New answers tagged

0

In my favorite CAS is has two identic forms integrate in the sense of an unbounded general integral. I can only do screenshots because the MathML is ill-posed. This can be looked up in books with integreal relationsships and representations like Abramovitch, Stegun. First a longer one: Now are either integration done automatically or looked up in formula ...


0

Perhaps it would be easier to start by finding the volume of a small slice of the sphere created by leaving only the portion of the sphere that lies in the range $z=a\dots b,\,b>a$. This section would essentially be a flat pancake with its symmetry axis along z-axis. What is the radius of that pancake? Call it $\rho$. From basic Pythagoras theorem $\rho=\...


3

I think your confusion originates from misunderstanding radians and degrees. $\pi$ is a number. It happens that the circumference of a circle is always $2\pi$ times the radius of the circle. So, in radians, an angle of $\pi$ does correspond to an angle of $180$ degrees. However, $\pi \neq 180.$ When you see $\pi$ appearing in an integral, it's probably ...


1

Let us write $$I=\int_{-\pi}^{\pi} \ln^2(\cos(x/2)) dx=4\int_{0}^{\pi/2} \ln^2 \cos y dy$$ Let $\cos y=t$, then $$I=4\int_{0}^{1} \frac{ln^2 t}{\sqrt{1-t^2}} dt =4\int_{0}^{1} \sum_{k=0}^{\infty} \ln^2 t ~ C_k ~(-1)^k~t^{2k} dt, C_k={-1/2,\choose k}$$ Let $t=e^{-u}$, then $$I=4\int_{0}^{1}\sum_{k=0}^{\infty} (-1)^k C_k u^2 e^{-(2k+1)u} =8\sum_{k=0}^{\infty} \...


1

With $t= \frac x2$ \begin{align} I& = \int\limits_{-\pi}^{\pi}{\log^2{(\cos{\frac{x}{2})}}dx}\\ &= 4\int_0^{\frac\pi2}\ln^2 (\cos t) dt= 2\int_0^{\frac\pi2}(\ln^2 (\cos t)+ \ln^2 (\sin t) )dt \\ &=\int_0^{\frac\pi2}\left( \ln^2 (\sin t\cos t)dt + \ln^2 \frac{\sin t}{\cos t} \right)dt =J+K\tag1 \end{align} where \begin{align} J &= \int_0^{\...


1

The limits of integration are almost correct, but your integrand is wrong (and lacking parentheses). You should have $$\rho\cos\phi\cdot \rho^{-3} \cdot \underbrace{\rho^2\sin\phi\,d\rho\,d\phi\,d\theta}_{dV}.$$ Now, the lower limit for $\rho$ is wrong, as you should have the equation $\rho\cos\phi = 5/2$, so $\rho=\frac52\sec\phi$. (Think about why what you ...


2

Usually, improper Riemann integration will correctly provide the expectation of a continuous random variable with an unbounded support. Where you will usually encounter problems with improper Riemann integrals is when the integral of interest does not converge absolutely. In this case the improper integral may have some value, but this value has limited ...


1

For $k=1$: $$I=\int_{0}^{1} f(x) \cos \pi x ~ dx ~~~(1)$$ Use: $\int_{0}^{a} f(x) dx=\int_{0}^{a} f(a-x) dx$, then $$I=\int_{0}^{1} -f(1-x) \cos \pi x ~dx~~~~~(2)$$ Add (1) and (2), to gt $$2I=\int_{0}^{1}[f(x)-f(1-x)] \cos \pi x ~dx~~~~(3)$$ $x <1-x, x \in (0,1/2)$, as $f(x)$ is decreasing function, we get $$f(x)>f(1-x), x \in (0,1/2), ~f(x)< f)1-x)...


1

$I$ can be negative. For instance, with $k=2$, take the function $g(x)=1$ if $x\le\frac34$, and $g(x)=0$ if $x>\frac34$. Then $$\int_0^1 g(x)\cos(2\pi x)dx=\int_0^\frac34\cos(2\pi x)dx=-\frac{1}{2\pi}$$ Of course, $g$ doesn't satisfy your conditions; but you can easily construct a continuous, differentiable, strictly decreasing $f$ that does satisfy them, ...


