Tag Info

New answers tagged

3

$$\int_0^{\infty} \! \!\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \! \int_0^{2\pi} \! \frac{e^{-r^2(cos^2\theta cos^2\phi +cos^2\theta sin^2\phi + sin^2\color{red}\phi) }}{\sqrt{r^2(cos^2\theta cos^2\phi +cos^2\theta sin^2\phi + sin^2\color{red}\phi)}}r^2cos\theta \, \mathrm{d}\phi\mathrm{d}\theta\mathrm{d}r$$ The red bits should be $\theta$ and $$\forall ...


3

Hint: $\cos^2 \theta \cos^2 \phi +\cos^2 \theta \sin^2 \phi = \cos^2 \theta (\cos^2 \phi + \sin^2 \phi) $.


2

Hint: if $g(x) \equiv 0$, this is just the expectation of $-h(Y)$ where $Y$ is a $N(x,1)$ random variable.


0

Ok to bad, the answer is again: No! For $s_n:=n\chi_{[0,\frac{1}{n}]}$ it follows $s_n\to 0$ but $I(s_n)\to 1$.


0

$$ \int_{\partial\Omega}\frac{\partial}{\partial n_y}\frac{1}{|x-y|}\,dS(y)=\int_{\partial\Omega} \nabla_{y}\frac{1}{|x-y|}\cdot\,d\overline{S}(y). $$ If $x \notin\Omega$, then the right side becomes, by the Divergence Theorem, $$ \int_{\Omega}\nabla_{y}^{2}\frac{1}{|x-y|^{2}}dV(y) = 0. $$ If $x \in \Omega$, let $\Omega_{\epsilon}$ be the region ...


0

We can use the following results $$\sum_{n=-\infty}^\infty(-1)^n\frac{1}{bn+a}=\frac{\pi}{b\sin\frac{a\pi}{b}}, \frac{1}{1-x}=\sum_{n=0}^\infty x^n $$ to evaluate the generalization. In fact \begin{eqnarray} \int_0^\infty\dfrac{x^{a-1}}{1+x^b}\ dx&=&\int_0^1\frac{x^{a-1}}{1+x^b}dx+\int_0^1\frac{x^{-a-1}}{1+x^b}dx\\ ...


6

We will prove that $$I=-\frac{\pi^4}{2880}.$$ Indeed, let $$ J=\int_0^{\pi/2}\log^2(\sin x)\log(\cos x)\tan x \,dx $$ It is easy to see that $$\eqalign{J&=\int_0^{\pi/4}\log^2(\sin x)\log(\cos x)\tan x \,dx+ \int_{\pi/4}^{\pi/2}\log^2(\sin x)\log(\cos x)\tan x \,dx\cr &=\int_0^{\pi/4}\log^2(\sin x)\log(\cos x)\tan x \,dx+ \int_{0}^{\pi/4}\log^2(\cos ...


1

Take a look in the abstracts of the 16 workshop in Bedlew (non coummutativ harmonic..) 6-12 of 7


6

I am afraid that what you did is wrong : you are not integrating a polynomial expression. Just for your curiosity,$$\frac{d}{dt}\Big(\frac{e^{t+t^2}}{t^2/2+t^3/3}\Big)=\frac{6 e^{t+t^2} (t+2) \left(4 t^2-3\right)}{t^3 (2 t+3)^2} \neq e^{t+t^2}$$ To compute $$I=\int e^{t}.e^{t^2} dt=\int e^{t^2+t} dt$$ first complete the square for the exponent and perform a ...


