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8

HINT: As $y^3+1=(y+1)(y^2-y+1),$ $$\dfrac{x^4+1}{x^6+1}=\dfrac{x^4-x^2+1}{x^6+1}+\dfrac{x^2}{x^6+1}=\dfrac1{x^2+1}+\dfrac{x^2}{x^6+1}$$ Set $x^3=u$ for the second part


7

Hint: break it into two pieces, one with $u < v$ and the other with $u > v$.


6

Someone changed the function to be plotted from the original function, but the below figure is for the equation as currently stated.


6

The General Case: Consider instead the integral \begin{align} J(\eta) &=\Re\int^{2\pi}_0x^\eta\ {\rm Li}_2(e^{ix})^2\ {\rm d}x\\ \end{align} for $\eta\in\mathbb{N}_0$. Deforming the contour around $z=0$, we get \begin{align} J(\eta) &=\Re\oint_{|z|=1}\frac{{\rm Li}_2(z)^2\ln^{\eta}(z)}{i^{\eta+1}z}{\rm d}z\\ &=\Re\int^1_0\frac{\left((\ln(x)+2\pi ...


5

I can confirm $I(1)$: Using the series expansion for Clausen's function we have $$ I(p)=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\int_0^{2 \pi}x^p\frac{\sin(n x)}{n^2}\frac{\sin(m x)}{m^2}\,\mathrm dx $$ now setting $p=1$. Furthermore we may use the orthogonality of sine $\int^{2 \pi}_{0}{\sin(n x)\sin(m x)}\,\mathrm dx=\pi\delta_{mn}$ and integrate by ...


5

We have: $$\sin^2(x) = \frac{1-\cos(2x)}{2}\tag{1}$$ hence Parseval's identity implies: $$ \int_{-\pi}^{\pi}\sin^4(x)\,dx = 2\pi\cdot\frac{1}{4}+\pi\cdot\frac{1}{4}=\color{red}{\frac{3\pi}{4}}.\tag{2}$$


5

The integral is one over the $2$-simplex $\Delta_2(2) = \{ (x,y ) : x, y \ge 0, x+ y \le 2 \}$. One standard trick to deal with integral over $d$-simplex of the form $$\Delta_d(L) = \{ (x_1, x_2, \ldots, x_d ) : x_i \ge 0, \sum_{i=1}^d x_i \le L \}$$ is convert it to one over the $d$-cuboid $[0,L] \times [0,1]^{d-1}$ through following change of variables ...


4

Here we have a 3D picture of the entire region ($x$ horizontal, $y$ depth, $z$ height): In blue we see the $x$-$y$ plane, while the pink part stretches both through the $y$-$z$ plane and the $x$-$y$ plane. This side is determined by the equation $x = y$, so the image can be a bit off-putting. Suppose we cut at $y = 2$, then we obtain: Here we see ...


3

Hint. We have, as $x \to 0$, $$ \ln(1+x^4)=x^4+\mathcal{O}(x^5) $$ giving $$ \frac{\ln^{\alpha}(1+x^4)}{x^4}=\frac1{x^{4-4\alpha}}+\mathcal{O}\left(\frac1{x^{3-4\alpha}}\right) $$ and $$ \frac{\ln^{\alpha}(1+x^4)}{x^4}\cos \frac1x=\frac1{x^{4-4\alpha}}\cos \frac1x+\mathcal{O}\left(\frac1{x^{3-4\alpha}}\right). \tag1 $$ Thus $$ ...


3

$$ \int\frac{dx}{x + \sqrt{1-x^2}}\ $$ Take $x=\sin\theta$, $$ \int\frac{\cos\theta}{\sin\theta + \cos\theta} \, d\theta $$ $$ \frac{1}{2}\int\frac{2\cos\theta}{\sin\theta + \cos\theta} \, d\theta $$ $$ \frac{1}{2}\int \frac{\cos\theta+\sin\theta}{\cos\theta+\sin\theta} \, d\theta+\frac{1}{2} \int \frac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}\, ...


