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1

For the case $a >1$, $$ \begin{align} \int_{0}^{2 \pi} \frac{1}{\cos^{2} (x)+a^{2}} \ dx &= \frac{1}{a^{2}} \int_{0}^{2 \pi} \frac{1}{1+ \frac{\cos^{2}(x)}{a^{2}}} \ dx \\ &= \frac{1}{a^{2}} \int_{0}^{2 \pi} \sum_{n=0}^{\infty} (-1)^{n} \frac{\cos^{2n}(x)}{a^{2n}} \ dx \\ &= \frac{1}{a^{2}} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{a^{2n}} ...


0

As I have learned from Gabor Tardos (and Peter Csikvari) this is well-known and can be proved easily using the submodularity of entropy. First we reduce the problem to the finite case by approximating $P$ by a collection of cubes. Then let us select a cube uniformly random and denote its $i$-th coordinate by $X_i$. We have $$\log ...


1

If anyone wants to see a complex analysis solution. Let $\gamma$ be the unit$\require{autoload-all}$ circle. This proof holds for all complex $a$ such that the integral exists. $$ I(a)=\int_0^{2\pi}\!\!\! \frac{\mathrm{d}t}{\cos^2(t)+a^2}$$ $$ \toggle{ \text{Set} \; x = e^{it}\quad\enclose{roundedbox}{\text{ Click for Information }} }{ \begin{align} x ...


2

HINT Make the substitution $u=\tan x$. The identity $1+\tan^2x \equiv \sec^2 x$ tells us that $$1+u^2 = \frac{1}{\cos^2x}\implies \cos^2 x = \frac{1}{1+u^2}$$ Moreover, if $u=\tan x$ then $\mathrm{d}u=\sec^2x~\mathrm{d}x$ which, again by $1+\tan^2x \equiv \sec^2 x$, gives $$\mathrm{d}x=\frac{1}{1+u^2}~\mathrm{d}u$$ If you make these substitutions then ...


0

It is worth to calculate the integral regarding the ellipse as a normal domain, from which: $$\begin{eqnarray*} I &=& \int_{-1}^{3}\int_{-2-3\sqrt{1-\left(\frac{x-1}{2}\right)^2}}^{-2+3\sqrt{1-\left(\frac{x-1}{2}\right)^2}}\frac{1}{1+x^2+y^2}\,dy\,dx.\end{eqnarray*}$$ Now exploiting the fact that ...


1

Defining $I$ as the definite integral, $$I:= \int\limits_{0}^{1}\left[\frac{\zeta{(2)}-2\log^2{2}}{1-x}-\frac{1}{x(1-x)}\left(2\operatorname{Li}_2{\left(\frac{1-\sqrt{1-x}}{2}\right)}-\log^2{\left(\frac{1+\sqrt{1-x}}{2}\right)}\right)\right]\mathrm{d}x,$$ prove: $$I=\frac52\zeta{(3)}-2\zeta{(2)}\log{2}-\frac43\log^3{2}.$$ Substituting ...


2

You can interpret the integral of a function as "area" only if the function is positive. Otherwise, you may interpret the result as the area under the positive part of the function minus the area above the negative part of the function. (N.B. I'm guessing you are talking about integrals, but your post is not very clear)


0

Differentiate both sides, we get $[x]^2 = 2$. Le us denote by $\{x\}$ the fractional part of the number $x$, so that $\{x\}=x-[x]$. It is clear that $0\le \{x\} <1$ and $[x]=x-\{x\}$. Thus, the equation becomes $$[x]^2=(x-\{x\})^2=2$$ that is $$x^2-2x\{x\}+\{x\}^2=2.$$ Setting $p=\{x\}$, and write $$x^2-2px+p^2-2=0.$$ Now, using the general law we have ...


1

If $g(x)=\int_x^1f(t)dt-2x$, then $g(0)=0$ and $g'(0)\ne0$, and consequently $g'$ keeps a constant sign on $(-\epsilon,+\epsilon)$ for some $\epsilon>0$. This proves that $g$ is strictly monotonous in this interval. Thus, the equation $g(x)=0$ has a unique solution in $(-\epsilon,+\epsilon)$ which is clearly $x_*=0$.


2

Basically, when you parametrize the line segment, the form of the vector is $$\vec{r}(t) = <x_0, y_0,z_0> + t<x_1, y_1,z_1>$$ Where $(x_0, y_0, z_0)$ is your initial point, and $(x_1, y_1, z_1)$ is your final point, and $t$ always have to be from $0\le t \le 1$ (do you see why?). Think back to lines and how you parametrized them. This is no ...