0

If you mean 2 dim. Fourier transform as mentioned in comment by reuns and in the format of $F(s,w)=\frac 1{2\pi}\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x,y)\exp(\!-\!isx\!-\!iwy)\:dx\:dy\;$ so $\;f(x,y)=2\pi\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} F(s,w)\exp(isx\!+\!iwy)\:ds\:dw\;$ and you really meant $f(x,y)=e^{(-x^2-y^2\!-xy)}\;$ then the ...


1

I presume the question is to find the expected value of the random variable with the given cumulative distribution function. The distribution function is a staircase function with finite jumps at $2$ points, namely $0,1$, so the random variable $X$ must be discrete and takes only two values $0,1$ (Bernoulli). So$$P(X=0)=F(0)-F(0^-)=1/2\\P(X=1)=F(1)-F(1^-)=1/...


1

As I wrote in a very early comment, it seems that you lost the squares and for $n=100$ the formula should be $$I=\sum ^{99}_{i=0}0.5\times \left( 1-4\left( \dfrac{i}{100}-\dfrac{1}{2}\right)^{\color{red}{2}} +1-4\left( \dfrac{i+1}{100}-\dfrac{1}{2}\right)^{\color{red}{2}} \right) \times \dfrac{1}{100}$$ Making it for general $n$ $$I=\sum ^{n-1}_{i=0}\frac 12\...


4

Note that the projectile’s flying time in the air is $T=\frac{2v\sin\theta}g$. The area under the trajectory is \begin{align} A =\int_0^T y(t)x’(t)dt=\int_0^T (vt \sin \theta - \frac{1}{2} gt^2)v\cos\theta dt=\frac{2v^4}{3g^2}\cos\theta\sin^3\theta \end{align} Then, set $dA/d\theta=0$ to get the optimal angle $\theta =\frac\pi3$. Thus, the maximal area is $$...


2

Find the range. Set $y=0$. One solution is $x=0$, divide by $x$, and solve for the other root. Integrate to find the area. $$A=\int _0^{x_{max}} y dx$$ Take the derivative with respect to $\theta$ and set it to $0$: $$\frac{dA}{d\theta}=0$$ For the given $\theta$ calculate the area.


4

Let $$J(a)=\int_{\frac{\pi}{20}}^{\frac{3\pi}{20}}\tanh^{-1}\frac{2a\cos2x}{1+a^{2}}dx$$ with $J(0)=0$ and \begin{align} J’(a) &= \int_{\frac\pi{20}}^{\frac{3\pi}{20}}\frac{2(1-a^2)\cos2x}{a^4+1-2a^2\cos4x}dx=\frac1{2a}{\left.\tan^{-1}\frac{2a\sin2x}{1-a^2}\right|_{\frac{\pi}{20}}^{\frac{3\pi}{20} } }\\ &=\frac1{2a}\tan^{-1}\frac{ \frac a{1-a^2}(\sin\...


0

Yes, the centroid of an $n$-dimensional pyramid or cone is always $1/(n+1)$ of the distance along the line from the centroid of its $(n-1)$-dimensional base to its apex. If the base is $B\subset\mathbb R^{n-1}$, the pyramid $P\subset\mathbb R^n$ with height $h$ can be parametrized as $$P=\{(1-t)(x_1,x_2,\cdots,x_{n-1},0)+t(0,0,\cdots,0,h)\mid(x_1,x_2,\cdots,...


1

A solution by Cornel Ioan Valean If we set $a_n=H_n/n$ and $b_n=\zeta(2)-H_{2n}^{(2)}$, and then apply Abel's summation, we arrive at $$\sum_{n=1}^{\infty}\frac{H_n}{n}\left(\zeta(2)-H_{2n}^{(2)}\right)$$ $$=\frac{1}{2}\underbrace{\sum_{n=1}^{\infty}\frac{H_n^2-H_n^{(2)}}{(2n+1)^2}}_{\displaystyle A}+\underbrace{\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{(2n+1)^2}}...


1

For the first part, you don't need MVT. The IVT already suffices: let $I=\int_0^1f(x)dx$ and $g(a)=\int_0^af(x)dx$. Since $f$ is positive and continuous, $g$ is strictly increasing and continuous. Also, $g(0)=0<\frac1nI\le g(1)$. Therefore, by IVT, there exists some $a_n\in[0,1]$ such that $g(a_n)=\frac1nI$. The value of $a_n$ cannot be zero because $g(0)&...


