New answers tagged

0

It's obvious that there exists $p(x)>1$ such that $$\frac1{(1+|x|)^{p(x)}}=\frac12\frac1{1+|x|}\quad\quad(x\ne0).$$


0

OK, I'm not super happy with what I have, but I'm posting in case it helps others find a better formula. $$\boxed{I(x)=\pi\log 2 -\frac 1 2 \int_x^1 \frac{K(\sqrt{1-t})-\frac \pi 2}{1-t}dt=\pi\log 2-\frac {\pi} 4 \sum_{n\geq 1}\left( \frac{(2n)!}{2^{2n}(n!)^2}\right)^2\frac{(1-x)^{n+1}}{n+1}}$$ where we identify the complete elliptic integral of the first ...


0

One such condition would be: there is a countable set $C$ such that$$(\forall x\in\mathbb R\setminus C):f(x)\geqslant g(x).$$


3

Before attacking the integral, I mention something about cubic theta function. The whole solution is quite self-contained if you accept the stated facts, the "footnote" contains more information. The three cubic theta functions are defined by $$\begin{aligned} a(q) &= \sum_{m,n} q^{m^2+mn+n^2}\\ b(q) &= \sum_{m,n} \zeta_3^{m-n} q^{m^2+mn+n^2}\\ c(q) ...


1

Let $u(x) = f'(x)$ and $v(x) = \sqrt{1+u(x)^2}$. So we want $u$ to have an elementary integral (so that we can write down $f$ on the assignment sheet) and $v$ to have an elementary integral (so our students can solve it.) In other words, we want functions $u$ and $v$, both with elementary integrals, so that $v^2 = 1 + u^2$. Rewrite this as $(v+u) (v-u) = 1$....


0

Your error lies in the computation of the intersection points of the curves. If $y^2=4x$ and $y=4x-2$, then $4x=(4x-2)^2$ indeed, but this is not an equivalence. It turns out that your region is the region below the graph of $2\sqrt x$ and above the graph of $-2\sqrt x$ (with $x\in\left[0,\frac14\right]$) plus the area of the region below the graph of $2\...


1

If you actually look at a graph of the required area you will see that it is quite difficult to find when integrating with respect to $x$. We can instead rearrange both equations to get $$x=\frac14y^2$$ $$x=\frac14y+\frac12$$ Then the graphs intersect at points where $y=-1$ and $y=2$ respectively so the area is given by $$\int_{-1}^2\left(\frac14y+\frac12\...


1

Use the trigonometric substitution $x(\theta)=\tan{\theta}$: $$\int\frac{1}{\sqrt{x^2+1}}\,dx= \int\frac{1}{\sqrt{[x(\theta)]^2+1}}x'(\theta)\,d\theta=\\ \int\frac{1}{\sqrt{\tan^2{\theta}+1}}\sec^2{\theta}\,d\theta= \int\frac{\sec^2{\theta}}{\sqrt{\sec^2{\theta}}}\,d\theta=\\ \int\frac{\sec^2{\theta}}{|\sec{\theta}|}\,d\theta. $$ Here, $-\frac{\pi}{2}<\...


1

Hint: Substituting $$x=\tan(t)$$ then we get $$x^2+1=\tan^2(t)+1=\frac{\sin^2(t)+\cos^2(t)}{\cos^2(t)}=\frac{1}{\cos^2(t)}$$ and $$dx=(\tan^2(t)+1)dt$$


2

For c=1, If we consider THE following integral $I=\int_{0}^{\infty}x^{a}e^{x^{b}}e^{-\lambda(e^{x^{b}}-1)}dy$ then by letting $y=x^{b}$ we get I=$\frac{e^{\lambda}}{b^{a+1}}\int_{1}^{\infty} (log~y)^{\frac{a}{b}}e^{-\lambda y}dy$ if we consider $\frac{a}{b}$ is an integer then $I=\frac{e^{\lambda}}{b^{a+1}} (\frac{a}{b})! E_{0}^{\frac{a}{b}}(\lambda)$ ...


0

As pointed out, this is really straight forward (if say, $a\lt0$): $\int_0^1 ab^ac^{-a-1}\operatorname dc=-b^a[c^{-a}]_0^1=-b^a$. If $a\gt0$, it appears to not converge.


2

We can write $$ \ln(1-x)=-\int_0^x \frac{dt}{1-t},\hspace{7mm}\operatorname{Li}_3(x)=\int_{0\leq t_1\leq t_2\leq t_3\leq x}\frac{dt_1\,dt_2\,dt_3}{(1-t_1)t_2t_3}. $$ We can multiply out $\ln(1-x)^3\operatorname{Li}_3(x)$ and break the result into a sum over the different possible orderings of the variables of integration. This will allow us to write $I$ as ...