2

Building on Surb's answer, any function that can be written as a sum of the form (with $M$ possibly infinite): $$\sum_{m=0}^M a_m \sin{((n+2m)x)}$$ will satisfy $\int_{0}^{\pi}f(x)\cos{(nx)}dx=0$. In addition, for $m\in\mathbb{N}$ and $m\neq n$, $\int_0^\pi \cos{(mx)}\cos{(nx)}=0$. So more generally, $$f(x)=c+\sum_{m=1}^{\infty}\left(a_m ...


1

Another example: Let $k \in \mathbb{N}$ then $$\int_{0}^\pi \sin(kx)\cos(nx)dx =\frac{-n\sin(\pi k)\sin(\pi n)+k(1-\cos(\pi k)\cos(\pi n))}{k^2+n^2}. $$ So for every $k = n + 2m, m \in \mathbb{N}$ this integral is zero.


0

I assume that $[t]=\lfloor t\rfloor$. For $x\ge0$, $$ \int_0^x\lfloor t\rfloor^2\,\mathrm{d}t $$ is convex (its derivative is non-decreasing), the equation $$ \int_0^x\lfloor t\rfloor^2\,\mathrm{d}t=2(x-1) $$ can have at most $2$ solutions. One solution is $x=1$ where both sides are $0$. For $2\le x\le3$, we have $$ \int_0^x\lfloor ...


1

Rewrite your equation as $$\sum_{n=1}^{[x]}n^2+\int_{[x]}^x[x]^2\,dx$$ This is $${[x]([x]+1)(2[x]+1)\over 6}+[x]^2\{x\}$$ Multiplying it all out we get $${1\over 3}[x]^3+\left({1\over 2}+\{x\}\right)[x]^2+{1\over 6}[x]$$ If this has the desired value, we get: $${1\over 3}[x]^3+\left({1\over 2}+\{x\}\right)[x]^2+{1\over 6}[x]=2([x]+\{x\}-1)$$ $$\iff ...


0

Draw a picture of this problem and solve for the CDF of Z. This will be a single integral of the area under the curve XY=Z on [0,1]x[0,1]. Differentiate the result to get the PDF.


3

Not only is $f$ constant, that constant is either $0$ or $1$. $$ \begin{align} \int_a^b\left[f(x)^2-f(x)\right]^2\,\mathrm{d}x &=\int_a^b\left[f(x)^4-2f(x)^3+f(x)^2\right]\,\mathrm{d}x\\ &=0 \end{align} $$ Thus, $(f(x)-1)f(x)=0$ for almost all $x\in[a,b]$. Since $f$ is continuous, we have either $f(x)=0$ for $x\in[a,b]$ or $f(x)=1$ for $x\in[a,b]$.


0

Say, $\displaystyle \int_{0}^{\log n_k}\frac{P(x)}{Q(x)}\,\mathrm{d}x=\frac{n_k}{\pi(n_k)}\quad \text{ for the subsequence } \,\{n_k\}_{k\in \mathbb{N}}\subset \mathbb{N}$. Then the subsequence $\{n_k\}$ has either infinitely many primes or infinitely many composites. Either way we have a subsequence $\{n_{k_{r}}\}$ of $\{n_k\}$, such that ...


1

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3

Here's at least one simplification. Since the integrand $f(x,y,z)=x^2+y^2+z^2$ and the region of integration $D$ are symmetric with respect to interchanging any of pair $x,y,z$, the integral simplifies to any of the following three choices: $$\begin{align} \iiint_{D}(x^2+y^2+z^2)\,\mathrm{d}x\mathrm{d}y\mathrm{d}z ...


7

By the Cauchy Schwarz Inequality, for any integrable function $f(x)$: $\displaystyle\left(\int_a^b f(x) \cdot f(x)^2\,dx\right)^2 \le \left(\int_a^b f(x)^2\,dx\right) \left(\int_a^b (f(x)^2)^2\,dx\right)$ $\displaystyle\left(\int_a^b f(x)^3\,dx\right)^2 \le \left(\int_a^b f(x)^2\,dx\right) \left(\int_a^b f(x)^4\,dx\right)$ But by the given conditions, we ...


3

One function that will produce the given graph is $$ f(x)= \sum_{k=1}^{\lfloor x \rfloor} k = \frac{\lfloor x \rfloor \cdot \lfloor x + 1 \rfloor}{2} $$ provided $x \geq 0$ and $f(x)=0$ if $x < 0$. Here $\lfloor x \rfloor$ denonotes the floor function, i.e., decimal truncation. If you restrict the domain of $f$ to set bounded from above then you can ...


0

Here's a thought: since your spiral has spacings between arms that are decreasing by less and less as the object spirals inwards, we can start your spiral with a formula that makes the spacings increase as a function of theta, like this: r = a +b*theta^1.1 or r = a + b*theta^1.2 Now this or some other ever-increasing formula should model the behavior ...


