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 Jun20 answered Compute: $\sum_{k=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{k^2n+2nk+n^2k}$ Jun20 comment infinite series involving harmonic numbers and zeta @Chris you mean double summation $\sum_n \sum_m$? I don't think so, because when treating this sum elementary $H_n$ is sum itself. Jun19 awarded Mortarboard Jun19 comment Question regarding Kuhn-Tucker multiplier Pleas use LaTeX syntax and dollar signs for math input. Jun19 revised Question regarding Kuhn-Tucker multiplier I've made some formulas look nicer. Jun19 suggested approved edit on Question regarding Kuhn-Tucker multiplier Jun19 comment Evaluating Integral with Residue Theorem @Synia sure, if the residues sum up to $0$ why not? Consider for example function $g(z) = \frac{1}{z} - \frac{1}{z-1}$. Then ${\rm Res}_{z=0} g(z) = 1$ and ${\rm Res}_{z=1} g(z) = -1$ so when integrating along adequate contour the integral will be $0$. Jun19 answered infinite series involving harmonic numbers and zeta Jun19 comment Integration by means of complex analysis @robjohn right. In the chat room I proposed to have $I'(t) = \int_0^{+\infty} \frac{\cos tx}{1+x^2}\, dx$ than $I(t) = \int_0^{+\infty} \frac{\sin tx}{x(1+x^2)}\,dx$ and ODE is $I''(t) - t^2 I(t) = - \frac{\pi}{2}, \; I(0)=0, I'(0) = \frac{\pi}{2}$ but there is a problem how to solve it in a nice way :) Jun19 comment Integration by means of complex analysis Let's take $$F(a) = \int_0^{+\infty} \frac{\cos (a+x)}{1+x^2} \, dx$$ then we have the desired ODE: $F''(a) + F(a) = 0$ But the problem is to choose initial conditions :| . Jun19 revised Evaluating Integral with Residue Theorem Fixing formating Jun19 answered Evaluating Integral with Residue Theorem Jun19 suggested approved edit on Evaluating Integral with Residue Theorem Jun19 answered Evaluating $\lim_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$ Jun19 revised How integrals are computed? Use LaTeX to write equations Jun19 suggested approved edit on How integrals are computed? Jun18 comment Integration by means of complex analysis @PeterTamaroff yes, you should also note that $F''(x)$ exists also only in P.V. sense. Jun18 comment Integration by means of complex analysis @PeterTamaroff I don't follow - what mistake? :) Jun18 awarded Autobiographer Jun18 comment Integration by means of complex analysis Where $$I_1'(u) = - \int_{- \infty}^{+\infty} \frac{\omega \sin (\omega u)}{1 + \omega^2} \, \mbox{d}\omega = - \int_{-\infty}^{+\infty} \frac{t \sin t}{u^2 + t^2} \, \mbox{d}t = - I_1(u)$$ sorry for two comments, but the characters limit... :)