New answers tagged

0

I am going to play with it and see if anything happens. Turns out that this can be represented in terms of two simpler integrals. I reached a point where I can't go further. I'll enter what I have done in hopes that it might be useful to someone else. First I'll make it easier to type, changing $\int_{\pi/2}^{\pi}\int_{\pi/2}^{\pi}J_{0}\left(\alpha ...


0

It can be computed by using the complex error function (aka the Faddeeva function): $$1-{\rm{erf}}(z)=e^{-z^2}w(iz)$$ Matlab and C packages for the Faddeeva function are available in the Matlab Central.


1

\begin{align*} C(x+n,y+n) &= \int_{-\infty}^\infty \frac{dt}{(1+t^2)^n(1+it)^x(1-it)^y} \\ &= \int_{-\infty}^\infty \frac{du/\sqrt{n}}{(1+u^2/n)^n(1+iu/\sqrt{n})^x(1-iu/\sqrt{n})^y} \qquad (\textrm{let }u=t/\sqrt{n}) \\ &= \int_{-\infty}^\infty \frac{1}{\sqrt{n}} e^{-u^2} + O(1/n) \qquad (\textrm{standard limit for }e) \\ &= ...


1

$$ y(t) = L^{-1} [ F(s) e^{-s} - F(s) e^{-2s}] = (t-1)u(t-1) - f(t-2)u(t-2)$$ $u(t-t_0)$ is the unit step function. It has the value $0$ for all values of $t<t_0$ and has the value $1$ for all values of $t>t_0$. Using this definition you can see that for $0<t<1$ all unit step functions vanish. Resulting in $y(t)=0$. For $1<t<2$ only ...


2

For $t<1$ we have $u(t-1)=0$ and for $t<2$ we have $u(t-2)=0$. So for $t<1$ we have $f(t-1)u(t-1)=0$ and $f(t-2)u(t-2)=0$ and then $y(t)=0$. For $t> 1$ we have $u(t-1)=1$ and for $t<2$ we have $u(t-2)=0$. So for $1<t<2$ we have $f(t-1)u(t-1)=f(t-1)$ and $f(t-2)u(t-2)=0$ and then $$y(t)=f(t-1)={1 \over 2} - e^{-(t-1)} + {1 \over 2} ...


0

They are using the Euler–Maclaurin formula which states that the approximation $$\sum_{i=n}^mf(i)\approx\int_n^mf(x)dx$$ holds (see here for further details) and combining it with the fact that $\sum_{i=N}^\infty f(i)$ is small for any convergent summation. So you can approximate the sum with it's "completed" infinite sum, and then approximate the infinite ...


2

Let us denote with $$f(n, m)=\int_0^\infty \frac{\log^n x}{(1+x^2)^m}\, {\rm d}x$$ It is quite straigtforward to note that: $$f(n, 1)=\int_{0}^{\infty} \frac{\log^n x}{1+x^2}\, {\rm d}x = \left\{\begin{matrix} 0 &, &{\rm odd} \\ 2n! \beta(n+1)&, &{\rm even} \end{matrix}\right.$$ Here $\beta$ denotes the Beta dirichlet function. On ...


2

We have: $$\begin{eqnarray*} \text{Li}_{-k}(x) = \sum_{n\geq 1}n^k x^n = \left.\frac{d^k}{dt^k}\sum_{n\geq 1}e^{nt} x^n\right|_{t=0}&=&\sum_{j=0}^{k}j!{k+1\brace j+1}\left(\frac{x}{1-x}\right)^{j+1}\\[0.2cm]&=&\frac{1}{(1-x)^{k+1}}\sum_{j=0}^{k-1}\left\langle k\atop j\right\rangle x^{k-j}\end{eqnarray*} $$ as a consequence of Worpitzky's ...