3

Here's one idea: $\cos(nt) \geq 1/2$ on $[0,\pi/3n]$, and is always $\geq -1$, so $$\int_0^\pi e^{-t} \cos(nt) \geq \int_0^{\pi/3n} e^{-t}/2 dt - \int_{\pi/3n}^\pi e^{-t} dt.$$ I don't know if this is tight enough for your purposes, however. If I needed this a little tighter I would repeat the same idea: cut off the region where $\cos(nt) \geq 1/2$ and ...


3

$I(1)$ can be evaluated in the following way also. One has to note that $$\int_0^{2\pi} x\sin(mx)\sin(nx)\,dx=\begin{cases} 0 & n \neq m \\ \pi^2 & n=m \end{cases}$$ To prove the above, write $$\int_0^{2\pi} x\sin(mx)\sin(nx)\,dx=\frac{1}{2}\int_0^{2\pi} x\cos((m-n)x) \,dx-\frac{1}{2}\int_0^{2\pi} x\cos((m+n)x)\,dx$$ and then use integration by ...


3

Just find two pairs of functions $f_1, g_1$ and $f_2,g_2$ such that $$\int f_1 = \int f_2, \quad\quad \int g_1 = \int g_2$$ but $$\int f_1 g_1 \neq \int f_2 g_2$$ Probably any two pairs you come up with will work.


3

Another way, Solution : \begin{align} \int_{-\infty}^{\infty}\frac{x^4+1}{x^6+1}\,\mathrm dx&=2\int_{0}^{\infty}\frac{x^4+1}{x^6+1}\,\mathrm dx\\[7pt] &=2\int_{0}^{\infty}\frac{x^4\,\mathrm dx}{x^6+1}+2\int_{0}^{\infty}\frac{\mathrm dx}{x^6+1}\tag{$\color{red}{❤}$}\\[7pt] ...


2

$$\begin{eqnarray*}\int_{\mathbb{R}}\frac{x^4+1}{x^6+1}\,dx&=&2\int_{\mathbb{R}^+}\frac{x^4+1}{x^6+1}\,dx=2\int_{0}^{1}\frac{x^4+1}{x^6+1}\,dx+2\int_{1}^{+\infty}\frac{x^4+1}{x^6+1}\,dx\\&=&4\int_{0}^{1}\frac{x^4+1}{x^6+1}\,dx=4\int_{0}^{1}\sum_{k\geq 0}(-1)^k\left(x^{6k}+x^{6k+4}\right)\,dx\\&=&4\sum_{k\geq ...


2

$$x=a\cos{t}\Longrightarrow dx=-a\sin{t}dt$$ $$x=0\Longrightarrow t=\dfrac{\pi}{2}$$ $$x=a\Longrightarrow t=0$$ Thus you can rewrite your integral as: $$\int_0^ay\ dx=\int_{\pi/2}^0(b\sin{t})(-a\sin{t})dt=\ldots=\dfrac{1}{4}\pi ab=\dfrac{1}{4}\mathcal{A}_{\text{ ellipse}}$$


2

Since: $$\frac{d}{dx}\sqrt{1+x^4} = \frac{4x^3}{2\sqrt{1+x^4}}$$ we have: $$ \int_{0}^{1}\frac{x^3}{\sqrt{1+x^4}}\,dx = \frac{\sqrt{2}-1}{2}.$$


2

To answer my own question, after looking at Convergence of the improper integral $\int_{0}^{\pi/2}\tan^{p}(x) \; dx$ I tried the substitution $u = i-r$ and got $$ \int_{a + bt^2}^{i} \frac{r \left(i-r\right)^{\frac{3}{2}\left(i-1\right)}}{\left(j+i-R-r\right)^{\frac{3}{2}\left(j+i-1\right)}}\mathrm{d} r =\\\frac{2\left(-a-b t^2+i\right)^{\frac{3 ...


2

$$\begin{eqnarray*} I &=& \frac{1}{n}\int_{0}^{n\pi}e^{-t/n}\cos t\,dt =\frac{1}{n}\sum_{j=0}^{n-1}\int_{j\pi}^{(j+1)\pi}e^{-t/n}\cos t\,dt\\&=&\frac{1}{n}\int_{0}^{\pi}\cos t\,e^{-t/n}\sum_{j=0}^{n-1}(-1)^j e^{-j\pi/n}\,dt\tag{1}\end{eqnarray*}$$ Now: $$\sum_{j=0}^{n-1}(-1)^j ...


