$$ and $${}_2F_1 \left(a,b;c;z\right)=(1-z)^{-b} {}_2F_1 \left(c-a,b;c;\frac{z}{z-1}\right)$$ $$\textit{proof}$$ Start by the integral representation $${}_2F_1 \left(a,b;c;z\right)=\frac{1}{\beta(b,c-b)}\int_0^1 \frac{t^{b-1}(1-t)^{c-b-1}}{(1-tz)^a}\, dt$$ By the the transformation $t\to 1-t$ $$\int_0^1 \frac{(1-t)^{b-1}t^{c-b-1}}{(1-(1-t)z)^a}\, dt=\int_0^1 \frac{(1-t)^{b-1}t^{c-b-1}}{(1-z+tz)^a}\, dt$$ Which can be written as $$\frac{(1-z)^{-a}}{\beta(b,c-b)}\int_0^1t^{c-b-1}(1-t)^{b-1} \left( 1-t\frac{z}{z-1} … Continue reading Posted in Hypergeoemtric function | | 1 Comment ## Integral representation of the digamma function using Abel–Plana formula$$ \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}$$The … Continue reading Posted in Digamma function | | 3 Comments ## Integral representation of Gauss Hypergeometric function proof$$\beta(c-b,b) \, _2F_1(a,b;c;z)=\int_0^1 \frac{t^{b-1}(1-t)^{c-b-1}}{(1-tz)^a}\, dt \textit{proof}$$Start by the RHS$$\int_0^1 t^{b-1}(1-t)^{c-b-1} \, (1-tz)^{-a}\, dt $$Using the expansion of $(1-tz)^{-a}$ we have$$\int_0^1 t^{b-1}(1-t)^{c-b-1} \sum_{k=0}^\infty\frac{(a)_k}{k!}\, (tz)^k $$Interchanging the integral with the series$$ \sum_{k=0}^\infty\frac{(a)_k}{k!}\, z^k … Continue reading

$$\sum_{k=1}^\infty \frac{H^{(p)}_k}{k^q}+\sum_{k=1}^\infty \frac{H^{(q)}_k}{k^p} =\zeta(p)\zeta(q)+\zeta(p+q)$$ $$\textit{proof}$$ Take the leftmost series and swap the finite and infinite sums $$\sum_{i=1}^\infty \,\sum_{k=i}^\infty\frac{1}{i^p} \frac{1}{k^q}=\sum_{i=1}^\infty \,\sum_{k=1}^\infty\frac{1}{i^p} \frac{1}{k^q}-\sum_{i=1}^\infty\frac{1}{i^p} \,\sum_{k=1}^{i-1} \frac{1}{k^q}$$ The second sum can be written as $$\sum_{i=1}^\infty\frac{1}{i^p} \,\sum_{k=1}^{i-1} \frac{1}{k^q} = \sum_{i=1}^\infty\frac{1}{i^p} \,\left(\sum_{k=1}^{i} \frac{1}{k^q}-\frac{1}{i^p}\right)$$ By separating and … Continue reading