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0

It seems not true in general. In $\mathbb{R}^2$, Consider $S$ as the set of all $(x,y)$ in $[0,1] \times [0,1]$ such that atleast one of $x$ or $y$ is irrational and $Q$ to be any rectangle containing $[0,1] \times [0,1]$ . These $S$ and $Q$ satisfies the given condition and $S$ contains no rectangle at all.


1

Use the residue theorem! We first assume that: $$\exp\sqrt{x}=z$$ Then your formula turns out to be: $$ \int_{1/\sqrt{2}}^{1} \frac{\arccos\frac{z}{\sqrt{2}}} {1-z^2} \,\mathrm{d}x = -\frac{\pi^{3}}{192} $$ The residue theorem gives $$ \int_{1/\sqrt{2}}^{1} \frac{\arccos\frac{z}{\sqrt{2}}} {1-z^2} \,\mathrm{d}x ...


0

$sin(x\cdot sint) = x\cdot sint - \dfrac{(x\cdot sint)^3}{3!} + ...$, and integrate term by term should give the answer.


1

Yes, the $Max$ function of two Riemann -integrable functions is Riemann -integrable. This is because the $Max$ of two a.e -continuous bounded functions is also a.e -continuous and bounded. For $f,g$ continuous, the function $Max${f,g} is continuous. This implies that (since a Riemann-integral must be a.e -continuous) , that $Max${f,g} is a.e -continuous; ...


0

Given $ \varepsilon> 0 $ is $ k \in \ N $ such that $\varepsilon >1/k$ consider a 'homogeneous' partition of $ [0,1] \times [0,1] $, i.e., all sub- rectangles of partition has area $1/k^2$, we know that $ m_B = 0$ and $ m_B = 1$ if $B$ intection the line $ y = x $ is not empty set, so $ S (f, P)-s (f, P) = \sum_ {B \ in P} (m_B-m_B) vol (B) = \sum_ ...


0

Note that the product of bounded functions is bounded, so since $f$ and $g$ are Riemann integrable, $fg$ is bounded too. Furthermore, if $f$ and $g$ are continuous at $x$, so is $fg$; hence we have the inclusion $$\{x : fg \text{ is not continuous at } x\} \subseteq \{x : f \text{ not continuous}\} \cup \{x : g \text{ not continuous}\}$$ Since $f$ is ...


0

Hints: (1) What could you say if $g(x)=\int_a^xy(t)dt$? (2) Note that $\int_{u(x)}^{v(x)}y(t)dt =\int_{a}^{v(x)}y(t)dt - \int_{a}^{u(x)}y(t)dt$.


5

Let $ \displaystyle I(z) = \int_{0}^{\infty}\frac{\arctan \frac{x}{z}}{e^{\pi x}-1} \ dx$. Then $$ I(z) = \int_{0}^{\infty} \int_{0}^{\infty}\frac{1}{e^{\pi x}-1} \frac{\sin xt}{t}e^{-zt} \ dt \ dx$$ $$ = \int_{0}^{\infty} \frac{e^{-zt}}{t} \int_{0}^{\infty} \frac{\sin tx}{e^{\pi x}-1} \ dx \ dt$$ $$ = \frac{1}{2}\int_{0}^{\infty} \frac{e^{-zt}}{t} ...


2

I think the best justification is that it leads into the general context of substitutions for multiple integration. For example, it can be shown that the Jacobian of a polar transformation $$T(r,\theta)=(r\cos \theta,r\sin \theta)$$ is given by $$\begin{vmatrix}\cos\theta&\sin\theta\\ ...


2

Any indefinite integral, which you can integrate to get a known function can be done just based on the fundamental theorem of calculus. All the other tricks like $u$-substitution, partial fractions, integration by parts only make this procedure easier, i.e., it helps us guessing the antiderivative.


0

$$\int(\ln(x))^ndx=x(\ln(x))^n-\int xd((\ln(x))^n)=x(\ln(x))^n-\int x\left(\dfrac{n(\ln(x))^{n-1}}{x}\right)dx\\ \implies \int(\ln(x))^ndx=x(\ln(x))^n-\int {n(\ln(x))^{n-1}}dx$$ $$\int x^ne^{ax}dx=\dfrac{1}{a}\int x^nd(e^{ax})=\dfrac{1}{a}e^{ax}x^n-\dfrac{1}{a}\int e^{ax}(nx^{n-1})dx$$