1

$|S(r)|$ is the number of $y$ not less than $r$, so $|S(r)|=\sum_{i=1}^n I_{\{y_i \geq r\}}$, here $I_{\{y_i \geq r\}}=1 $ if $y_i \geq r$, $$\int_0^\infty |S(r)|dr=\int_0^\infty\sum_{i=1}^n I_{\{y_i \geq r\}} dr=\sum_{i=1}^n\int_0^\infty I_{\{y_i \geq r\}} dr=\sum_{i=1}^n y_i =1.$$


0

You have $u=x/\sqrt{2}$ and $v= (3x+y)/\sqrt{2}$. The $u$-axis is the set of points where $v=0$, and that is where $3x+y=0$. That's a line whose slope is $-3$. The $v$-axis is where $u=0$, and that's the same as $x=0$, so it's the same as the $y$-axis. Your problem is that the $u$- and $v$-axes are not at right angles to each other; and also a change ...


1

Substitute $x=\frac{1}{u}$. Then $\frac{1}{x}dx = \frac{1}{x}\frac{-1}{u^2}du = -\frac{1}{u}du$. So $$\int_{1}^\infty \frac{\{u\}}{u\lfloor u\rfloor}du = \int_{1}^\infty\left(\frac{1}{\lfloor u\rfloor}-\frac{1}{u}\right) du = \gamma$$


0

The mapped region is a rectangular region that is defined by the equations: For $y = x^2, u = 1$, $y = 2x^2, u = 2$. For $x = 2y^2, v = 2$, and $x = 3y^2, v = 3$. You can sketch the mapped region in $uv$-coordinates.


2

We can try the harmonic analysis path. Since: $$\log(2\cos x)=\sum_{n=1}^{+\infty}\frac{(-1)^{n+1}}{n}\cos(2nx),$$ $$\log(2\sin x)=-\sum_{n=1}^{+\infty}\frac{\cos(2nx)}{n},\tag{1}$$ we have, as an example: $$\int_{0}^{\pi/2}\log^3(2\sin x)dx=-\frac{3\pi}{4}\zeta(3)$$ since $$\int_{0}^{\pi/2}\cos(2n_1 x)\cos(2n_2 x)\cos(2n_3 x)\,dx = ...


7

If we set $$ f(x)=\prod_{n=0}^{+\infty}(1+r^{2n}x^2) $$ we have: $$\int_{0}^{+\infty}\frac{dx}{f(x)}=\pi i\sum_{m=0}^{+\infty}\operatorname{Res}\left(f(z),z=\frac{i}{r^m}\right)=\frac{\pi}{2}\sum_{m=0}^{+\infty}\frac{1}{r^m}\prod_{n\neq m}(1-r^{2n-2m})^{-1}\tag{1}$$ but since $$ \prod_{n=0}^{+\infty}(1-x^n ...


3

\begin{align} \int\frac{\sqrt{\cos 2x}}{\sin x}\ dx&=\int\frac{\sqrt{\cos^2x-\sin^2x}}{\sin x}\ dx\\ &\stackrel{\color{red}{[1]}}=\int\frac{\sqrt{t^4-6t^2+1}}{t^3+t}\ dt\\ &\stackrel{\color{red}{[2]}}=\frac12\int\frac{\sqrt{u^2-6u+1}}{u^2+u}\ du\\ &\stackrel{\color{red}{[3]}}=\int\frac{(y^2-6y+1)^2}{(y-1)(y-3)(y+1)(y^2+2t-7)}\ dy\\ ...


1

If in the last integral you substitute $t=\frac{1}{\sqrt{2}}\cosh z$, you end with: $$ I = \frac{1}{\sqrt{2}}\int\frac{1}{\frac{\cosh^2 z}{2}-1}dz=-\operatorname{arctanh}(\sqrt{2}\tanh z)=-\operatorname{arctanh}\left(\sqrt{2-\frac{1}{t^2}}\right).$$


0

We can use the geometric series $\frac{1}{1-x}=\sum_{k=0}^\infty x^n$ for $|x|<1$ to evaluate: \begin{eqnarray} \int_0^\infty\frac{1}{1+x^n}dx&=&\int_0^1\frac{1+x^{n-2}}{1+x^n}dx\\ &=&\sum_{k=0}^\infty(-1)^k\int_0^1(1+x^{n-2})x^{nk}dx\\ &=&\sum_{k=0}^\infty(-1)^k\left(\frac{1}{nk+1}+\frac{1}{nk+n-1}\right)\\ ...


