# Tag Info

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It's only in cartesian (rectangular) coordinates that $A = \iint_\Omega{\rm d}x\,{\rm d}y$ denotes the area of the region $\Omega$. When we change to a different coordinate-system then the Jacobian describes how the area of a small element of size ${\rm d}x'\,{\rm d}y'$ in the new coordinates is related to the corresponding area of size ${\rm d}x\,{\rm d}y$ ...

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This is similar to the answer from anomaly, but I hope to make it easier to understand by moving the essential point out of a little subscript and into the text. The first equation in the question, $A=\iint dx'\,dy'$ is fine when the integral is over $A$, as you said. But the next bit, $\dots=\iint G\,dx\,dy$ requires that the integral be over a different ...

0

Notice that $$-\log\left(1-\frac1kx\right)=-\log\left(\frac1k\left(k-x\right)\right)=-\log(k-x)-\log\left(\frac1k\right),$$ and that $k$ is constant with respect to $x$, which shows that your antiderivatives differ by a constant, namely $-\log\left(1/x\right)$. For more information look up the fundamental theorem of calculus.

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Since $$\frac1k=-\left(\frac{N_t}{k}\right)'\implies -\int\frac{-\frac1k}{1-\frac{N_t}k}dN_t=-\log\left(1-\frac{N_t}k\right)+C=\log\frac k{k-N_t}+C$$

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As written, the sum does not make sense because the $\delta$ function is zero unless $x=(n-a)^{2/3}$ and infinite there. If you make it $1$ at that point your sum has at most one term for a given $a$ and $x$. It does make sense to write $$F(x,a)=\int_0^N f(x,n).\delta (x - {(n-a)^{2/3}})dn=\begin {cases} f(x,x^{3/2}+a) &0 \le x^{3/2}+a \le N \\0&\... 0 You can use \displaystyle A=\int_{-\pi/3}^{\pi/3}\frac{1}{2}\left((\cos\theta)^2-(1-\cos\theta)^2\right)d\theta=2\int_0^{\pi/3}\frac{1}{2}\left((\cos\theta)^2-(1-\cos\theta)^2\right)d\theta \displaystyle\hspace{.16 in}=\int_0^{\pi/3}(2\cos\theta-1)\,d\theta 1 \textbf{Hint:} Using polar coordinates gives \displaystyle\int_{\pi/3}^{3\pi/4}\int_{\sec\theta}^{2\sec\theta}\frac{1}{r}\cdot rdr\,d\theta=\int_{\pi/3}^{3\pi/4}\sec\theta\, d\theta 1$$I=\int_{1}^{2}\int_{-y}^{\frac{y}{\sqrt{3}}}\frac{1}{(x^2+y^2)}dxdy=\int_{1}^{2} \frac{1}{y}\tan^{-1}\left(\frac{x}{y}\right)\Big{|}_{-y}^{\frac{y}{\sqrt{3}}} dy=\frac{5\pi}{12}\ln 2$$0 By the symmetry (u,x) \mapsto (-u,-x), the integral over the region$$ |x| < 2 \qquad \text{and} \qquad u < x $$is the same as the integral over the region$$ |x| < 2 \qquad \text{and} \qquad u > x $$and adding them together gives$$ 2 \int_{-2}^2 \int_{-\infty}^{x} f(u) \, \mathrm{d}u\, \mathrm{d}x = \int_{-2}^2 \int_{-\infty}^{+\infty} ...

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This is ugly, but may show how, even having no clue what one is looking for, it's still possible to try something (and possibly arrive at the answer, albeit not in the most elegant way). The important thing being: if you have no clever idea, do the non-clever systematic cumbersome thing. It might work. I don't like numbers, so instead of $2$ I used $a$. The ...

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Switching the order of integration, we have $$\int_{-2}^2\int_{-\infty}^xf(u)dudx=\int_{-2}^2\int_u^2 f(u)dxdu+\int_{-\infty}^{-2}\int_{-2}^2f(u)dxdu$$ $$=\int_{-2}^2(2-u)f(u)du+4\int_{-\infty}^{-2}f(u)du$$ $$=2\int_{-2}^2f(u)du+4\int_{-\infty}^{-2}f(u)du$$ $$=4\int_{-2}^0f(u)du+4\int_{-\infty}^{-2}f(u)du$$ $$=4\int_{-\infty}^0f(u)du$$ $$=4\times\frac{1}{2}... 1 It is not referring to the area you show. By definition of "area enclosed" in polar coordinates, it should be the area enclosed for the same range of \theta, in other word I think the author is referring to the area on the right hand side:$$A=\int_{-\pi/2}^{\pi/2}\left[\frac{1}{2}(2(1+\cos\theta))^2-\frac{1}{2}2^2\right] d\theta=\int_{-\pi/2}^{\pi/2}...