0

Solving improper integral $$\int \frac{dx}{\sqrt{x(1-x)}}=\int \frac{dx}{\sqrt{(\frac{1}{2})^2-(x-\frac{1}{2})^2}}$$ (by completing square method) $$=\sin^{-1}(\frac{x-\frac{1}{2}}{\frac{1}{2}})+C$$ $$=\sin^{-1}(2x-1)+C$$ Hope it helps:)


5

It's quite common for such integrals to use algebraic identities in order to solve them, see also here. We can use the first one in the link from above, namely: $$6ab^2=(a+b)^3+(a-b)^3-2a^3\Rightarrow I=\int_0^1 \frac{\ln(1-x) \ln^2(1+x)}{x} dx$$ $$=\frac16\int_0^1 \frac{\ln^3(1-x^2)}{x}dx+\frac16\int_0^1 \frac{\ln^3\left(\frac{1-x}{1+x}\right)}{x}dx-\...


0

that is a great question but there is a splendid Q , I will post for knowlodge $$I=\int_{0}^{\frac{\pi }{4}}x^2 cotxdx\ \ \ \ \ \ \ \ \ by\ parts\ we\ have\\ \\ I=x^2ln(sinx)\tfrac{\frac{\pi }{4}}{0}-2\int_{0}^{\frac{\pi }{4}}xln(sin(x))dx\\ \\ =-\frac{\pi ^{2}ln(2)}{32}+\int_{0}^{\frac{\pi }{4}}(2ln(2)+2\sum_{k=1}^{\infty }\frac{cos(2kx)}{k})xdx\\ \\ =\...


0

$$I=\int_{0}^{1}\int_{0}^{1}\frac{ln(x)-ln(y)}{x-y}dxdy\\ \\ \\$$ $$=\int_{0}^{1}\int_{0}^{1}\frac{ln(\frac{x}{y})}{x-y}dxdy\ \ \ , let\ \ \frac{x}{y}=t\\ \\$$ $$=\int_{0}^{1}\int_{x}^{\infty }\frac{ln(t)}{t(t-1)}dt=[\int_{x}^{\infty }\frac{xln(t)}{t(t-1)}dt]\tfrac{1}{0}+\int_{0}^{1}\frac{ln(x)}{x-1}dx\\ \\ =\int_{1}^{\infty }\frac{ln(t)}{t(t-1)}dt+\int_{0}^...


5

That substitution has no meaning since $x\mapsto x(x-a-b)$ is not injective. You can not express $x$ in uniqe way with $u$: $$x= {a+b\pm \sqrt{(a+b)^2+4u}\over 2}$$ So what is $$\int _{-ab}^{-ab} {f(x)\,du \over 2x-a-b}?$$ Is it choice for $x$ with + or -?


0

$$\int_{0}^{\infty }\frac{ln(1+x^2)ln^2x}{1+x^2}dx\\ \\ let\ I(a)=\int_{0}^{\infty }\frac{ln^2(x)ln(1+a^2.x^2)}{1+x^2}\\ \\ \therefore I'(a)=\int_{0}^{\infty }\frac{2ax^2ln^2(x)}{(1+a^2x^2)(1+x^2)}dx=\frac{2a}{1-a^2}\int_{0}^{\infty }\frac{ln^2(x)}{1+a^2x^2}-\frac{ln^2}{1+x^2}dx\\ \\ let\ \ G=\int_{0}^{\infty }\frac{ln^2(x))}{1+a^2x^2}dx\ \ \ \ ,\ \ but \ we ...


7

Considering the algebraic identity \begin{align*} &(a-b)^3b = a^3b - 3a^2b^2 + 3ab^3 - b^4 = -2a^3b +3(a^3b+ab^3) -3a^2b^2 -b^4\\ &\Longrightarrow \ \ \ 2a^3b = -{b^4 \over 2} -{b^4 + 6a^2b^2\over 2} + 3(a^3b+ab^3) - (a-b)^3b \end{align*} with $a = \ln(1-x)$ and $b= \ln (1+x)$ it follows that \begin{align*} 2\int_0^1 {\ln^3(1-x)\ln(1+x)\over x}dx =&...


2

I'll just show an idea that avoids those type of sum, but skip the calculations. You might also have better ideas to solve them. For start we will denote $a=\ln(1-x)$ and $b=\ln(1+x)$ and use the following identity: $$a^2=\frac12 (a+b)^2+\frac12(a-b)^2-b^2$$ $$\Rightarrow I=\frac12 \underbrace{\int_0^1 \frac{\ln x\ln^2(1-x^2)}{1+x}dx}_{I_1}+\frac12\...