1

I will assume that $f(a)$ and $f(b)$ are positive, because your formula isn't really adapted for negative values of $f$. Fitting a quadratic hyperbola (is that a legitimate name?) works. Namely, fit a line to the values $f(a)^{-1/2}$ and $f(b)^{-1/2}$, then raise the equation of the line to power $-2$. The equation is $$g(x) = ...


0

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} ...


1

Note that $$ \sqrt{(b-x)(x-a)}=\sqrt{\left(\frac{b-a}{2}\right)^2-\left(x-\frac{b+a}{2}\right)^2}\tag{1} $$ Therefore, we can substitute $$ x=\frac{b+a}{2}-\frac{b-a}{2}\cos(\theta)\quad\text{so that}\quad\sqrt{(b-x)(x-a)}=\frac{b-a}{2}\sin(\theta)\tag{2} $$ Furthermore, by the definition of $y$, $$ ...


5

Use the substitution: $x = \tan\theta$. The integral is then equal to: $$I= \int_{0}^{\pi/4} \ln(1+\tan\theta) \ d\theta (*)$$ Also,we know the property: $$\int_{0}^{b} f(x) \ dx = \int_{0}^{b} f(b-x) \ dx$$ so we have $$I = \int_{0}^{\pi/4} \ln\biggl(1+\tan\Bigl(\frac{\pi}{4}-\theta\Bigr)\biggr) \ d\theta = \int_{0}^{\pi/4} ...


1

Let $v(x) = A + Bx + Cx(x-1)$. Then you want $v(x)$ and $g(x)$ to have the same values at $x=0$, $x=1$, and $x=2$: \begin{align} g(0) &= v(0) = A + B\cdot 0 + C\cdot 0\cdot(0-1) \\ g(1) &= v(1) = A + B\cdot 1 + C\cdot 1\cdot(1-1) \\ g(2) &= v(2) = A + B\cdot 2 + C\cdot 2\cdot(2-1). \end{align} Can you take it from here? EDIT: These ...


0

Try using $x=\tan\theta$ for the substitution. Then use the second result you have mentioned in your question with new limits. You may want to use the formula for $\tan (A+B)$ at some point.


5

The formula of the are of the surface given as a graph of the function $z=f(x,y)$ over the region $(x,y) \in D$ is $$A(S)=\iint_D \sqrt{1+f_x^2+f_y^2}dA$$ In this case $D$ is the disk of radius $1$ with center at $(0,0)$: $D=\{(x,y): x^2+y^2 \leq 1\}$ $$z=f(x,y)=xy$$ $$f_x=y, f_y=x$$ So, we have the following: $$A(S)=\iint_D ...


0

Roughly speaking, the only working method to decide whether an integral converges is to use estimates. If you can bound the (positive) integrand function from above by an integrable function, you are done. If you can estimate it from below by a non-integrable (positive) function, again you are done. However, this approach is often useless in exercises, ...


4

Write the inequality as $$\frac{1}{\mu(E)} \int_E \log f\,d\mu \leqslant \log \frac{1}{\mu(E)}.$$ Jensen's inequality gives you $$\exp \left(\frac{1}{\mu(E)}\int_E \log f\,d\mu\right) \leqslant \frac{1}{\mu(E)}\int_E f\,d\mu.$$ The remaining part should be clear.


3

$$x^8+1=x^8+2x^4+1-2x^4=(x^4+1)^2-2x^4=(x^4+\sqrt{2}x^2+1)(x^4-\sqrt{2}x^2+1)$$ $$x^4 \pm \sqrt{2}x^2+1= x^4+2x^2+1 -(2 \pm \sqrt{2})x^2=(x^2+1 -\sqrt{2 \pm \sqrt{2}}x)(x^2+1 +\sqrt{2 \pm \sqrt{2}}x)$$ Therefore $$x^8+1=(x^2+1 -\sqrt{2 + \sqrt{2}}x)(x^2+1 +\sqrt{2 + \sqrt{2}}x)(x^2+1 -\sqrt{2 - \sqrt{2}}x)(x^2+1 +\sqrt{2 - \sqrt{2}}x)$$ Now the partial ...