1

Since we have $$n(n-1)\cdots (n-2k+1)=\frac{n!}{(n-2k)!}$$ and $$\begin{align}&(2n-1)(2n-3)\cdots (2n-2k+1)\\&=\frac{(2n-1)!}{(2n-2k)!}\cdot\frac{1}{(2n-2k+2)(2n-2k+4)\cdots (2n-2)}\\&=\frac{(2n-1)!}{(2n-2k)!}\cdot \frac{1}{2^{k-1}(n-k+1)(n-k+2)\cdots (n-1)}\\&=\frac{(2n-1)!}{(2n-2k)!}\cdot ...


0

If $g(x) \ge 0$, $\lim_{x \to -\infty} g(x) = 0$, and $\int_{-\infty}^{\infty} g(t) dt = \infty $, then $f(x) =\int_{-\infty}^x g(t) dt $ is such a function. (added a bit later) If you want all the derivatives to be monotonic, impose that restriction on $g$.


0

I am just a master student writing a thesis in that direction, but maybe you find it helpful nonetheless. One can show that the Bohr almost-periodic functions are the closure of the trigonometric functions in the supremum norm, i.e. $$ \mathcal{A}:= \overline{\{ \sum_{1\leq j \leq n} a_i e^{i \nu_j x} : n \in \mathbb{N}, a_i\in ...


2

Let $\phi:\>t\mapsto\phi(t)$ $(-\infty<t<\infty)$ be a bump function satisfying $\phi(t)\equiv1$ for $|t|\leq\epsilon$ and $\phi(t)\equiv0$ when $|t|\geq2\epsilon$. Then $$f(x_1,\ldots, x_m):=\prod_{k=1}^m\phi(x_k)$$ satisfies your requirements. Such a $\phi$ can be obtained as follows: Start with $$\psi(t):=e^{-1/t}\qquad(t>0)$$ and $:\equiv0$ ...


1

Using contour integration one can show that \begin{align} &\int_{-\infty}^\infty \frac{\sin \pi x}{x \cdot (1-x)\ldots (n-x)}dx\\ &=\text{Im}\left\{\pi i \left(\underset{z=0}{\text{res}}-\sum_{k=1}^n \underset{z=k}{\text{res}}\right)\frac{e^{\pi i z}}{z \cdot (1-z)\ldots (n-z)}\right\}\\ &=\pi\left(\frac{1}{n!}-\sum_{k=1}^n\frac{(-1)^k}{k\cdot ...


1

So we want to study the behaviour for a large $t$ of: $$ \sum_{n\in\mathbb{Z}}'\int_{n^2/t}^{(n+y)^2/t}\frac{\cos u}{u}\,du = 2\sum_{n\in\mathbb{Z}}'\int_{n}^{n+y}\cos(v^2/t)\frac{dv}{v}$$ that by the dominated convergence theorem and the Weierstrass product for the sine function approaches: $$ ...


2

If you want $c_{\infty}, c_{-\infty}$ both in $(0,\infty)$, I suggest $$ f(x) = \frac{x^n}{a+x^ne^x} $$ with $a$ chosen large enough to avoid dividing by zero in case of odd $n$. Then for $x\to\infty$ we have $$ \frac{f(x)}{e^{-x}} = \frac{x^n}{ae^{-x}+x^n} \to 1 $$ and for $x\to\infty$ $$ \frac{f(x)}{x^n} = \frac{1}{a+x^ne^x} \to \frac1a $$ since ...


0

This is my incomplete try (my definition is probably wrong and seems not getting the answer) $$ \begin{align} &~ C'(u,v) \\ = &~\int_0^1 C_{\theta}(u,v)\theta^{\alpha}d\theta \\ = &~\int_0^{|v-u|} \min(u,v)\theta^{\alpha}d\theta + \int_{|v-u|}^{1-|u+v-1|} \max(u+v-1,0)\theta^{\alpha}d\theta + \int_{1-|u+v-1|}^1 \frac {u+v-\theta} ...


0

The function $f(x)=\begin{cases}-1 \text{ if }x<0 \\ 1 \text{ if } x\geq 0 \end{cases}$ can not be the solution of a differential equation, not even if we consider the weak derivative


1

As $x\to\infty$, the series representation of the hypergeometric function is the right approach (since then the argument of $_2F_1$ goes to $0$). We have $$f(x\to\infty)=x^{2m}\sum_{k=0}^{\infty}\frac{\left(-m\right)_k (-1)^kx^{-2k}}{k!(2k+1)}.$$ In particular, the leading asymptotic term is just $x^{2m}$. As $x\to 0$, use this formula to transform $f(x)$ ...