2

The standard way of proving this identity is to write $x^{-x} = e^{-x\log x}$ and then expand $e^{-x\log x} = \sum_{n=0}^{\infty} \frac{(-x\log x)^{n}}{n!}.$ Now the integral is $$\int_{0}^{1} \sum_{n=0}^{\infty} \frac{(-x\log x)^n}{n!} \mathrm{d}x = \sum_{n=0}^{\infty} \int_{0}^{1}\frac{(-x\log x)^{n}}{n!}\mathrm{d}x= \sum_{n=0}^{\infty} ...


2

The sequence for $A(x)$ is $a_0, a_1, a_2, \cdots, a_n, \cdots$ $$A(x) = a_0x^0 + a_1x^1 + a_2x^2 + \cdots + a_n x^n +\cdots$$ $$\int_0^x{A(t)} = a_0\frac{x^1}{1} + a_1\frac{x^2}{2} + a_2\frac{x^3}{3} + \cdots a_{n-1}\frac{x^n}{n}+\cdots$$ So the sequence for the integral of $A(x)$ is: $$0, \frac{a_0}{1}, \frac{a_1}{2}, \frac{a_2}{3}, \cdots, ...


2

$$ \int_0^t \int_0^s \frac{\min(u,v)}{uv} \, dv \, du $$ The suggestion given by "Ant" and by Robert Israel will work. I would add this: What you get may depend on whether $s<t$ or $t<s$. To divide the region into two parts as suggested, you need to look at that. Suppose $s<t$. Then the region where $u\le v$ is where $0\le u\le v\le s$, and the ...


2

Let $$ R_n=\dfrac{1}{n}\sum\limits_{i=0}^{n-1}\int_{i}^{i+1}f(\dfrac{i}{n})g(x)dx $$ First we prove that $$ \lim\limits_{n\to\infty}R_n=\int_{0}^{1}f(x)dx\int_{0}^{1}g(x)dx $$ Since $f(x)$ is uniform continuous on $[0,1]$, $\forall \epsilon>0,\space\exists N>0, \forall n>N$, let $\delta=\dfrac{1}{n},\space \forall x_1,x_2\in [0,1], ...


1

First $$ \int_{S_{n}}^{S_{n+1}}\dfrac{dx}{x\ln(x)}\leq\int_{S_{n}}^{S_{n+1}}\dfrac{dx}{S_{n}\ln(S_{n})}=\dfrac{1}{S_{n}\ln(S_{n})}\int_{S_{n}}^{S_{n+1}}dx=\dfrac{a_{n+1}}{S_{n}\ln(S_{n})} $$ for $S_{n+1}\geq S_n$ and $x\ln(x)\downarrow$. So $$ ...


1

I get that the integral goes like $\frac{\pi(1-e^{-\pi})}{n^2}$ for even $n$. Here is the annoying long derivation. I'll assume that $n$ is even for now and write $n = 2m$. First, split the integral into $m$ parts. $\begin{array}\\ I(n) &=I(2m)\\ &=\int_0^\pi e^{-t}\cos n\, t\ dt\\ &=\int_0^\pi e^{-t}\cos (2mt) dt\\ ...


1

It seems your fundamental difficulty is with definitions of probability concepts, not so much with the process of integration. If $X \sim Unif(50, 150),$ then the density function $f_X(x)$ has three parts: (i) For $x < 50,\;f_X(x) = 0;\;$; (ii) for $50 \le x \le 150,\;f_X(x) = 1/100 = 0.01;$ and (iii) for $x > 150,\;f_X(x) = 0.$ If you want to ...


1

Your integral with limit $\int_{r_e}^{r_o}$ will produce $\frac{q^2}{8\pi r_e}-\frac{q^2}{8\pi r_o}$ (in fact this follows from $\int_{r_e}^{r_o}=\int_{r_e}^\infty-\int_{r_o}^\infty$), so it does make sense to replace the infinite upper limit with a finite one (which would however feel arbitrary), but even then you cannot replace the lower limti with $0$.


1

hint What about the change of variable $1+x^4=u$ such that $4x^3dx=du$ So the integral becomes $$\frac{1}{4}\int_1^2 u^{-\frac{1}{2}}du$$ Can you take it from there



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