0

Hint:It easy cheek that $2x^4+2x^2-1=\frac{1}{2}(2x^2+1+\sqrt{3})(2x^2+1-\sqrt{3})$


2

Hint $$\displaystyle\int\dfrac{dx}{2x^4+2x^2-1}=2\displaystyle\int\dfrac{dx}{4x^4+4x^2-2}=2\displaystyle\int\dfrac{dx}{(2x^2+1)^2-3}$$


2

(Partial answer) For your question that there does not exists a function $g$ increasing on $[0,1]$ such that for all $0\leq a<b\leq 1$ we have $$g(b)-g(a)\geq \frac{\sqrt{b}-\sqrt{a}}{\sqrt{b}+\sqrt{a}}$$ you can argue as follows. First, changing $g$ to $g(x)-g(0)$ if necessary, we can suppose that $g(0)=0$. Then if $b>a=0$, we get $g(b)\geq 1$. Now ...


0

I just want to confirm that the formula you found was correct. Here's how I justify (derive) it. Suppose the big hollow sphere $S$ is centered at the origin. It turns out that the area of $S \cap ([a, b] \times \mathbb R \times \mathbb R)$ is equal to $2 \pi R(b - a)$ for $-R \le a \le b \le R$. (This was quite surprising to me the first time I discovered ...


1

I do not know if this solution is what you are looking for! First notice that $$ (t/\lambda-x)^2 \geq 0 \implies -(t/\lambda-x)^2 \leq 0 \implies e^{-(t/\lambda-x)^2}\leq 1. $$ and $$ |e^{-2\pi i w t}| =1. $$ So we have $$ |F_{\lambda}(x, w)| \leq \int_{\mathbb{R}}e^{-t^2}dt < \infty. $$


0

If the path is a straight line from $0$ to $1+i$ you can do the following: $$ \int_C f(z) dz = \int_\alpha^\beta f(\gamma (t)) \cdot \gamma '(t) dt $$ Now lets describe $C$ as $\gamma (t) = t + it, 0 \le t \le 1$ and $\gamma '(t) = 1+i$ and evaluate: $$ \int_0^1 it^2 \cdot (1+i) dt $$ I hope this can give you a push in the right direction. If your path is ...


2

To sketch the function, note that On $[0,2],$ $f$ is constant and so its integral will be of the form $kx,$ and the graph of $F$ will be a straight line between $(0,0)$ and $(2,8)$ (you have already shown that $F(0) = 0$ and $F(2) = 8$). On $[2,6]$, $f$ is of the form $cx+d$ for negative $c$ and so the integral of $f$ will be a quadratic, and so the graph ...


1

$$m=y_1-y_2,\ t=y_1\implies g(y_1-y_2)\mathbf 1_{0\lt y_2\lt y_1\lt1}=g(m)\mathbf 1_{0\lt m\lt t\lt1}$$


2

Part A: $$F(0) = 0$$ $$F(2) = 8$$ $$F(4) = 12$$ $$F(6) = 8$$ $$F(10) = 0$$ because $F$ below the $x$-line has to be calculated negative. Part B: local maximum is at $x=4$ and local minimums are at $x=0$ and $x=10$ Part C: Your are right. It's increasing for $x(0,4)$ and decreasing for $x(4,10)$ Part D: If we look at the graph we know that: $$f(x) = 4 \ ...


