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7

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 ...


1

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 $|z|<1$.


1

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

$$ \lim_{n\rightarrow \infty}\frac{c_n}{c_{n+1}} = \lim_{n\rightarrow \infty}\frac{(kn+1)(kn+2)\cdots(kn+k)}{(n+1)^k} = \lim_{n\rightarrow \infty}k^k\frac{(n+\frac1k)}{n+1}\cdots\frac{(n+1)}{n+1} = k^k. $$


1

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



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