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Let $\displaystyle \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathcal I=\int_0^\infty\frac{1}{1+x^6} \,\mathrm dx$ \begin{align} I&=\frac{1}{2}\left[ \int_0^\infty \frac{(1-x^2+x^4)+x^2+(1-x^4)}{(1+x^2)(1-x^2+x^4)} \,\mathrm dx \right]\tag{1}\\ &= \frac{1}{2}\left[\int_0^\infty \frac{1}{1+x^2} \,\mathrm dx + \int_0^\infty ... 6 To answer your questions (1) Integrating along a contour means integrating along a parametrized path, i.e. given a parametrization \gamma: [0,1]\to \Bbb C which has the property that \gamma([0,1])=C, the contour, integrating along the path means to compute\int_C f=\int_0^1(f\circ\gamma)(t)\gamma'(t)\,dt$$(2) z is the argument of the function, ... 5 The cosine function does not vanish on the semicircle as R \to \infty; in fact, it does the opposite. You need to either 1) take the real part of e^{i x} in the upper half plane, or 2) use \cos{x} = (e^{i x}+e^{-i x})/2 and use both the upper and lower half planes, respectively. 5 Only your last line is incorrect. What you should write is$$z^2 = \frac{1\pm\sqrt{-3}}{2}$$5$$z^6=-1=e^{(2n+1)\pi i}$$where n is any integer$$\implies z=e^{\dfrac{(2n+1)\pi i}6}=\cos\dfrac{(2n+1)\pi}6+i\sin\dfrac{(2n+1)\pi}6$$where 0\le n\le 5 Top region of the plane, \implies the ordinate has to be >0 \implies\sin\dfrac{(2n+1)\pi}6>0\implies0<\dfrac{(2n+1)\pi}6<\pi\iff0<2n+1<6\implies-.5< n<2.5 \implies ... 4 Solve \cos z = w, let y=e^{iz}. Then \cos z = \frac 12\left(y+y^{-1}\right). So y+\frac{1}{y} = 2w, or y^2-2wy+1=0 or$$y=\frac{2w\pm\sqrt{4w^2-4}}{2}= w \pm \sqrt{w^2-1}$$So iz = \log\left(k\pi\pm \sqrt{k^2\pi^2-1}\right). So:$$z = i\log\left(k\pi\pm \sqrt{k^2\pi^2-1}\right)$$Since (k\pi+ \sqrt{k^2\pi^2-1})(k\pi- \sqrt{k^2\pi^2-1})=1, ... 4 Rough answer.$$\cos z=\frac{e^{iz}+e^{-iz}}{2}$$So we must solve:$$\frac{e^{iz}+e^{-iz}}{2}=\pi k.$$After some algebra, we get$$e^{2iz}-2\pi ke^{iz}+1=0.$$Let w=e^{iz}, the equation is now quadratic in w. So, by quadratic formula,$$w=\pi k\pm\sqrt{\pi^2k^2-1}.$$Substituting e^{iz} back for w and solving for z gives ... 4 C contour:the upper half of the circle$$f\left( z \right) = \frac{{e^{iz} }}{{1 + z^2 }}$$\begin{array}{l} \oint\limits_C {\frac{{e^{iz} }}{{1 + z^2 }}dz} = \int\limits_{ - R}^{ + R} {\frac{{e^{ix} dx}}{{1 + x^2 }}} + \int\limits_\Gamma {\frac{{e^{iz} dz}}{{1 + z^2 }}} = \pi e^{ - 1} \\ \Rightarrow \int\limits_{ - R}^{ + R} {\frac{{\cos \left( ... 3 The support is contained in a interval of the form [-m,m] for m large enough. The function is uniformly continuous on A=[-m-1,m+1] since A is compact. Take \epsilon >0. From uniform continuity on A for there exists \delta>0 such that for |x-y|< \delta it implies that  |f(x) - f(y)| < \epsilon. We prove continuity on ... 3 how can this be done WITHOUT complex analysis? \quad All integrals of the form ~\displaystyle\int_0^\infty\frac{x^{k-1}}{(x^n+a^n)^m}dx~ can be evaluated by substituting x=at and u=\dfrac1{t^n+1} , then recognizing the expression of the beta function in the new integral, and lastly employing Euler's reflection formula for the \Gamma function ... 3 Using \boldsymbol{\pi\csc(\pi z)} Since \pi\csc(\pi z) has residue (-1)^n at z=n for n\in\mathbb{Z}, we will use the contours$$ \gamma_\infty=\lim\limits_{R\to\infty}Re^{2\pi i[0,1]}\qquad\text{and}\qquad\gamma_0=\lim\limits_{R\to0}Re^{2\pi i[0,1]} $$To sum over all n\in\mathbb{Z} except n=0, we use the difference of the contours, which ... 2 Hint Use the ratio test to prove that the radius of convergence is R=1. Use the Dirichlet's test to prove the convergence for |z|=1 and z\ne1, and for z=1 use the Leibniz criterion to prove the convergence. 2 There is no pole at z=-1; it is merely a branch point. The Bromwich contour from which the ILT may be found must be deformed so as to avoid this branch point, like this: You may show that the integrals over C_2, C_4, and C_6 all vanish. The result is, letting z=-1+e^{i \pi} u on C_3 and z=-1+e^{-i \pi} u on C_5,$$\int_{c-i \infty}^{c+i ...

