Tag Archives: Integral

Creating Difficult integrals by the residue theorem

Theorem Let $f$ be analytic function in the unit circle $|z|\leq 1$  such that $f\neq 0$ . Then $$\int^{2\pi}_0f(e^{it})\,dt =2\pi \, f(0)$$ $$\textit{proof}$$ Since the function $f$ is analytic in and on the … Continue reading

$$\int^\infty_0\frac{\cos(x)}{\sqrt{x}}\,dx =\int^\infty_0\frac{\sin(x)}{\sqrt{x}}\,dx =\sqrt{\frac{\pi}{2}}$$ $$\textit{solution}$$ Consider the following function $$f(z)=z^{-1/2}\,e^{iz}$$ Where we choose the principle root for $z^{-1/2}=e^{-1/2\log(z)}$. By integrating around the following contour $$\int_{C_r}f(z)\,dz+\int_{r}^R f(x)\,dx+\int_{\gamma}f(z)\,dz+\int^{iR}_{ir}f(x)\,dx = 0$$ Taking the integral around the small quarter circle with $r\to 0$ $$\left| … Continue reading Posted in Contour Integration | | Leave a comment Integration related to gamma function using rectangle contour [Ex 9] Watson’s complex integration$$\int^{\infty}_{-\infty}e^{-x^2}\,\cos(2ax)dx=e^{-a^2}\sqrt{\pi}proof$$Integrate the following function$$f(z) = e^{-z^2}$$Use the following contour Note that the function is entire, hence$$\int^{L}_{-L} e^{-t^2}\,dt+\int^{L+ai}_{L} e^{-t^2}\,dt+\int^{-L}_{-L+ai} e^{-t^2}\,dt+\int^{-L+ai}_{L+ai}e^{-t^2}\,dt=0$$For the forth integral use the substitution $x= t-ai$$$\int^{-L+ai}_{L+ai}e^{-t^2}\,dt=\int^{-L}_{L}e^{-(x+ai)^2}\,dx=-e^{a^2}\int^{L}_{-L}e^{-x^2}\,e^{-2iax}dx$$Take … Continue reading Posted in Contour Integration, Gamma function | | Leave a comment Contour integration for a rational function of cos and cosh$$ \int^{\infty}_{-\infty} \frac{\cos(ax)}{\cosh(x)} \,dx = \pi \, \mathrm{sech} \left( \frac{\pi a}{2}\right)\textit{proof}$$Consider$$f(z) = \frac{e^{iaz}}{\sinh(z)}If we integrate around a contour of height $\pi$ and stretch it to infinity we get By taking T \to \infty … Continue reading Posted in Contour Integration | | Leave a comment General formula for an integral involving powers of logarithms \int\limits_0^1 \dfrac{\log^{m} (1+x)\log^n x}{x}\; dx = (-1)^{n+1}(n!) (m)! \sum_{\{m-1\}} \sum_{k=1}^\infty \frac{(-1)^k}{k^{n+2}} \prod^{l’}_{j=1}\frac{(-1)^{i_j}}{(i_j)!} \left( \frac{H_{k-1}^{(r_j)}}{r_j}\right)^{i_j} \textit{solution} Stirling numbers of the first kind might be useful here, Consider m! \sum_{k=m}^\infty (-1)^{k-m} \left[k\atop m\right] \frac{x^k}{k!} = \log^m(1+x) \int\limits_0^1 \dfrac{\log^m (1+x)\log^n x}{x}\; dx = … Continue reading | | Leave a comment 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 Posted in Contour Integration | | Leave a comment 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 Posted in Uncategorized | Tagged , , , , | 2 Comments 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
$$\sum_{k=1}^\infty\frac{H^{(p)}_k}{k^q} = \zeta(p)\zeta(q) +(-1)^{p}\frac{1}{ (p-1)!}\int^1_0\frac{\mathrm{Li}_q(x)\log(x)^{p-1}}{1-x}\,dx$$ $$\textit{proof}$$ Note that $$\psi_0(a+1)= \int^1_0\frac{1-x^a}{1-x}\,dx$$ By differentiating with respect to $a$ , $p$ times we have $$\psi_p(a+1) = \frac{\partial}{\partial a^p}\int^1_0\frac{1-x^a}{1-x}\,dx$$ $$\psi_p(a+1) = -\int^1_0\frac{x^a\log(x)^{p}}{1-x}\,dx$$ Let $a =k$ $$\psi_{p-1}(k+1) = -\int^1_0\frac{x^k\log(x)^{p-1}}{1-x}\,dx$$ Use the relation to polygamma H^{(p)}_k … Continue reading