Monthly Archives: February 2017

Contour method for shifted logarithm branch

Prove  \( a,b,c,d >0 \) $$\int^\infty_0 \frac{\log(a^2+b^2x^2)}{c^2+d^2x^2}\,dx = \frac{\pi}{cd} \log \frac{ad+bc}{d}$$ Consider the function $$f(z) = \frac{\log(a-ibz)}{c^2+d^2z^2}$$ We need the logarithm with the branch cut \( y<-\frac{a}{b} , x =0 \) . Note that this corresponds to $$\log(a+ibz) = \log\sqrt{(a+y)^2+b^2x^2}+i\theta … Continue reading

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Solving Euler sums using Contour integration

Prove that $$\sum_{n=1}^\infty \frac{H_n}{n^2} = 2\zeta(3)$$ $$\textit{proof}$$ Consider the function $$f(z) = \frac{(\psi(-z)+\gamma)^2}{z^2}$$ Note that \( f \) has poles at non-negative integers By integration around a large circle \( |z| = \rho \) Note that $$\oint f(z)\,dz = 2\pi … Continue reading

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Integral of rational function with cosine hyperbolic function using rectangle contour

$$ \int^{\infty}_{-\infty} \frac{1}{(5 \pi^2 + 8 \pi y + 16y^2) }\frac{\cosh\left(y+\frac{\pi}{4} \right)}{\cosh^3(y)}dy=\frac{2}{\pi^3}\left(\pi \cosh\left(\frac{\pi}{4} \right)-4\sinh\left( \frac{\pi}{4}\right) \right)$$   $$\textit{proof}$$ Consider $$f(z) = \frac{\sinh(z)}{z \sinh^3(z-\pi/4)}$$ If we integrate around a contour of height \( \pi \) and stretch it to infinity we … Continue reading

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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

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Solving an integral using Dogbone contour

Prove that $$\int^{1}_{0} \sqrt{x}\sqrt{1-x}\,dx = \frac{\pi}{8}$$ $$\textit{proof}$$ Consider the function $$f(z) = \sqrt{z-z^2} = e^{\frac{1}{2}\log(z-z^2)}$$ Consider the branch cut on the x-axis $$x(1-x)\geq 0\,\, \implies \, 0\leq x \leq 1 $$ Consider \( w= z-z^2 \) then $$\log(w) = \log|w|+i\theta,\,\, … Continue reading

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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

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Bromwich contour integration of the gamma function

$$\frac{1}{2\pi i}\int^{c+i\infty}_{c-i\infty}\Gamma(a+t)\Gamma(b-t) s^{-t}\,dt= \frac{\Gamma(a+b)}{(1+s)^{a+b}}s^a$$ $$\textit{proof}$$ Consider the following function $$f(z) = \Gamma(z+a)\Gamma(b-z) s^{-z}$$ Suppose that \(a,b \in \mathbb{R} \) and \(a < b \). Note that the Gamma function has a pole of order 1 at each non-positive integer where … Continue reading

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Triple integral with sines and cosines

Find the integral $$\begin{align}\int^\infty_0 \int^\infty_0 \int^\infty_0 \frac{\sin(x)\sin(y)\sin(z)}{xyz(x+y+z)}(\sin(x)\cos(y)\cos(z)\\ + \sin(y)\cos(z)\cos(x) + \sin(z)\cos(x)\cos(y))\,dx\,dy\,dz \end{align}$$   $$\textit{solution}$$ This can be rewritten as $$3\small\int^\infty_0 \int^\infty_0 \int^\infty_0 \frac{\sin^2(x)\sin(y)\cos(y)\sin(z)\cos(z)}{xyz(x+y+z)}\,dx\,dy\,dz$$ Now consider $$\small F(a) = 3\int^\infty_0 \int^\infty_0 \int^\infty_0\frac{\sin^2(x)\sin(y)\cos(y)\sin(z)\cos(z) e^{-a(x+y+z)}}{xyz(x+y+z)}\,dx\,dy\,dz$$ Taking the derivative $$\small F'(a) = -3\int^\infty_0 \int^\infty_0 … Continue reading

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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

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