0

For the first question define $I_{n}=\int(\ln(x))^{n}dx$ then by integrating by parts we have: $$I_{n}=x(\ln(x))^{n}-n\int(\ln(x))^{n-1}dx=x(\ln(x))^{n}-nI_{n-1}$$ For (b) we define $I_{n}=\int x^{n}e^{ax}dx$ then by integrating by parts we have: $$I_{n}=\frac{1}{a}x^{n}e^{ax}-\frac{n}{a}\int x^{n-1}e^{ax}dx=\frac{1}{a}x^{n}e^{ax}-\frac{n}{a}I_{n-1}$$


2

If by Type this or that integrals you mean changing the order of integration, you'd get $$0\le x\le 1\;,\;\;0\le y\le x^2\;\;\longleftrightarrow\;\; 0\le y\le 1\;,\;\;\sqrt y\le x\le 1$$


3

$$\int\frac x{\sqrt{x^2+k^2}}dx=\frac12\int\frac{(x^2+k^2)'}{\sqrt{x^2+k^2}}dx=\frac122\sqrt{x^2+k^2}+C=\sqrt{x^2+k^2}+C$$


3

Hint: Let $u=x^{2}+k^{2}$. Then $du=2xdx$ since $k$ is a constant.


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

The argument looks fine, although the notation is somewhat sloppy: for instance I would write the first step as $$\int_0^{\infty} f(x)\cos(nx)\,dx = \cos(nx)\int_0^x f(y)\,dy\ \biggr\vert_{x=0}^{\infty} + n\int_0^{\infty} f(x)\sin(nx)\,dx.$$ The claim becomes false if $f(x)$ is locally integrable but not $L^1$, i.e. if $\lim_{x\to \infty} \int_0^x ...


2

Try taking the indefinite integral of both sides instead: \begin{align*} \frac{ds}{s} &= \mu \, dt \\ \int \frac{1}{s} \, ds &= \int \mu \, dt \\ \ln|s| &= \mu t + C &\text{for some constant $C \in \mathbb R$}\\ e^{\ln|s|} &= e^{\mu t + C}\\ |s| &= e^{\mu t} \cdot e^C \\ s &= \underbrace{(\pm e^C)}_{A}e^{\mu t} \\ s(t) &= ...


1

Zarrax's method is certainly the one intended, but you could note that an explicit antiderivative is available: $$ F(t) = t \ln\left(1 + \dfrac{1}{t^2}\right) + 2 \arctan(t) $$ and find the limits of $F(t)$ as $t \to 0$ and $t \to \infty$.


3

Use that $\ln(1 + {1 \over t^2}) = \ln(t^2 + 1) - 2\ln(t)$. The first term is bounded near $t = 0$, so the problem reduces to showing that the integral of $\ln(t)$ converges near $t = 0$.


3

One way of viewing the hyperbolic functions is that they are "abbreviations" for combinations of exponential functions, so whatever can be done using hyperbolic functions can equally be done using exponentials - albeit in a less concise manner. Similarly, inverse hyperbolics are really just log functions in various combinations. What this means is that if ...


2

You might think about a diagram like this: region between the white planes for x, region between the pink planes for y, region between the cyan planes for z. The solid is a bit easier to see now, demarcated with blue points


3

You can find the limits of integration by solving the inequalities. But in this case, the inequalities are already in solved form, if you integrate in the order $x,y,z$ (listed from innermost to outermost), so there's nothing to do. If you wanted to integrate in the order $z,y,x$, we could solve in this other direction. In this case, it's pretty easy: we ...


0

Write the equation for the plane using cylindrical coordinates: $y+2z=2\implies \rho\sin{\phi}+2z=2$. Solving for $z$ we have, $$z(\rho,\phi)=\frac{2-\rho\sin{\phi}}{2},~~\text{where}~0\leq\rho\leq1,~0\leq\phi\leq2\pi.$$ Then the surface parametrization is $$\vec{r}(\rho,\phi)=\langle\rho\cos{\phi},\rho\sin{\phi},\frac{2-\rho\sin{\phi}}{2}\rangle.$$


0

\begin{align} \frac \rho C & = \sec\theta \\[10pt] \sqrt{\left(\frac\rho C\right)^2 -1} & = \sqrt{\sec^2\theta-1}= \tan\theta \\[10pt] \frac{d\rho}{C} & = \sec^2\theta\,d\theta \end{align} \begin{align} \int{1 \over \sqrt{\left(\dfrac{\rho}{C}\right)^2 - 1}} \, d\rho & = \int\frac{1}{\tan\theta} \Big( C\sec^2\theta\,d\theta\Big) = C\int ...