1

Integrating by parts, $$ \begin{align} \int \frac{\ln (x)}{1-x^{2}} \ dx &= \ln(x) \ \text{arctanh} (x) - \int \frac{\text{arctanh}(x)}{x} \ dx \\ &= \ln (x) \ \text{arctanh} (x) - \frac{1}{2} \int \frac{\ln (1+x)}{x} \ dx + \frac{1}{2} \int \frac{\ln (1-x)}{x} \ dx \\ &= \ln(x) \ \text{arctanh}(x) + \frac{1}{2} \text{Li}_{2}(-x) - ...


2

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0

Your DEQ is inariant to the Lie stretching group $x'=\lambda x$, $y'=\lambda y$ and stabilizers for this group are $\mu=\frac{y}{x}$ and $\nu=y'$. Since $$ x\frac{d\mu}{dx}=\nu-\mu$$ and in stabilizer form your DEQ is $$\nu=1+\mu$$ then $$x\frac{d\mu}{dx}=1$$ $$\mu=lnx+lnC=ln(Cx)$$ Substituting for $\mu$ yields the answer $$y=xln(Cx)$$


1

$$ G = G_1 \cap G_2 \cap G_3 $$ where $G_1$ is a sphere of radius 4 around the origin. $G_2$ and $G_3$ describe half spaces each, meaning $\mathbb{R}^3$ cut into two halves by a plane. A plane because e.g. for $E_2$: $x+y+z = 1$ is a 2 dimensional affine space, it is the plane through $(1,0,0)^T$, $(0,1,0)^T$ and $(0,0,1)^T$. Normal form: $$ (x, y, ...


1

To elaborate on @rogerl's comment, integration by parts holds when $f$ and $g$ are regular and of finite variation. For an example of why it is not sufficient for $f$ and $g$ to just be differentiable, see this reference.


1

Your second integration by parts has an extra minus sign. Overall, what you are trying to do wouldn't yield anything useful, because you write essentially this: $$\int fg' = [fg]-\int f'g = [fg]- \left([fg]-\int fg'\right).$$ So unless you suddenly can produce an antiderivative of $\ln (1+x^2)$, your double integration by parts doesn't help.


0

Note that $g(x)=2 \ln(x) + 2$. To find the roots of $f(x)$, you can integrate by parts to find $f$ explicitly. The derivative of $f$ is $g=2(\ln(x)+1)$, and from this you can find where the function is increasing or decreasing and where the critical points lie. Differentiating again gives $f^{\prime \prime}(x)=\frac{2}{x}$, which lets you find the ...


0

So i am going to answer my own question; I hope it will be helpful for others . Basically being $\psi$ a distribution it is defined as acting on a test function through a scalar product or analogously integrated over its spatial variable $<\psi|\phi>=\int dr \psi\phi$. Moreover one has to consider $\psi$ and $\psi^*$ independend one from the other ...


0

Since $(Nu)(x,y) =\mu(x,y)+f(x,y,I_\theta^r u(x,y),u(x,y)), (Nu)(x,y)=u(x,y)$by fixed point theorem, where $I_\theta^r$ is the left-sided mixed Riemann-Lioville integral of order r, $r=(r_1,r_2)\in(0,\infty)\times(0,\infty), \theta\in(0,0)$ and $u\in L^1(J),L^1(J)$ is the space of Labesgue-integrable functions from $J=[0,a]\times[0,b]$ into $R^n.$ ...


0

We consider the square $Q = [0,x]^2$, because of symmetry, we will look only at the lower left subsquare $Q_1 := [0,\frac x2]^2$. $Q$s center is $(\frac x2, \frac x2)$. For a point $y \in Q_1$, with $y_1 \le y_2$, its distance to the sides is $y_1$, its distance to the center is $\sqrt{(y_1-\frac x2)^2 + (y_2 - \frac x2)^2}$. We have \begin{align*} ...


0

You may find my answer irrelevant, but I hope it will somehow help you. May I assume $$F(x,z)=\int zf_x(z)dz=\int z(\int g(x,z)dx)dz$$ and $$G(x,z)=\int (\int zg(x,z)dz)dx$$. (which are 2-variable functions) Then we have:$$\frac{\partial}{\partial x}\frac{\partial F}{\partial z}=\frac{\partial}{\partial z}\frac{\partial G}{\partial x}$$ or ...


1

we suppose that the center of circle is $(a,b)$ and the radius is $r$. then we have the equation of the circle: $$(x-a)^2+(y-b)^2=r^2=C$$ here $C$ is a constant. Because the circle pass the two points $(1,1)$ and $(-1,-1)$, then we have the equations: $$\left\{\matrix{(1-a)^2+(1-b)^2=C\\(1+a)^2+(1+b)^2=C}\right.$$ solve it we have ...