-2

A natural choice of constant $k$ considering continuity can be found. EDIT 1: If $u(x)/v(x)$ is a constant, then the derivative vanishes $$ \dfrac{v(x) u'(x) - v'(x) u(x) }{v(x)^2 } =0 $$ so that by Quotient Rule $$ \dfrac{u(x)}{ v(x)} = \dfrac{u'(x)}{ v'(x) }$$ which when applied to the quotient here $$ k= \frac{\ln x}{x} = \frac{1/x}{1} $$ Cross ...


2

Hint If we change variables to $y = \frac{1}{x}$ and rearrange, we produce $$y \log y = -k , $$ which has precisely the form that the original question remarks can be handled by the Lambert W function. NB that many people (myself included) would not consider the Lambert W function to be closed.


2

When z is real and tends to $\infty$ (that is, $z>0$) the second part of this double series is dominant, that is, is exponentially bigger that the first part and then, in asymptotic sense, you can avoid it to obtain $$\mathbf{M}(a,1,z)\sim \frac{e^z z^{a-1}}{\Gamma(a)}\sum_{s=0}^{+\infty}\frac{\bigl((1-a)_s\bigr)^2}{s!z^s}$$ so, the complex terms in ...


2

The function $G(x)$ has a simple expression in terms of erf function. $$ G(x) =\lim_{n\to \infty} \sum_{i=1}^n g_i(x)=\frac{1}{\sigma\sqrt{2\pi}}\lim_{n\to \infty} \frac{1}{n}\sum_{i=1}^n\exp \left(-\frac{(x-i/n)^2}{2 \sigma^2}\right)\\ =\frac{1}{\sigma\sqrt{2\pi}}\int_0^1 \exp \left(-\frac{(x-y)^2}{2 \sigma^2}\right) dy=\frac{1}{2} ...


0

The complementary error function can't be written in terms of elementary functions. You could rewrite it in terms of the integral of the Gaussian density, use integration by parts, and then a table for $\Phi$ (the aforementioned function). That or rewrite it in terms of the gamma function, or in terms of the appropriate ODE and boundary conditions, and use ...


0

Based on answer of @C. Dubussy , I want to generalize his answer, But I dont know, how to prove it. If we want to expand this equation: $$f:={(x_1^n+x_2^n+\cdots+x_k^n)}\, .$$ We can use this formula: $$f=(\sum_i^1{x_i})(x_1^{n-1}+x_2^{n-1}+\cdots+x_k^{n-1})- (\sum_{i_1\neq i_2}^2{x_{i_1}x_{i_2}})(x_1^{n-2}+x_2^{n-2}+\cdots+x_k^{n-2})$$ $$+(\sum_{i_1\neq ...


3

Both claims are wrong: For $x=y=2$ $$\frac{\Gamma(x+y)}{\Gamma(xy)-1}=\frac{6}{6-1}=\frac{6}{5}$$ For $x=y=1.5$ $$\frac{\Gamma(x+y)}{\Gamma(xy)-1}\approx 15.0372439$$


2

Because $x^4+y^4+z^4$ is a bit complicated, we try to go back to lower powers. You can see for example that $$x^4+y^4+z^4=(x+y+z)(x^3+y^3+z^3)-(xy+yz+xz)(x^2+y^2+z^2)+xyz(x+y+z).$$ Hence $$x^4+y^4+z^4=U(x^3+y^3+z^3)-V(x^2+y^2+z^2)+UW.$$ But $$x^2+y^2+z^2=(x+y+z)^2-2xy-2yz-2xz=U^2-2V$$ and $$x^3+y^3+z^3=(x+y+z)^3-3(xy+yz+xz)(x+y+z)+3xyz=U^3-3UV+3W.$$ All ...



Top 50 recent answers are included