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No, that's only for ordinary series. You can see by the root test that $$\lim_{n\to\infty}\left|\sqrt{n}z^n\right|^{1/n}=\lim_{n\to\infty}n^{1/2n}|z|=|z|$$ converges absolutely when this limit is $<1$, i.e. when $|z|<1$.

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Contour integration may save you, but here I want to present some real-analytic method. Let $I$ denote the integral. With the change of variable $t = \pi x$, we have $$I = \frac{1}{\pi^{3}} \int_{-\infty}^{\infty} \frac{dx}{(x^{2} + 1)^{2} \cosh(\pi x)}.$$ 1. Preliminary Now for an $L^{2}$ function $f$, we denote its Fourier transform by $$... 1 The simplest way of solving this equation is the method based on DeMoivre's Formula that Lab Bhattacharjee outlined. That said, you can make your method work. You found the roots z \pm i by setting the factor z^2 + 1 equal to zero. As Rasolnikov and 5xum noted, you should have obtained$$z^2 = \frac{1 \pm \sqrt{-3}}{2}$$when you set the factor ... 1 I think you've got it, all you need to do is to combine these into a plot. I guess you want something like this? mapping edit *error in the picture, it should be the upper half of the ellips. Because when$$\gamma = \frac{\pi}{2}t + i \quad\text{then}\quad \sin(\Gamma) = -\cosh 1 \sin\frac{\pi}{2}t +i\sinh 1 \cos\frac{\pi}{2}t$$Meaning if t = 0 then ... 1 If you just need to find some domain D in which f and g are to live, the unit disk will do nicely. When |z|<1, the real part of 1-z^2 is positive. In the right half plane the principal branch of square root is holomorphic (it is defined there as re^{i\theta}\mapsto \sqrt{r}e^{i\theta/2} for -\pi/2<\theta<\pi/2), allowing for a direct ... 1$$ \lim_{n\rightarrow \infty}\frac{c_n}{c_{n+1}} = \lim_{n\rightarrow \infty}\frac{(kn+1)(kn+2)\cdots(kn+k)}{(n+1)^k} = \lim_{n\rightarrow \infty}k^k\frac{(n+\frac1k)}{n+1}\cdots\frac{(n+1)}{n+1} = k^k. 

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$\cos z = \Re(e^{iz})$ and $\Re$ is a linear operator, so $\Re\int = \int\Re$.

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