1

It is false that "the difference between the left-hand sum and right-hand sum always gets smaller as the number of subdivisions gets larger". Here's a counter example. Take the function defined on the interval $[-1,1]$ by $$f(x) = \begin{cases}-x+1 & x\in[0,1] \\ x+1 & x\in[-1,0] \end{cases}\;.$$ Consider breaking it into two equal subdivisions, ...


0

There is something questionable in the wording (about the bounds of the integral). So, two interpretations are presented below :


2

You know the Gauss' divergence theorem says that $\int_V \nabla \cdot \vec{A} \rm dV = \int_S \vec{A} \cdot d \vec{S}$ Now taking $\vec{A}= f \hat{c}$, where $\hat{c}$ is a constant vector, you can prove: $\int_V (\nabla f)\rm dV = \int_S f d \vec{S}$ Now telling f tends zero on the surface will not be sufficient to say that left hand side is zero. You ...


0

Provided the function is smooth enough, I'd say yes: the volume integral of its gradient is a vector that has three components; to calculate, say, the first component, you can integrate the derivative with respect to $x$ over $x$ first and, taking into account the boundary conditions, obtain zero. EDIT (7/9/2014): I guess some extra conditions may be ...


1

This is difficult to do without using measure theory! The result is known as Arzela's Bounded Convergence Theorem. An elementary proof proof can be found here.


3

Hint: define $Q(x) = x^2/2$. $$x\phi(x) = \frac 1{\sqrt{2\pi}}Q'(x)\exp(-Q(x)) $$


0

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2

http://en.wikipedia.org/wiki/Hoeffding%27s_inequality gives a way to bound the difference between $\frac{1}{\sqrt{3}}$ and the given integral. If we take $X_1,\ldots,X_n$ as indipendent random variables with a uniform distribution over $[0,1]$, then $X_1^2,\ldots,X_n^2$ are bounded indipendent random variables with density function $$ ...


3

Hint: $$xy'-y=x$$ $$\frac{xy'-y}{x^2}=\frac{1}{x}$$ $$\left(\frac{y}{x}\right)'=\frac{1}{x}$$


1

First, split the integral and change the order of integration as $$ I = \int_0^{2}\int_0^{2} x^2\sqrt{1+4x^2+4y^2} \hspace{2mm} dxdy - 2\int_0^{2}y\int_0^{2}x \sqrt{1+4x^2+4y^2} \hspace{2mm} dxdy = I_1+I_2 . $$ For the inner integral of $I_1$ you can use integration by parts with $u=x$ and for the inner integral of $I_2$ use the substitution $u=x^2$.


0

Starting with $$\int_0^4\int_0^4 xy\sqrt{1+x^2+y^2}\;\text{d}y\,\text{d}x$$ we can remove the x from the first integral because technically it's a constant inside the inner integral. $$\int_0^4x\int_0^4 y\sqrt{1+x^2+y^2}\;\text{d}y\,\text{d}x$$ Let $u= 1+x^2+y^2$ and $\text{d}u_y=2y\;\text{d}y$. Note that when $y=0, u=1+x^2$ and when $y=4, u=17+x^2$. ...


1

I got it : First substitute : $x^2+y^2+1 = u$, Note that $x$ is constant. $ $ Then substitute $17+x^2 = u$ and $1+x^2 = p$ in two different integrals that you get after that. $ $ You should get the answer $$\dfrac{1-578 \sqrt{17}+1089 \sqrt{33}}{15} $$


2

Here's a graph of the region that you are to revolve about the "horizontal" axis $ \ y \ = \ -1 \ $ [the red line]. The outer radius of your "washers" will be measured "upward" from that line to the curve $ \ \ln(x+1) \ $ [the "upper" curve] , so you will have $ \ r_{outer}(x) \ = \ \ln(x+1) \ - \ (-1) \ $ ; you will have something similar for the inner ...


3

Are we allowed to have badly discontinuous functions and irregular subdivisions? If so, let's look at left and right endpoint sums for $\displaystyle\int_{0}^{1}f(x)\,dx$ where $f(x) = \begin{cases}1 & x \in \mathbb{Q} \\ 0 & x \not\in \mathbb{Q}\end{cases}$. Subdivide $[0,1]$ into $2n$ intervals: $[x_0,x_1], \ldots, [x_{2n-1},x_{2n}]$ where ...


0

Hint: $$\int xy\sqrt{1+x^2+y^2} \,dy = \frac{1}{3} \left(x^2+y^2+1\right)^{3/2} + C$$


2

A concise answer is as follows. As the number of subdivisions get larger, if the function is integrable, then the left-hand and the right-hand sums reach the integral of that function over the given interval; in other words, they become the same as the number of subdivisions goes to infinity.



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