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8 votes
Accepted

How to evaluate the integral $\int_0^{\infty} \frac{\sin(k \ln(t))}{\sqrt{t}} \left\{\frac{1}{t}\right\} \, dt $

Firstly, let's replace the sine function by a complex exponential, so that your integral corresponds to the imaginary part of the following integral : $$ \int_0^\infty \frac{\exp(k\ln t)}{\sqrt{t}} \...
Abezhiko's user avatar
  • 10.4k
3 votes

how to find a good change of variables to solve these kinds of multi-integral questions?

You do not need to solve for $x$ and $y$. These problems are usually constructed very carefully (and artificially) to work out just right. Remember that the Jacobian you need is the reciprocal of the ...
Ted Shifrin's user avatar
3 votes
Accepted

how to find a good change of variables to solve these kinds of multi-integral questions?

When it comes to changes of variables, one possible approach is to determine formulas for $x(u,v)$, $y(u,v)$, compute the Jacobian, and apply the change of variables theorem as you outline. In some ...
Rob's user avatar
  • 7,297
2 votes

Proving an inequality on a monotonic sequence of positive real numbers

Case 1. Let $q=e^{-z_M}.$ Then $$\sum_{i=1}^Me^{-iz_i}>\sum_{i=1}^Mq^i={q(1-q^M)\over 1-q}$$ Therefore $$ p={1-q^M\over \sum_{i=1}^Me^{-iz_i}}<{1-q\over q}=e^{z_M}-1$$ When $z_i\nearrow z_M$ we ...
Ryszard Szwarc's user avatar
2 votes

Formally find the domain of convergence of $\sum_{p\in P}x^p$ where P is the set of Prime numbers

Can't you note that $$\Big|\sum_{p \in P} x^p\Big| \quad \le \quad \sum_{p \in P} |x|^p$$ $$\le \quad \Big(\sum_{p \in P} |x|^p + \sum_{n \in \mathbb{N} \setminus P} |x|^n\Big) \quad = \quad \sum_{n \...
Mike's user avatar
  • 21.2k
1 vote

Derivative of an Infinite Fraction

$y=\frac{x}{x^2+y}$ can be written as $y^2+x^2y-x=0$. Solving this for $y$ gives $$y=\frac{-x^2\color{red}+\sqrt{x^4+4x}}{2}$$ Since $$\begin{align}&(x^2+y)(x+y^2) \\\\&=\frac{x^2+\sqrt{x^4+4x}...
mathlove's user avatar
  • 145k
1 vote

What is the degree of the following differential equation?

The ODE is not polynomial, not even if that criterion is only applied to the highest order derivative. Thus the idea of "degree" does not apply. The main branch of the inverse or arcus ...
Lutz Lehmann's user avatar
1 vote
Accepted

Integrating "mixed" integrals such as $\int_{10}^\infty x^a\cos(t\log(x))\ \tanh(x/2) \,\mathrm dx$

As the comments pointed out we want a lower bound of $0$ for a nice solution (otherwise the integral most likely does ot have a closed formula). Furthermore then we requiere $a \in (-2,-1)$ for ...
M.E.W.'s user avatar
  • 431
1 vote
Accepted

Find the point $P$ on an ellipse such that $\overline{AP} + \overline{BP}$ is minimum for given points $A$ and $B$

Given $p = (x,y,z)'$ eliminating $t$ in the equation $$ p = v_0+v_1\cos t+v_2\sin t $$ with the given values $$ \cases{ v_0 = (2,3,7)'\\ v_1 = (1,0.5,0.5)'\\ v_2 = (-2,6,2)\\ p_A = (5,10,2)'\\ p_B=(1,-...
Cesareo's user avatar
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1 vote
Accepted

An estimate involving Poisson Kernel

We have $$ u(x,y)={4y\over \pi}\int\limits_{x_0-x}^{x_0-x+h}{1\over t^2+y^2}\,dt \\ ={4\over \pi}\int\limits_{(x_0-x)/y}^{(x_0-x+h)/y}{1\over t^2+1}\,dt $$ The interval of integration contains $0$ ...
Ryszard Szwarc's user avatar
1 vote

Solve this integral:$\int_0^\infty\frac{\arctan x}{x(x^2+1)}\mathrm dx$

Let $x=\tan y$, then $$ \begin{aligned} \int_0^{\infty} \frac{\arctan x}{x\left(x^2+1\right)} d x & =\int_0^{\frac{\pi}{2}} \frac{y}{\tan y\left(\tan ^2 y+1\right)} \cdot \sec ^2 y d y \\ & =\...
Lai's user avatar
  • 22.4k
1 vote

Solving the Riccati equation with constant coefficients $y^\prime = a y^2 + b y + c$

This is a separable ODE. $$\frac{dy}{dx}=ay^2+by+c\quad\implies\quad dx=\frac{dy}{ay^2+by+c}$$ $$x=\int \frac{dy}{ay^2+by+c}dx=\frac{2}{\sqrt{4ac-b^2}}\tan^{-1}\left(\frac{2ay+b}{\sqrt{4ac-b^2}} \...
JJacquelin's user avatar
  • 66.8k
1 vote

Solving the Riccati equation with constant coefficients $y^\prime = a y^2 + b y + c$

Set $y=pt+q$ and bring the RHS to the form $r(1\pm t^2)$. By identification $a(pt+q)^2+b(pt+q)+c=r(1\pm t^2)$ yields $q=-\dfrac b{2a}, r=c-\dfrac{b^2}{4a},p=\sqrt{\left|\dfrac ra\right|}$ and the sign ...
Yves Daoust's user avatar

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