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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}...

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$\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}} \... 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} $$around s = 0. One may prove that (see Theorem ... 2 \newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \... 1 It is a bit an overkill, but since for any a,b>0 we have:$$ \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 NoticeS_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)^...

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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{(...

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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-functionQ_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)\,{\...

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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\...

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

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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 $$xsi(x)+\cos x = -\cos x (1- \frac{2!}{x^2} +\frac{... 1 The result you're thinking of does not work for \ln x since it takes nonreal values at x<0. However, note that \mathfrak{R}(\ln x) is even on \mathbb{R}\setminus\{0\} (where \mathfrak{R}z is the real part of z) 0 \newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \... 1 Hint. We have$$ \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{...

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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 ...

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$S(x)$ 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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