2

Given the formula for the area in polar coordinates and the symmetry of the configuration, the area you want to compute is just: $$2\pi + \int_{\pi/2}^{\pi}\left[2\left(1+\cos\theta\right)\right]^2\,d\theta =\color{red}{5\pi-8}.$$

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I can give a heuristic derivation of the results found numerically by @RaymondManzoni and @Arentino. I will try to make things more rigorous the next days First of all we observe the invariance of the integral under the transformation $x \leftrightarrow y$ (i relabel $u\rightarrow x, v\rightarrow y$). Therefore we can write by symmetry $$I_n=\color{... 1 In order to recognize that f(x) simply equals \frac{1}{2\cos x}, you just have to notice that$$ 2\cos(x)\left(\cos(5x)+5\cos(3x)+10\cos(x)\right)=\left(\cos(6x)+6\cos(4x)+15\cos(2x)+10\right)\tag{1}$$as a consequence of$$ 2\cos(x)\cos(nx) = \cos((n+1)x)+\cos((n-1)x).\tag{2}$$Then$$ \int\frac{dx}{\cos x}=\int\frac{\cos x}{1-\sin^2 x}\,dx = C-\frac{1}...

0

No matter what continuous function $a$ is, there is small enough $\Delta t$ to make this a reasonable approximation. However, there are "arbitrarily bad" continuous functions that need arbitrarily small $\Delta t$ to make this reasonable. A class of situations where you can quantify this is when $a$ is continuously differentiable near $a$; in this case this ...

2

Consider the integral $$I= - \int_0^1 \frac{\ln(1-x^2)}{\sqrt{1-x^2}} (\sin^{-1} x)^4 \,dx.$$ Since $$(\sin^{-1} x)^4 = \frac32 \sum_{n=1}^{\infty} \cfrac{2^{2n} H_{n-1}^{(2)}}{n^2 \binom{2n}{n}} \,x^{2 n} \tag{1}$$ and $$-\int_0^1 \frac{\ln(1-x^2)}{\sqrt{1-x^2}} x^{2n}\,dx= \frac{\pi}{2} \binom{2n}{n} \frac{(H_n + 2\ln2)}{2^{2n}}, \tag{2}$$ we have $$\... 0 Aakash Kumar, I think that you were near a simplest form. You can use the binomial formula so as to express u^n + \frac{1}{u^n} in function of T = u + \frac{1}{u} and its powers (this change of variable is a classical technic for finding roots of symmetric polynomials). Using this inside \dfrac{ u^5 +\frac{1}{u^5}+ 5 (u^3+ \frac{1}{u^3}) + 10(u+ \frac{... 0 The Wolfram Alpha solution is from before the question was corrected. (The \cos{2x} in the denominator was formerly \cos{x}). Note that the coefficients at the top are 0, 1, 5, 10, 0 and the coefficients at the bottom are 0+1, 1+5, 5+10, 10+0. This gives the following idea: Substituting \frac12(e^{nix}+e^{-nix}) for \cos{nx}, we get$$\frac{ e^{...

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The first thing to show is that the decomposition is unique. That is, if $f$ is continuous on $\mathbb{R}$ has such a representation, then $d$ and $k$ are unique ($k$ is unique as an element of $L^1[0,\infty)$.) Equivalently, if $f=d+\int_{0}^{\infty}e^{ixt}k(t)dt$ is the $0$ function on $\mathbb{R}$, then $d=0$ and $k=0$ as an element of $L^1[0,\infty)$. ...

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Generalized Differential for polynomials: Given $$f\left(x\right)=\sum\limits_{\epsilon>0}^{1}\left(\sum\limits_{i\in\mathbb{Z}}^{}a_{i}x^{i+\epsilon}\right)$$ then $$f^{(k)}\left(x\right)=\sum\limits_{\epsilon>0}^{1}\left(\sum\limits_{i\in\mathbb{Z}}^{}\frac{\Pi\left(i+\epsilon\right)}{\Pi\left(i+\epsilon-k\right)}a_{i}x^{i+\epsilon}\right)$$ As ...