1

A different manner to answer to the question is to show a graphical representation to the function $f(x)$ for various values of $n$ and a graphical representation of the respective integrals $F(X)$. ( The old saw that “One little picture says more than a long speech”) This makes more understandable the behavior of the function and integral for $n$ tending ...


-1

\begin{equation} \begin{split} f(x) &= \int^{\ln2}_0 (3e^u - e^{2u} - 2)\sin nu \,du\\ &=3\int^{\ln2}_0 e^u\sin nu \,du-\int^{\ln2}_0e^{2u}\sin nu \,du-2\int^{\ln2}_0\sin nu \,du\\ &=3\int^{\ln2}_0 e^u\sin nu \,du-\int^{\ln2}_0e^{2u}\sin nu \,du+\frac{2}{n}(\cos(n\ln2)-1) \end{split} \end{equation} The first two integrals can be solven by using ...


2

Your intuition is correct: for the first part, it is equal to $\lfloor \frac{x}{\pi} \rfloor \int_{0}^{\pi}\sin (t) dt$, which you can compute. The second one is equal to $\int_{0}^{x-\pi \lfloor \frac{x}{\pi} \rfloor}\sin (t) dt$. As pointed out in comments, we can argue like this since we are looking at $|\sin|$. Morally, $\int_0^{\lfloor ...


2

This one is perhaps easier than the arbitrary function $f(x)$ in the example in the comments. First note that: $$ \begin{aligned} &\int_0^1 \int_0^1 \ldots \int_0^1 \sin \bigg(\frac{x_1+x_2+\ldots+x_n}{n}\bigg)\,dx_1 \,dx_2 \ldots \,dx_n\\&=\Im{\left[\int_0^1 \int_0^1 \ldots \int_0^1 \text{exp} \bigg(i\frac{x_1+x_2+\ldots+x_n}{n}\bigg)\,dx_1 \,dx_2 ...


1

Note that $f$ is bounded on $[\epsilon,1]$, so you just need to show that the integral of a bounded function which vanishes outside the Cantor set is zero. Use the fact that the Cantor set is covered by a finite union of closed intervals of arbitrarily small measure.


1

It looks to me that your problem is: $$\iiint_{D}x^{2}\:\mathrm{d}x\:\mathrm{d}y\:\mathrm{d}z,$$ Where $D$ is the region described by: $x^{2}+2y^{2}+z^{2}\leq 2$. Which is a triple integral, rather than the double integral that you describe, however, the process is much the same. We first describe our transformation from Cartesian co-ordinates into the new ...


0

\begin{align} \frac{1}{1+\sin{(2\, \theta)}\, \cos{\alpha}} &=\frac{1}{1+2\sin{\theta}\cos{\theta}\, \cos{\alpha}}\\ &= \frac{\sec{(\theta)}^2}{\sec{(\theta)}^2+2\tan{\theta}\, \cos{\alpha}}\\ &= \frac{\sec{(\theta)}^2}{1+\tan{(\theta)}^2+2\tan{\theta}\, \cos{\alpha}}\\ &= ...


1

Parts (ii), (iii), and (v) are pure linear algebra. We are given a linear map $$L:\quad C^1(I)\to C^0(I),\qquad f\mapsto f'+a\> f\ .$$ Solving the ODE $(2)$ means finding the $f$'s in $C^1(I)$ with $Lf=b$, where $b\in C^0(I)$ is given in advance. Denote the set of these $f$'s by ${\cal S}$. (ii) When $f$, $f_0\in{\cal S}$ then $L(f-f_0)=Lf ...


1

Well, your function does not depend on $\phi$ for the very beginning therefore it is correct that you have no dependancy in the end :) Some more details that I noticed: "$f(\theta, \phi) = u_0$ for the range $[0 \le \theta < \pi/2]$" <- but you have to integrate over range $[0 \le \theta < \pi]$, what's the $f(\theta, \phi)$ for $[\pi/2 \le \theta ...


0

$\int_{0}^{\frac{\pi}{2}}\frac{1}{2(\sin(2\theta)\cos(\alpha)+1)}d\theta=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\frac{1}{2\sin(\theta)\cos(\theta)\cos(\alpha)+1}d\theta=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\frac{\sec^{2}(\theta)}{2\tan(\theta)\cos(\alpha)+\sec^{2}(\theta))}d\theta$ ...


