New answers tagged

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\,\...


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.


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

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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 ...


0

This belongs to a special case of Emden-Fowler equation. And luckily we can find its general solution in http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=333.


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

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 ...


0

The true Fabius function is no-where analytic. This implies a lack of an analytical function describing it. The linked page a has a discussion concerning a function that approaches the Fabius function when taken to infinity. (though it gets reasonably close at around n=20): Recursive Integration over Piecewise Polynomials: Closed form?


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)}\...


2

The given series admits a closed-form in terms of the poly-Stieltjes constants. The poly-Stieltjes constants arise in the context of finding the Laurent series expansion of the poly-Hurwitz zeta function $$ \begin{align} \zeta(s\mid a,b)= \sum_{n=1}^{+\infty} \frac{1}{(n+a)^{s}(n+b)}, \tag1 \end{align} $$ around $s = 0$. One may prove that (see Theorem ...


2

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1

It is a bit an overkill, but since for any $a,b>0$ we have: $$ \int_{0}^{+\infty}x\sin(ax)e^{-bx}\,dx = \frac{2ab}{(a^2+b^2)^2} \tag{1}$$ it happens that: $$ \sum_{k\geq 1}\frac{2k}{(k^2+c^2)^2}=\int_{0}^{+\infty}\sum_{k=1}^{+\infty}\frac{x\sin(cx)}{c}e^{-kx}\,dx = \frac{1}{c}\int_{0}^{+\infty}\frac{x\sin(cx)}{e^x-1}\,dx\tag{2} $$ and if $|c|\leq 1$ we ...


3

You may observe that $$\frac{2k}{\left(k^{2}+c^{2}\right)^{2}+\frac{1}{4}+c^{2}}<\frac{2k}{\left(k^{2}+c^{2}\right)^{2}}\tag{1} $$ and $$\frac{2k}{\left(k^{2}+c^{2}\right)^{2}+\frac{1}{4}+c^{2}}=\frac{1}{\left(k-\frac{1}{2}\right)^{2}+\frac{1}{4}+c^{2}}-\frac{1}{\left(k+\frac{1}{2}\right)^{2}+\frac{1}{4}+c^{2}} $$ hence, if we take the sum in $(1)$ we ...


1

There's Fourier expansion: $$\operatorname{am} (u,k)= \frac{\pi u}{2K}+\sum_{n=1}^{\infty} \frac{\sin \frac{n\pi u}{K}}{n\cosh \frac{n\pi K'}{K}} $$ where $K(k)=F \left( \frac{\pi}{2},k \right)$ and $K'=K(\sqrt{1-k^2})$. By the ways, its better to use symbolic software to ease your work. For example, in Mathematica JacobiAmplitude[u,m] for $\...


1

Complete the square with respect to $k$ to get\begin{align}\sum_{k,n\in\mathbb{Z}}q^{3\big(k+\tfrac{1-n}2\big)^2+\tfrac14+\tfrac{9n^2}4-\tfrac{3n}2}&=\sum_{k,n\in\mathbb{Z}}q^{3\big(k+\tfrac{1-n}2\big)^2+\tfrac14(3n-1)^2}\\ &=\sum_{n\in\mathbb{Z}}q^{\tfrac14(3n-1)^2}\sum_{k\in\mathbb{Z}}q^{3\big(k+\tfrac{1-n}2\big)^2} \end{align}then split the sum by ...


1

Notice $$S_n^{(j)} = \int_0^1 \left(1+\frac{x}{n}\right)^n x^{j-1} dx = \int_0^1 e^{n\log\left(1+\frac{x}{n}\right)} x^{j-1} dx = \int_0^1 e^{x - \frac{x^2}{2n} + \frac{x^3}{3n^2} + O(n^{-3})} x^{j-1}dx\\ = \int_0^1 e^x \left[1 - \frac{x^2}{2n} + \frac{8x^3+3x^4}{24n^2} + O(n^{-3})\right]x^{j-1} dx $$ Compare with expansion in question, we get $$(-1)^j(\...


0

I'm assuming we're working over the real numbers. Recall that a quadratic space is a vector space equipped with a quadratic form, typically written (q, V), and that the orthogonal sum of two quadratic spaces $(q_1, V_1)\perp (q_2, v_2)$ is the direct sum of the $V_1, V_2$ as vector spaces but equipped with the new quadratic form $q = q_1+q_2$. By ...


0

I solved it in the following manner: First, I prove that $n/p\to 0$. This comes from: $$n/p = \frac{n\log(\frac n{n-2})}{\log(n)} \to \frac 2 \infty = 0$$ Then the denominator is dealt with, as $\Gamma$ is continuous. As for the numerator, the fact that $n/p\to 0$ along with the fact that $\Gamma$ is analytic around $1$ gives me $\left(\Gamma(1 + 1/p)\right)^...


1

Here are some aspects to master such expressions: Let's denote the last expression with $G(x)$. \begin{align*} G(x)&=-A\frac14\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\frac43)}x^{3n-1} -\frac{B}{\sqrt{3}}\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\frac43)}x^{3n-1}\\ &\quad+\left[A+\frac{B}{\sqrt{3}}\right]\sum_{n=0}^\infty\frac{(...


3

The Bessel Function of the first kind and order $p$ has series representation given by $$J_p (x) = \sum_{n=0}^\infty \frac{(-1)^n}{n! \,\Gamma(n+p+1) } \left( \frac x 2 \right)^{2n+p} $$ For $p=1/2$, we find $$\begin{align} J_{1/2} (x) &= \sum_{n=0}^\infty \frac{(-1)^n}{n! \,\Gamma(n+3/2) } \left( \frac x 2 \right)^{2n+1/2} \\\\ &=\sqrt{\frac {...