1

Interpreting your place as $\Bbb C$, on a radius-$a$ loop centred on the origin $z=a\exp i\theta$, so $$\oint_L\frac{dr}{r^2}=\frac{i}{a}\int_0^{2\pi}e^{-i\theta}d\theta=0.$$


4

Not an answer, but an extended comment for now. This hypergeometric function is a special case, and some complicated quadratic and cubic transformations apply to it. See this like for reference: https://dlmf.nist.gov/15.8. The formulas 15.8.25 and 15.8.26 both apply here. However, the most interesting one is so called Ramanujan’s Cubic Transformation (15....


2

There is an implicit assumption that you may have overlooked here. In physics, the convention is that the initial position is zero, i.e. $z(0) = 0$. Therefore, you should first write your integral as, $$z(t) = Z + v_0t+\frac{v_e(m_0-qt)}{q} \left[ \ln\left(1-\frac{qt}{m_0}\right)-1 \right]-\frac{gt^2}{2} \tag{1}$$ Then, find out $Z$ by setting $z(0)=0$, ...


2

Your calculation \begin{align}\frac{v_e\Big(m_0-qt\Big)\Big(\ln(1-\frac{qt}{m_0})-1\Big)}{q}&=v_e\Big(\frac{m_0}{q}-t\Big)\Big(\ln\big(1-\frac{qt}{m_0}\big)-1\Big)\\&=v_e\Big(\frac{m_0}{q}-t\Big)\Big(\ln\big(1-\frac{qt}{m_0}\big)\Big)-v_e\Big(\frac{m_0}{q}-t\Big)\\&=v_e\Big(\frac{m_0}{q}-t\Big)\Big(\ln\big(1-\frac{qt}{m_0}\big)\Big)+v_et-v_e\Big(...


1

Hint: $\int ln \, (1-at)dt=-\frac 1 a\int ln \, s \, ds$ where $s=1-at$ and $\int ln \, s \, ds=sln\, s -s+C$. See my comment below for determining $C$.


0

Note that you have $y^2=9x $ and $y= \frac{3x^2}{8}$ thus you need to have $$9x = (\frac{3x^2}{8})^2 $$ to start with.


1

There is an error in the calculation of the upper bound, where a square operation is overlooked. The correct equation for the bounds is, $$9x = \left(\frac{3}{8}x^2\right)^2$$ which yields $x_1=0$ and $x_2=4$. As a result, $$\int_0^4 \left( 3\sqrt{x} - \frac{3}{8}x^2 \right) = 8 $$


1

You have the right idea in that you have to first find the points of intersection of the two curves, but you are forgetting that one curve was given as $y^2$ and the other was given as $y$. Try rewriting the first curve as $y=\pm3\sqrt{x}$, $x\geq 0$, and try again. Your overall method is correct!


0

As JG123's comment stated, you just flipped the $A$ and $B$ in your calculations. Note you have $$\begin{equation}\begin{aligned} \frac{5x-5}{3x^2 -8x -3} & = \frac{A}{3x+1} + \frac{B}{x-3} \\ \frac{5x-5}{3x^2 -8x -3} & = \frac{A(x-3) + B(3x + 1)}{(3x + 1)(x-3)} \\ 5x-5 & = A(x-3) + B(3x + 1) \end{aligned}\end{equation}\tag{1}\label{eq1}$$ This ...


0

In the way you have written the partial fractions, $A=1, B=2$ doesn't work. Check the numerator: for instance, you get $7x$ instead of $5x$. As pointed out in one of the comments, the textbook has probably swapped the factors, which explains why you get swapped values.


0

Since $\ f^1, f^2, \dots, f^n\ $ and $\ y\ $ are all in $\ L_\mu^2(X)\ $, so is $\ y-\sum_\limits{i=1}^n\beta^i f^i\ $, and \begin{eqnarray} \int_{x \in X} (y(x)-\sum_{i=1}^n \beta^if^i(x))^2 d\mu(x)&=& \lVert y - \sum_{i=1}^n \beta^if^i\rVert^2\\ &=& \lVert y\rVert^2 + \sum_{i=1}^n\sum_{k=1}^n \langle f^i,f^j\rangle \beta^i\beta^j - 2\sum_{i=...