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Without looking at your work, it is impossible to know where exactly your mistake is but I get the same thing both times: Using $u = x-2/3$ we see \begin{align*} \int^1_0 x\left(x-\tfrac 23 \right)^3 dx &= \int^{1/3}_{-2/3} \left(u+\tfrac 23 \right)u^3 du \\ &= \left[\frac{u^5}{5} + \frac{u^4}{6} \right]^{u=1/3}_{u=-2/3} \\ &= \frac{1}{5\cdot ...

1

Well, I would consider using the fact that there exists such an $F$ to be using Cauchy's theorem, since the usual proof of this for arbitrary holomorphic functions on simply connected domains uses Cauchy's theorem. But in this case you can avoid that by just writing down such an $F$ explicitly: $F(z)=\log(z-a)$, for some branch of $\log$ defined in the ball ...

1

Let $f(x)=\frac{x^2}{\log^{\sqrt{x}}(x)}$. Then, we have \begin{align} \int_2^\infty f(x)\,dx&=\int_2^\infty \frac{x^2}{\log^{\sqrt{x}}(x)}\,dx \end{align} Enforcing the substitution $x\to x^2$ yields \begin{align} \int_2^\infty \frac{x^2}{\log^{\sqrt{x}}(x)}\,dx&=2\int_\sqrt{2}^\infty \frac{x^5}{\log^x(x^2)}\,dx\\\\ &=2\int_\sqrt{2}^\... 0 Let us define: $$I_3(a) := \int\limits_{[0,1]^3} \sqrt{x+\sqrt{y+\sqrt{z+a}}} dx dy dz$$ Then by using elementary integration we have the following result: \begin{eqnarray} I_3(a) &=& \frac{32}{31} \left( \sum\limits_{k=0}^3 |[\begin{array}{r} 3 \\ k \end{array}]|(-1)^{3-k} \frac{(u^+)^{\frac{9}{2}+k} - (u^-)^{\frac{9}{2}+... 0 \int_{-\infty}^xe^{ax}{_1}F_1(-\alpha;-\beta;-\lambda x)~dx =\int_\infty^{-x}e^{-ax}{_1}F_1(-\alpha;-\beta;\lambda x)~d(-x) =\int_{-x}^\infty e^{-ax}{_1}F_1(-\alpha;-\beta;\lambda x)~dx =\int_{-x}^\infty\sum\limits_{n=0}^\infty\dfrac{(-\alpha)_n\lambda^nx^ne^{-ax}}{(-\beta)_nn!}~dx =\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{(-\alpha)_n\lambda^... 1 For the limit, notice that f (x) is bounded below by y=0. Next realize that for x  sufficiently large we have that \ln x>x^{1/3} hence for large x  we have0 <f (x)<g (x)=\frac {x^2}{x^{\sqrt {x}/3}}=x^{2-\sqrt {x}/3}$$Then notice that for x>81 we have that 2-\sqrt {x}/3<-1 (it equals -1 at x=81 and is decreasing from ... 1 AB is just the name of a set. Specifically, it's the set of all points (\cos t, \sin t), where t ranges from 0 to \pi/2. In set builder notation it'd be$$AB = \{ (\cos t, \sin t) \colon t \in [0,\pi/2] \}$$Why it's called AB, and not maybe just A (or some other one-letter name), is a mystery to me, but oh well. I guess it's because of ... 2 The solution to the problem \Delta u = 0 for x \in B(0,a) subject to u=\begin{cases} 2 \theta & 0<\theta<\pi \\ 0 & \pi<\theta<2 \pi \end{cases} for x \in \partial B(0,a) satisfies u(0)=\frac{1}{2\pi} \int_0^{2\pi} u(a,\theta) d \theta = \frac{1}{2 \pi} \int_0^\pi 2 \theta d \theta = \frac{\pi}{2}. This follows from the mean ... 1 Almost, but not quite. I'm seeing a lot of notational issues and other issues around here which could lead to trouble. Instead, say: u = f(x), \qquad dv = g(x) \, dx du = f'(x)\, dx, \qquad v = \displaystyle \int_0^x g(t) \, dt + v(0)\qquad  (this v is explained later in the post) Then we have$$ \int_0^a f(x) g(x) \, dx = \left[f(x) \left(\int_0^...

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