1

Factorize $x^{2n}+1= \prod_{k=1}^{2n}(x-x_k)$ with $x_k= e^{i a_k},\>a_k=\frac{(2k-1)\pi}{2n}$. Then $$\frac{x^n}{x^{2n}+1}=-\frac1{2n}\sum_{k=1}^{2n}\frac{x_k^{n+1}}{x-x_k}$$ and integrate \begin{align} &\int \frac{x^n}{x^{2n}+1}dx \\=& -\frac1{2n}\sum_{k=1}^{2n}\int \frac{x_k^{n+1}}{x-x_k} dx = -\frac1{2n}\sum_{k=1}^{2n} x_k^{n+1} \ln(x-x_k)\\ =...


2

Lt us integrate by slicing. The slice at height $z$ is defined by: $$\left(x+\frac{z}{2}\right)^2+\left(y+\frac{z}{2}\right)^2\leq\frac{1}{2}$$ This inequation defines a disk with radius $\sqrt{\frac12}$ Therefore the area of the slice at height $z$ is constant, equal to: $$S(z)=\pi \frac{1}{2}$$ giving a volume: $$V=\int_0^3 S(z)dz=\frac{3\pi}{2}$$


1

Starting from your last inequation on x,y,z, you can factorize to get a cannonical expression : $$\left(x+\frac{z}{2}\right)^2+\left(y+\frac{z}{2}\right)^2\leq\frac{1}{2}, 0\leq z\leq3$$ So you have : $$0\leq z\leq3 \\ -\frac{z}{2}-\frac{1}{\sqrt{2}} \leq x \leq -\frac{z}{2}+\frac{1}{\sqrt{2}} \\-\frac{z}{2}-\sqrt{\frac{1}{2}-\left(x+\frac{z}{2} \right)^2}\...


5

Define the transformation $S$ by $$S(x,y,z)=(x+y+z,x-y,z)=(u,v,w)$$ Then the image of your set $T$ under $S$ is the truncated cylinder $$S(T)=\{u^2+v^2 \leq1\}\cap\{0\leq w\leq3\}\cap \mathbb{R}^3$$ It's easy to see how $$\frac{\partial(x,y,z)}{\partial(u,v,w)}=-\frac{1}{2}$$ Therefore $$\int_Tdxdydz=\int_{S(T)}\Bigg|\frac{\partial(x,y,z)}{\partial(u,v,w)}\...


1

If I am not mistaken, this is just changing the order of integration (which has to be justified by integrability conditions): \begin{align} \int (f*s)(x)w(x) dx & = \int\int f(y-x)s(y)dy\, w(x)dx\\ & = \int\int f(y-x)w(x)dx\, s(y)dy\\ & = \int (f\star w)(y)s(y)dy. \end{align}


2

When rotating about $x$-axis, if $b>a>0$, the integral to solve is $$V=\pi \int_a^b \frac{\arctan x}{x^2+1} \, dx=\frac{1}{2} \pi \left(\arctan^2 b-\arctan^2 a\right)$$


1

Actually it is necessary to add one hypothesis so that the result holds. We need to assume that the application shift $\tau_y : \Omega \rightarrow \Omega$ is $\mathbb{P}$ preserving for any vector $y \in \mathbb{R}^3$. Using this stationarity hypothesis, we see that : $$\mathbb{E}\left[ \bar{\partial}_i v u \right]= \mathbb{E}\left[ \bar{\partial}_i (v \circ ...


1

We have the limits of $x : \frac{\pi}{2}$ to $\frac{5\pi}{2}$ and the limits of $y: 1$ to $\sin x$ We need to convert these such that $y$ lies between $2$ constant limits and $x$ has limits which involve $x$ as a function of $y$. Now, from $x = \frac{\pi}{2}$ to $\frac{5\pi}{2}$, the highest value $y$ can achieve is $\boxed{1}$ and the lowest value is $\sin\...


1

The equations for the lines in terms of $u$ and $v$ are $u=0,v=0$ and $u+v=1$. So the limits are $0 < v < 1-u$ and $0<u<1$.


