Hot answers tagged

3

One may replace $t$ by $it$, getting $$ e^{2itx+t^2}=\sum_{n=0}^\infty (-1)^ni^n\frac{t^n}{n!}H_n(x) $$ then one may take the real part: $$ e^{t^2}\cos2xt=\sum_{n=0}^\infty \frac{t^{2n}}{(2n)!}H_{2n}(x) $$ where we have assumed that $t,x$ are real numbers.


2

A primitive of $\tan(2x)$ is clearly $-\frac{1}{2}\ln(\cos(2x))$. This primitive is undefined for $x=-\frac{\pi}{4}$. Thus the integral is not defined.


2

It is enough to compute asymptotics for $$ S_n = \sum_{k=1}^{n}\sqrt{k} = \frac{2}{3}\,n^{3/2}+\frac{1}{2} n^{1/2}\color{red}{-\frac{\zeta\left(\frac{3}{2}\right)}{4\pi}}+O\left(\frac{1}{\sqrt{n}}\right)\tag{1}$$ and $$ T_n = \sum_{k=0}^{n}\sqrt{k+\frac{1}{2}} = \frac{1}{\sqrt{2}}\left(S_{2n+1}-\sqrt{2}\,S_n\right)\tag{2} $$ by summation by parts to find the ...


2

We may exploit Frullani's theorem to get an integral representation of our series. $$\begin{eqnarray*}S=\sum_{n\geq 1}\frac{\log(n+1)-\log(n)}{n}&=&\int_{0}^{+\infty}\sum_{n\geq 1}\frac{e^{-nx}-e^{-(n+1)x}}{nx}\,dx\\ &=&\int_{0}^{+\infty}\frac{1-e^{-x}}{x}\left(-\log(1-e^{-x})\right)\,dx\\&=&\int_{0}^{1}\frac{x\log x}{(1-x)\log(1-x)}\...


1

$\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{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \...


1

Following this: http://eqworld.ipmnet.ru/en/solutions/ode/ode0310.pdf Let $y = f(x), w = \frac{x}{y}y', z=\frac{x^3}{y^3}$, then $$w'_x = \frac{1}{y}y'+\frac{x}{y}y''-\frac{x}{y^2}(y')^2,$$ $$w'_x = w_z'\frac{dz}{dx}=w'_z (3\frac{x^2}{y^3}-3\frac{x^3}{y^4}y'_x),$$ $$xw_x'=3w_z'(z-zw),$$ $$xw_x'=\frac{x}{y}y'+\frac{x^2}{y}y''-\frac{x^2}{y^2}(y')^2=w+z-w^2,$$...


1

The fundamental fact is that, intuitively, $sin(X) \approx X$ when $X$ is small. Here you replace $X$ with $\frac{sin(x) + tan(x)}{2}$ and $\frac{sin(x) - tan(x)}{2}$ and that's how you get rid of the outter $sin$. Formally, you can say that $sin(x) \sim_{0} x$ meaning that $sin(x) = \epsilon(x) x$ or equivalently $x = \epsilon'(x)sin(x)$ where $\epsilon(x)$...


1

For small angles($ \theta$ $\rightarrow$ $ 0$), we can say $sin \theta=\theta$..(i) For $ \theta$, $\rightarrow$ $ 0$, $sin\theta \rightarrow0$ $tan\theta \rightarrow0$ $sin\theta + tan\theta \rightarrow0$ Using ..(i), $sin\frac{(sin\theta + tan\theta)}{2}=\frac{(sin\theta + tan\theta)}{2}$


1

With the notation $\log = \ln$, the following inequality is tied to your question: $$ n\log n - n < \log(n!) < n\log n$$. The right inequality is clear as $n! < n^n$, and the left uses Riemann sum, namely: $$\log(n!) = \log 1 + \log 2 + \cdots + \log n > \displaystyle \int_{1}^n \log xdx= n\log n - n$$. Thus put them together, the answer is $...


1

Through the formal substitutions $x=\arctan u$ and $u=v^{1/4}$ we get: $$I=\int_{0}^{\pi/4}\tan(2x)\,dx = \int_{0}^{\pi/4}\frac{2\tan(x)}{1-\tan^2(x)}\,dx = \int_{0}^{1}2u(1-u^4)^{-1}\,du \tag{1}$$ from which: $$ I = \frac{1}{2}\int_{0}^{1}v^{-3/4}(1-u)^{-1}\,du = \frac{1}{2}\,B\left(\frac{1}{4},0\right)=\frac{\Gamma\left(\frac{1}{4}\right)\Gamma(0)}{2\,\...


1

Put $$S_n=\sum_{k=1}^n (\frac{3}{2\sqrt{2}}\sqrt{2k}-\frac{1}{\sqrt{2}}\sqrt{2k+1}-\frac{1}{\sqrt{2}}\sqrt{2k-2})$$ and $\displaystyle T_n=\sum_{k=1}^n\sqrt{j}$. Using that $\displaystyle T_{2n+2}=\sqrt{2}T_n+\sqrt{2}\sqrt{n+1}+\sum_{j=1}^n \sqrt{2j+1}$, you get a formula for $S_n$ using $T_n$ and $T_{2n+2}$. Now using this answer Euler-Maclaurin Summation ...


1

Just an addendum, maybe some of this will be useful to you: $$\int_0^1 \frac{(1-x) \log (1-x)}{x \log x} \, {\rm d}x=$$ $$=1.2577468869\dots= \gamma_{1}(1,0) - \gamma=\int_0^1 \ln (1-x) \ln \left(\ln \left(\frac{1}{x}\right)\right) \, dx-\gamma=$$ $$=\sum_{k=1}^{\infty} \frac{\ln (k+1)}{k(k+1)}=\sum_{k=1}^{\infty} \frac{\ln (1+\frac{1}{k})}{k}=\sum_{k = 2}...


1

$\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{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \...



Only top voted, non community-wiki answers of a minimum length are eligible