1

By definition the function $\phi$ maps from $\mathbb{R}$ to $\mathbb{R}$. Hence there is no ambiguity in taking derivatives: there is only one variable and one direction. Furthermore the notation $\phi'(x)$ is an abbreviation for $$\bigg(\frac{\partial \phi}{\partial t}(t)\bigg) \bigg|_{t=x}.$$ First the function gets differentiated, then you plug in ...


1

notice that: $$I(x)=\int_{-1}^1\frac{e^{-xu}}{\sqrt{1+u^2}}du$$ $$I'(x)=-\int_{-1}^1\frac{ue^{-xu}}{\sqrt{1+u^2}}du$$ now applying IBP you get: $$I'(x)=2\sinh(xu)-x\int_{-1}^1e^{-xu}\sqrt{1+u^2}du$$ so we can say that: $$I(x)=2\int_0^x\sinh(v)dv-x\int_0^x\int_{-1}^1e^{-uv\sqrt{1+u^2}}dudv$$


1

Term $H(a-|x|)$ defines a sphere at origin with radius $a$. Term $H(b-|x+y|)$ defines a sphere at $y$ with radius $b$. You are basically asking for a volume of intersection. Which is: $$ \frac{\pi}{12|y|}\Big(a+b-|y|\Big)^2\Big(y^2+2|y|(a+b)-3(a-b)^2\Big)H(a+b-|y|) $$


2

Case 1 : $p>0$ In this case, the integral is divergent since for large enough $x$ $$x^p(\ln x)^q>x^{p\over 2}$$ Case 2: $p<-1$ In this case, the integral is convergent since for large enough $x$ $$x^p(\ln x)^q<x^{p-1\over 2}$$ Case 3: $-1<p<0$ In this case, the integral is divergent since for large enough $x$ $$x^p(\ln x)^q>x^{p-1\...


2

Hint: put $y=\ln x$. You get $\int_{\ln 2} ^{\infty} e^{(1+p)y} y^{q} dy$. Can you handle this? The answer is either $p <-1$ or $p=-1$ and $ q<-1$.


1

By changing the limits the given integral is equal to $$\frac{1}{2}\int_{u=-1}^1\int_{v=1}^{2-|u|}\cos\left(u\right)e^{v}dvdu+\frac{1}{2}\int_{u=1}^3\int_{v=1}^{|u-2|}\cos\left(u\right)e^{v}dvdu$$ where $1/2$ is due to the Jacobian of the transformation. Please check the new limits by making a drawing of the domain in the $uv$ plane and by comparing it with ...


0

Hint In that specific case, you don’t need to change the variables. Just develop $\cos(x+y)$ using usual trigonometric formula and use $e^{y-x}=e^ye^{-x}$. You then have to integrate maps with $x,y$ as separated variables.


2

\begin{align} \sum_{n=1}^\infty\frac{(-1)^{n-1}H_n}{(2n+1)^3}&=\sum_{n=1}^\infty(-1)^{n-1} H_n\int_0^1\frac12x^{2n}\ln^2 x\ dx\\ &=-\frac12\int_0^1\ln^2x\sum_{n=1}^\infty(-x^2)H_n\\ &=\frac12\int_0^1\frac{\ln^2x\ln(1+x^2)}{1+x^2}\ dx\tag{1} \end{align} . Lets evaluate the integral : \begin{align} I&=\int_0^1\frac{\ln^2x\ln(1+x^2)}{1+x^2}\ ...


1

Let $$I=\int_{0}^{p} \exp[-(\ln x- a)^2/(2b^2)] ~dx.$$ Let $x=e^t$, then $$I=\int_{0}^{\ln p} \exp[-(t-a)^2/(2b^2)+t]~ dt= \int_{0}^{\ln p} \exp[-(t^2-2at-2b^2 t+a^2)/(2b^2)] dt=$$ $$ I=\exp[((a+b^2)^2-a^2)/(2b^2)] \int_{0}^{p} \exp[-(t-(a+b^2))^2/(2b^2)].$$ Let $(t-a-b^2)(/b\sqrt{2})=u$, then $$\Rightarrow I= \exp[((a+b^2)^2-a^2)/(2 b^2)] \int_{-(a+b^2)/(...


6

$$\frac{d}{dt}\int_0^tf(x,t)dx=\int_0^t\frac{\partial}{\partial t}f(x,t)dx~+~f(t,t)$$ Leibniz Integral Rule (Differentiation under the integral sign): Let $f(x, t)$ be a function of $x$ and $t$ such that both $f(x, t)$ and its partial derivative $\frac{\partial f}{\partial x}$ are continuous in $t$ and $x$ in some region of the $(x, t)$-plane, ...