1

You need to show that there is linear subspace $U\subset C^1(I,\mathbb{C})$ and a vector $v_0\in C^1(I,\mathbb{C})$ such that every solution of $(2)$ lies in $v+U$. We claim that $U$ is a set of solutions of $(1)$, $v_0=f_0$. Indeed, by paragraph $(ii)$ we already know that every solution of $(2)$ is a sum of $f_0$ and some solution of $(1)$. Thus we ...


11

Let the desired integral be denoted by $I$. Noting that $1-(2t-1)^2=4t(1-t)$ we see that $$ I=4\int_{\square}\frac{-\log(1-(2x-1)^2)+\log(1-(2y-1)^2)}{(2x-1)^2-(2y-1)^2}dxdy\quad\hbox{with $\square=[0,1]^2$}. $$ But, for $t\in(-1,1)$, we have $$ -\log(1-(2t-1)^2)=\sum_{n=1}^\infty\frac{(2t-1)^{2n}}{n} $$ So the integrand becomes, for $(x,y)\in(0,1)^2$, ...


0

Or you could simply do the following: $$ \text {Area} = 9\times 4 -\int_1^9 \sqrt x dx $$


0

Note \begin{align}\int{\sqrt{26x-x^2}dx}& = \int{\sqrt{169-(x^2-26x+169)}dx} \\ &=\int{\sqrt{13^2-(x-13)^2}dx} \end{align} It can be evaluated using $x-13=13\sin t,0\leq t \leq \pi/2$.


0

Using Trigonometric Substitution, $$2ax-x^2=a^2-(x-a)^2$$ Set $$x-a=a\sin\theta$$ Here $a=13$


1

Hint: First notice that $26x-x^{2}=-(x^{2}-26x)=-((x-13)^{2}-169)=169-(x-13)^{2}$


1

From the picture it is clear that if you want to change the order of integration you should integrate over the region D. Region D can be divided in two parts: For D1 we have: integrate from $x=1$ to $x=y^2$, and $y\in(1,3)$, and for D2 we integrate from $x=1$ to $x=9$ and $y\in(3,4)$. Calculus for D1: $$ D_1=\int^3_1 dy \int^1_{y^2} dx $$ and for ...


0

I think it looks fine. Just mention $U$ is upper sum and $L$ is lower sum.


1

I'd suggest using the following theorem: $$ \psi_1(x) = \frac{1}{2\pi i} \int_{c - i\infty}^{c + i\infty} \frac{x^{s + 1}}{s(s+1)} \left( \frac{-\zeta'(s)}{\zeta(s)} \right) \mathrm{d}s$$ where $c > 1$. A proof of this equality can, for example, be found in Complex Analysis by Stein and Shakarchi. It is on page 191 being proposition 2.3 of chapter 7. ...


5

I assume you mean $$\int_0^{\infty} \dfrac{dx}{1+x^2} = 2 \int_0^1 \dfrac{dx}{1+x^2}$$ This is true because $$\underbrace{\int_1^{\infty} \dfrac{dx}{1+x^2} = \int_1^0 \dfrac{-du/u^2}{1+1/u^2}}_{x = 1/u} = \int_0^1 \dfrac{du}{u^2+1}$$ Hence, $$\int_0^{\infty} \dfrac{dx}{1+x^2} = \int_0^{1} \dfrac{dx}{1+x^2} + \int_1^{\infty} \dfrac{dx}{1+x^2} = \int_0^{1} ...


1

$$\int_{0}^{\infty} \frac {1}{1+x^2} \, \mathrm{d}x=\lim_{n\,\to\,\infty}\left({\arctan n}\right)-\arctan0={\pi\over 2}$$ It is not true that $${\pi\over 2}=\pi$$ unless someone made a big mistake and math is all wrong. Perhaps you meant $$\int_{0}^{1} \frac {1}{1+x^2} \, \mathrm{d}x=\int_{1}^{\infty} \frac {1}{1+x^2} \, \mathrm{d}x={\pi\over 4}$$


0

It is best practice to use a dummy variable in the integrand itself. For example, we could write $$\int_0^x{t\,dt}=\frac{x^2}{2}$$ It does not really make sense to write something like $$\int_0^x{x\,dx}$$ because $x$ is a variable which is changing as the integral is computed. $x$ starts at $0$ and moves toward... toward $x$? That said, I think most people ...


0

Do you have equally spaced values available? ~~-> Simpson or similar Can you fix the values where to compute the function? ~~-> use one of the Gaussian methods Wildly varying function? ~~-> some adaptative method



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