0

As gammatester told the initial integral is related to Marcum Q-function $$Q_M(\alpha,\beta)=\frac{1}{\alpha^{M-1}}\int_\beta^\infty x^Me^{-\frac{x^2+ \alpha^2}{2}}\mathrm{I}_{M-1}(\alpha x)dx$$ in the following manner: $$\begin{eqnarray} \int_{0}^{b}xe^{-\,{x^{2} + z^{2} \over 2\sigma^2}} {\rm I}_{0}\left(\vphantom{\large A}xz \over \sigma^{2}\right)\,{\...


1

For computing the principal values of such integrals, you may exploit the identities: $$ \mathcal{L}(\text{ci}(x)) = -\frac{\log(1+s^2)}{s^2}\,\qquad \mathcal{L}(\text{si}(x))=\frac{1}{s}\arctan\left(\frac{1}{s}\right)\tag{1} $$ given by the Laplace transform and differentiation under the integral sign. For instance, since $$ \mathcal{L}^{-1}\left(\frac{1}{q+...


4

New: In fact more or less the same argument shows that given $E\subset\Bbb R$ there exists $f:\Bbb R\to\Bbb R$ such that $E$ is the set where $f$ fails to have a limit if and only if $E$ is an $F_\sigma$ (a countable union of closed sets). First the answer to the OP, where $E=\Bbb Q$: Say $g(x)=\sin(1/x)$ for $x\ne0$, $g(0)=0$. Let $r_1,\dots$ be an ...


0

I think I've found a suitable [see EDIT below] way to generalize the fibonorial in the reals! (I've taken inspiration from an answer to this question.) The first terms of the fibonorial sequence $\mathfrak{F}(n)$ are: $$\mathfrak{F}_7(n)=\color{red}{1},1,2,6,30,240,3120,65620$$ where $\mathfrak{F}_7(n)$ is just $\mathfrak{F}(n)$ interrupted after the ...


3

If we set, using standard notations, $$ f(x) = -\frac{\pi x}{2}+\cos(x)+x\,\text{Si}(x) = x\,\text{si}(x)+\cos(x) \tag{1}$$ we have: $$ f'(x) = -\frac{\pi}{2}+\text{Si}(x) = -\int_{x}^{+\infty}\frac{\sin t}{t}\,dt \tag{2}$$ and since $f(0)=1$, $$ f(x) = 1-\int_{0}^{x}\int_{u}^{+\infty}\frac{\sin t}{t}\,dt\,du =1-\int_{1}^{+\infty}\frac{1-\cos(t x)}{t^2}\,dt\...


6

Hint: Use integration by parts to show that $$-si(x)=\frac{\cos(x)}{x}-\int_x^{+\infty}\frac{\cos(t)}{t^2}dt=\frac{\cos(x)}{x}+\frac{\sin(x)}{x^2}-2\int_x^{+\infty}\frac{\sin(t)}{t^3}dt$$


3

Define $$Si(x) = \int^x_0 \frac{sin t}{t} dt$$ By the relation $Si(x) = \frac{\pi}{2} + si(x) $ And the asymptotic series expansion of $si(x)$: $$si(x) = -\frac{\cos x}{x} (1- \frac{2!}{x^2} +\frac{4!}{x^4} -O(1/x^6))-\frac{\sin x}{x} (\frac{1}{x}- \frac{3!}{x^3} +\frac{5!}{x^5} -O(1/x^7)) $$ Therefore $$xsi(x)+\cos x = -\cos x (1- \frac{2!}{x^2} +\frac{...


1

The result you're thinking of does not work for $\ln x$ since it takes nonreal values at $x<0$. However, note that $\mathfrak{R}(\ln x)$ is even on $\mathbb{R}\setminus\{0\}$ (where $\mathfrak{R}z$ is the real part of $z$)


0

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1

Hint. We have $$ \frac{\partial}{\partial x} Γ(s,x)=-x^{s-1}e^{-x} $$ then, by the chain rule, we get $$ \begin{align} \frac{d}{d x}\left(Γ(1+d,A-c \ln x)\right)&=\frac{-c}{x}\cdot\left.\frac{\partial}{\partial t} Γ(s,t)\right|_{(s,t) =(1+d,A-c \ln x)} \\\\&=\frac{c}{x}\cdot(A-c \ln x)^de^{-(A-c \ln x)} \\\\&=c\:x^{c-1}e^{-A}(A-c \ln x)^d. \end{...


3

Extending what Zach466920 said: $$\partial_t[k(t,n)]=F(n+1)\cdot k(t,n+1)\tag1$$ We can develop a power series solution for this. Let $$k(t,n)=\sum_{m=0}^\infty \kappa(n, m)\ t^m$$ We equation $(1)$ as: $$\sum_{m=1}^\infty m\ \kappa(n, m)\ t^{m-1} =\sum_{m=0}^\infty(m+1)\ \kappa (n, m) \ t^m =F(n+1)\cdot\sum_{m=0}^\infty \kappa(n+1, m)\ t^m$$ Which gives ...


1

$S(x)$ is an entire function, it does not have any logarithmic singularity: $$ S(x)=\sum_{n\geq 1}\frac{x^n}{n\cdot n!}=\int_{0}^{x}\frac{e^t-1}{t}\,dt \tag{1}$$ also since $\frac{e^t-1}{t}$ is an entire function. The RHS of $(1)$ clearly depends on the exponential integral and clearly is not an elementary function.



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