2

Your $k$th summand equals $\dfrac{t}{k(k+t)}.$ We can thus write the series as $$\sum_{k=1}^{\infty}\frac{t}{k(k+t)}.$$ This series converges uniformly on $[a,b]\subset (-1,\infty)$ iff the sum from $k=2$ to $\infty$ converges uniformly on $[a,b].$ For this range of $k,$ we can use Weierstrass M: Fix $-1<a<b.$ Then for $k\ge 2$ and $t\in [a,b]$ we ...


1

$\DeclareMathOperator{\Res}{Res}$ $\newcommand{\i}{\mathrm{i}}$ $\newcommand{\d}{\mathrm{d}}$ This integral is a good candidate for residue calculus. Without loss of generality, set $a=1$ and ignore the factors of $\pi$. Then $$\begin{split}\int_{-\infty}^{\infty}\frac{\d x}{(x^2+1)((x-t)^2+1)}&=2\pi\i\left(\Res_{x\to \i}+\Res_{x\to t+\i}\right)\frac{\...


7

I mean, you can just compute this using partial fraction decomposition. $$I=\int_{-\infty}^\infty f(x)f(t-x)dx=\frac{a^2}{\pi^2}\int_{-\infty}^\infty\frac{1}{x^2+a^2}\frac{1}{(t-x)^2+a^2}dx$$ We write $$\frac{1}{x^2+a^2}\frac{1}{(t-x)^2+a^2}=\frac{Ax+B}{x^2+a^2}+\frac{Cx+D}{(t-x)^2+a^2}$$ Recombining these, our equations read $$1=(Ax+B)(x^2-2xt+t^2+a^2)+...


2

$f(x)=2x$ whenever $f$ is continuous at $x$. Since $f$ is increasing it is continuous except at a countable number of points Hence it is continuous on a dense set. Given any $x$ choose an increasing sequence $(x_n)$ and a decreasing sequence $(y_n)$ both tending to $x$ such that $f$ is continuous at each $x_n$ and at each $y_n$. Then $2x_n=f(x_n)\leq f(x) \...


1

It is not uniformly convergent on $\mathbb R$. Suppose it is. Then there exists $N$ such that $N_2 >N_1 >N$ implies $\sum\limits_{k=N_1}^{N_2} \frac t {k(k+t)} <1$ for all $t$. Put $t=N_2$ and observe that $\frac {N_2} {k(k+N_2)} \geq \frac 1 {2k}$ whenever $N_1 \leq k \leq N_2$. This gives $\sum\limits_{k=N_1}^{N_2} \frac 1 {2k} <1$ whenever $...


0

If you want to see what the region looks like, proceed as follows: Adding $y$ to both sides of the first set of inequalities gives $$y\le x\le 1+y,$$ and subtracting $y$ from sides of the second set gives $$2-y\le x\le 3-y.$$ Then it becomes apparent that the concerned region is defined by the system $y\le x, x\le 1+y, 2-y\le x$ and $x\le 3-y.$ These are ...


4

By Lebesgue differentiation theorem, we have $F' = f$ almost everywhere. So we have $f(x) = 2x$ almost everywhere. In particular, for at least two different points $x_1 < x_2$, we have $f(x_1) = 2 x_1$ and $f(x_2) = 2 x_2$, so $f$ can't be non-increasing. Now, if $f(x_0) \neq 2 x_0$ for some $x_0$, then either $f(x_0) < 2 x_0 - \varepsilon$ or $f(x_0) ...


0

In fact, the limit equalities that you wrote are valid if $f$ is supposed to be continuous. The function $$f(x)=\begin{cases} 2x & x \notin I \\ 0 & x \in I \end{cases}$$ where $I=\{1/n \ ; \ n \in \mathbb N\}$satisfies the requested equality but is not increasing. The function $f(x)=2x$ for all $x \in \mathbb R$ is the only increasing one ...


0

Here is an independent solution: \begin{align} I&=\int_0^1\frac{\ln y\ln^2(1+y)}{y}\ dy\overset{y=\frac{1-x}{x}}{=}\int_{1/2}^1\frac{\ln(1-x)\ln^2x-\ln^3x}{x(1-x)}\ dx\\ &=\underbrace{\int_{1/2}^1\frac{\ln(1-x)\ln^2x}{x}\ dx}_{IBP}-\int_{1/2}^1\frac{\ln^3x}{1-x}\ dx-\underbrace{\int_{1/2}^1\frac{\ln^3x}{x}\ dx}_{-(\ln^42)/4}+\int_{1/2}^1\frac{\ln(1-x)...


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