4

HINT Use the Fundamental Theorem of Calculus Part 1 which states $$\frac{d}{dx} \int^{h(x)}_{g(x)} f(t) \, dt = f(h(x))\cdot h'(x) - f(g(x))\cdot g'(x)$$ HINT 2


3

The first equality below uses the Fundamental Theorem of Calculus. The chain rule is used accordingly. \begin{align} \frac{dG}{dx} = \frac{1}{(x^2)^2+4} \cdot \frac{d}{dx}x^2=\frac{2x}{x^4+4} \end{align}


2

Hint: Use the Chain Rule and the Fundamental Theorem of Calculus.


1

find the perimeter of the intersection (circle) $p(r)$ and then let $A(r)=\int_0^r p(x)\,dx$ It's not that simple. Area (in 3 dimensions) is generally tricker to compute than volume (also in 3 dimensions), similarly to how length (in 2 dimensions) is harder to deal with than area (in 2 dimensions). When we have an object that fills up a solid chunk of ...


1

As already remarked in the comments above, it is important to have at least the sketch of a graph for such "change of variable" problems in order to sort out the vertices of the parallelogram, which is our domain $ \ \mathsf{D} \ $ . Since the boundaries of the region are straight lines, and the function we wish to integrate over $ \ \mathsf{D} \ $ is ...


1

Here is how you advance. $$ \int_{0}^{\pi/4}\int_{0}^{2\sin \theta}rdrd\theta + \int_{\pi/4}^{\pi/2} \int_{0}^{2\cos \theta} rdrd\theta. $$ Note: To find the point of intersection of both circles solve the two equations 1) $$ r=2\sin \theta,\quad r=2\cos \theta. $$ 2) $$ \iint_D dxdy = \iint_D r dr d\theta. $$


0

Got it. $$g(y)=\frac{\chi_{[2,\infty)}(y)}{\sqrt{y}\log y}.$$


2

Since you have $x^2 + y^2$, I suggest using polar coordinates. Since $$r^2 = x^2 + y^2$$ $$1 < r^2 < 4$$ So, $$1 < r < 2$$ To find $\theta$ $$\pi/6 < \theta < \pi/3$$ Don't forget to multiply $$rdrd\theta\ \text{in your double integral}$$


2

As it's a little complicated with finding the indefinite integral of $$\int\int(x^2+y^2)^{3/2}dydx$$ I would suggest using polar, as reversing the order of integration isn't really going to make a huge difference. Also the graph is a huge indicator of that.


1

What is going here is that when exponents are divided, they subtract each other. So: $$\large\frac{x - \sqrt{x}}{x^{1/3}}$$ $$= \large x^{1-1/3} - x^{1/2-1/3}$$ $$= \large x^{2/3} - x^{1/6}$$


1

$$\frac{{x - \sqrt x }}{{{x^{\frac{1}{3}}}}} = \frac{{{x^1}}}{{{x^{\frac{1}{3}}}}} - \frac{{{x^{\frac{1}{2}}}}}{{{x^{\frac{1}{3}}}}} = {x^{1 - \frac{1}{3}}} - {x^{\frac{1}{2} - \frac{1}{3}}} = {x^{\frac{2}{3}}} - {x^{\frac{1}{6}}}$$


1

When dividing, the exponents subtract. So $$\frac{x - \sqrt{x}}{x^{1/3}} = \frac{x^1}{x^{1/3}} - \frac{x^{1/2}}{x^{1/3}} = x^{1 - 1/3} - x^{1/2 - 1/3} = x^{2/3} - x^{1/6}$$


0

Let $R_n$ by rotor energy at step $n$ (read: $R_n$ is $\text{newRotorEnergy}$ and $R_{n-1}$ is $\text{oldRotorEnergy}$ ). Then we can define the following recurence relation: $$R_{n} = R_{n-1}a + b$$ where $$b = -\text{rotorMass}/10 + \text{liftTorque}$$ and $$a = 1 - \frac{1}{4000\text{rotorMass}} - \text{inductionTorque}$$ where $\text{inductionTorque}$ ...



Top 50 recent answers are included