1

You want $$\mathbb P(0.2 \lt X)=\int\limits_{0.2}^{+\infty} f_X^{\,}(x)\, dx =\int\limits_{0.2}^1 \left(\frac32 x^2 +\frac12\right)\, dx$$ as the lower limit (from the question) is $0.2$ and upper limit (from the support) is $1$


2

Let $\cos \theta =c$, $$S=\sum_{n=0}^{\infty} \frac{P_n(c)}{n+1}= \sum_{n=0}^{\infty} \int_{0}^{1} P_n(c) t^n dt= \int_{0}^{1} \frac{dt}{\sqrt{1-2ct+t^2}}=\int_{0}^{1} \frac{dt}{\sqrt{(t -c)^2+1-c^2}}$$ $$\implies S=ln[2((t-c)+\sqrt{(t-c)^2+s^2})|_{0}^{1}=\ln\frac{1-c+\sqrt{2(1-c)}}{(1-c)}=\ln \frac{1+\sin(\theta/2)}{\sin(\theta/2)}.$$


0

The answer is more no than yes. There are plenty definite integrals giving an answer which is a function of $\pi$. In particular yours, $$2\int_{-1}^1\sqrt{1-x^2}\,dx=\left.\left(x\sqrt{1-x^2}+\arcsin(x)\right)\right|_{-1}^1=\pi.$$ But this does not bring you any closer to the numerical value of $\pi$. There are many ways to obtain a desired number of ...


0

I came up with my own solution: We have that $$f(w) = \int_{\mathbb{R}^2}\log \frac{1}{|w-z|}h(z)\mathrm{d}x\, \mathrm{d}y = \int_{\mathbb{R}^2}\log \frac{|w|}{|w-z|}h(z)\mathrm{d}x\, \mathrm{d}y $$ by the specified condition. Now we may assume that $h$ has support in $B(0,R)$ and therefore for large enough $w$ $$|f(w)w|\leq \int_{B(0,R)}|w|\log \frac{|w|}{|...


2

Since $f$ is supposed to be continuous the values of $f(y)$ for $x <y<x+\Delta x$ are all close to $f(x)$. So we can approximate $\int_x^{x+\Delta x} f(y)dy$ by $\int_x^{x+\Delta x} f(x)dy$ (in which the integrand is a constant as far as integration w.r.t. $y$ is concerned) and the value of this integral $f(x)\Delta x$.


2

The inequality can be written as $$\frac{1}{\mu(E)}\int_E|f|\le\|f\|$$ (Put $|f|^p$ instead, if you wish.) That the average of a function on a subset need not be less than the total function (or even the total average) is obvious: Take $f(x)=\chi_{[0,1/2)}+2\chi_{[1/2,1)}$ and $E=[1/2,1)$, $$2>\frac{2+1}{2}$$


1

If $k$ has units $[ \frac{1}{m}]$ then the kroncker delta for k has units $[m]$. note that for any function f, $\int_{\mathbb{R}} f(x)\delta(x)dx = f(0)$ so if $dx$ has units of length and $f$ is, say, unitless, see that $\delta(x)$ must have units of $\frac{1}{length}$. Your intuition could be that under a change of units $x \to \alpha x$, we get $\delta(\...


0

You 'run' a definite integral over a set. In a way you can think of a ' definite integral' as a procedure which takes in a subset(S) of $\mathrm{R}$ and tells the area of a function underneath the curve and x-axis in that interval. When you do a change of variables, since you are changing the function you are integrating, you must also change the set you ...


1

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0

If $F$ is conservative, then the integral is zero. This is because $C$ is a closed curve (it begins and ends at the same point). As you said, the integral is independent of path. If you go all the way around the ellipse, then you end at the same point you started. You could instead take a different path, which just does nothing (I mean the path just sits at ...


0

Note that conjugation preserves vectors, so the integrand and hence $\mathbf{I}$ are pure imaginary. For pure quaternions $\mathbf{a}$ and $\mathbf{b}$, $$ \mathbf{ab}=-\mathbf{a}\cdot\mathbf{b}+\mathbf{a}\times\mathbf{b}. $$ Thus, writing $``\ln q "=\mathbf{v}$ we have $$ (\ln q)\mathbf{I}-\mathbf{I}(\ln q)=2\mathbf{v}\times \mathbf{I}=\exp(t\mathbf{v})\...


1

As @player3236 notes, in the problem's original notation $xdx=-\frac{\cos\theta}{\sin^3\theta}d\theta$. With $s=\sin\theta$,$$x^2=\int 2xdx=-\int 2s^{-3}ds=s^{-2}+C=\csc^2\theta+C.$$I'll leave you to find $C$.


1

The domain of integration is a tetrahedron with vertices at $(0,0,0),(3,0,0),(0,2,0),(0,0,3)$. The integrand is symmetric with respect to permuting variables. Thus we are allowed to rotate the domain of integration so that slices perpendicular to the $z$-axis are right isosceles triangles and the integral is equal to $$\int_0^2\int_0^{(6-3z)/2}\int_{(6-3z-2y)...


1

The domain of integration is $D$, the set of $(x,y,z)$ with $x,y,z\ge 0$ and $2x+3y+2z\le 6$. Well, to have a simpler, symmetric formula, we can change variables. For instance $X=2x/6$, $Y=3y/6$, $Z=2z/6$. Then the domain $D$ is transformed into the simplex (cone) $E$ of all $(X,Y,Z)$ with $X,Y,Z\ge 0$ and $X+Y+Z\le 1$. Formally, $x=3X$, $y=2Y$, $z=3Z$, so $...


1

No it's not. Consider for example $$f(x)=\sum_{n=1}^{\infty} n \chi_{[\frac{1}{n^3+1}, \frac{1}{n^3})}$$ for $x\in (0,1)$ and $f\equiv 0$ elsewhere. Then $f$ is in $L^1$, and lets say it has norm $\infty>K>0$. Choose $n$ large enough so that $n>K$. Then for $E=[\frac{1}{n^3+1}, \frac{1}{n^3})$ you have that $$\int_E f > \int_E K$$


0

In your attempt, the statement: $$ |F(y)-F(t)| = \left|\int_a^b(f(x,y) - f(x,t))dx\right|=0 $$ doesn't follow from what you assumed, that is, neither the form of $F$ nor that it is not uniformly continuous. Thus you can't get the contradiction $0\ge \epsilon$. By definition of continuity of $F$, let $\epsilon >0$. We need to show there is some $\delta>...


1

$$ \int_{-X}^{X} \; \mathrm{e}^{\mathrm{i} (k + (-k))x} \,\mathrm{d}x = 2X \text{.} $$


2

Area on XY plane is bound by $x^2 + y^2 \leq 1, y \geq 0, x + y \geq 0$ This is a sector of the circle $x^2 + y^2 \leq 1$ bound between positive $X$-axis and line $y = -x$ in the second quadrant. This comes from the fact that $y \geq 0$ so part of the circle in third and fourth quadrant of $XY$ plane is not included. $x + y \geq 0$ is true for quarter of the ...


1

Using spherical coordinates, you would have to split up $K$ into two regions, $$K_1=\left\{(r,\theta,\phi)\mid 0\le r\le2,0\le\theta\le\frac{3\pi}4,0\le\phi\le\frac\pi6\right\}$$ $$K_2=\left\{(r,\theta,\phi)\mid0\le r\le\sqrt{\csc\phi},0\le\theta\le\frac{3\pi}4,\frac\pi6\le\phi\le\frac\pi2\right\}$$ (where $x=r\cos\theta\sin\phi$, $y=r\sin\theta\sin\phi$, ...


2

This is much easier to solve in cylindrical coordinates. $$x=r\cos\theta\\y=r\sin\theta\\z=h$$ Then the limits for $r$ are $0$ and $1$, the limits for $\theta$ are from $-\frac\pi4$ to $\frac{3\pi}4$, and the limits for $h$ are $0$ and $4-r^2$. With these, $$V=\int_{-\frac\pi4}^{\frac{3\pi}4}d\theta\int_0^1dr\cdot r\int_0^{\sqrt{4-r^2}}dh$$ Note see comment ...


-1

$ \int_b^a u(x) dx= - \int_a^b u(x) dx.$


0

$f(x)$ (NOT $F(x)$) is the function being integrated. Here that is $f(x)= \sqrt{16- x^2}$. The interval from $0$ to $b$ is being divided into $n$ intervals of equal length so each has length $\frac{4}{n}$. The values of $x$ at the end of each interval are $\frac{4}{n}$, $2\cdot\frac{4}{n} = \frac{8}{n}$, $3\cdot\frac{4}{n}=\frac{12}{n}$, $i\cdot\frac{4}{...


0

For $n \in \mathbb N$ let $P_n=\{\frac{4j}{n}: j=1,2,...,n\}$ a partition of the interval $[0,4]$ and let $f(x)=\sqrt{16-x^2}$. Then the sum in the image $= \sum_{j=1}^nf(\frac{4j}{n}) \cdot \frac{4}{n}.$ Then we have: $$\sum_{j=1}^nf(\frac{4j}{n}) \cdot \frac{4}{n} \to \int_0^4 f(x) dx$$ as $n \to \infty.$


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