New answers tagged

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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}} \...


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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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We have: $$G(s)=F(s-2) = \frac{1}{3s-2+\sqrt{s^2-4}} $$ so, in order to tackle this problem like the previous one, we should find the coefficients of the Taylor series at $x=0$ of: $$ H(x) = G\left(\frac{1}{x}\right) = \frac{x}{3-2x+\sqrt{1-4x^2}}=\sum_{n\geq 1}g_n x^n\tag{1} $$ to deduce (have also a look at Ramanujan's master theorem): $$ (\mathcal{L}^{-1}...


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Finally, here is the solution. Step 1: Expressing the product of two Meijer G function using an identity from http://functions.wolfram.com/07.34.16.0003.01 \begin{align} G_{2,2}^{1,2}\left(x\Bigg\vert\begin{matrix}1,1\cr 1,0\end{matrix}\right)G_{1,2}^{2,0}\left(2\alpha\sqrt{ab}x\Bigg\vert\begin{matrix}1/2\cr v,-v\end{matrix}\right) = G_{0,0:2,2:1,2}^{0,0:...


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By replacing $t$ with $e^{i\theta}$, we want to prove the identity: $$ e^{ix\sin\theta} = \sum_{n\in\mathbb{Z}} J_n(x)\,e^{in\theta} \tag{1}$$ that (by Fourier inversion) is equivalent to proving that: $$ J_n(x) = \frac{1}{2\pi}\int_{0}^{2\pi}\exp\left(ix\sin\theta-n\theta\right)\,d\theta. \tag{2}$$ The fastest way is probably to notice that both your series ...


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Here comes a big hint: Start with the expansions $$ \begin{aligned} e^{xt/2}&=1+\frac{xt}{2}+\frac{1}{2!}\Bigl(\frac{xt}{2}\Bigr)^2+\frac{1}{3!}\Bigl(\frac{xt}{2}\Bigr)^3+\cdots\quad\text{and}\\ e^{-x/2t}&=1-\frac{x}{2t}+\frac{1}{2!}\Bigl(\frac{x}{2t}\Bigr)^2 -\frac{1}{3!}\Bigl(\frac{x}{2t}\Bigr)^3+\cdots. \end{aligned} $$ Multiply them, and look ...


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At first note, that $i=\sqrt{-1}$ is the imaginary unit in (9) and we see a plain multiplication with $i$. We start with the representation (5) and use (7) to obtain \begin{align*} y&=x^{\frac{1}{2}}Z_{\frac{1}{3}}\left(\frac{2}{3}ix^{\frac{3}{2}}\right)\\ &=x^{\frac{1}{2}}\left[AJ_{\frac{1}{3}}\left(\frac{2}{3}ix^{\frac{3}{2}}\right) +B\left(\...


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We can generalize the integral by manipulating the Laplace transform of $J_{n}(bx)$, namely $$ \int_{0}^{\infty} J_{n}(bx) e^{-sx} \, dx = \frac{(\sqrt{s^{2}+b^{2}}-s)^{n}}{b^{n}\sqrt{s^{2}+b^{2}}}\ , \quad \ (n \in \mathbb{Z}_{\ge 0} \, , \text{Re}(s) >0 , \, b >0 )\tag{1}. $$ (See this question for a derivation of $(1)$ using contour integration.) ...


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It is not difficult to find the generating function for the values of the $\zeta$ function over the positive odd integers: $$ f(x)=\sum_{n\geq 1}\zeta(2n+1) x^{2n} = -\gamma-\frac{1}{2}\left[\psi(1-x)+\psi(1+x)\right] \tag{1}$$ and: $$ \sum_{n\geq 1}\frac{\zeta(2n+1)}{n(2n+1)}=2\sum_{n\geq 1}\zeta(2n+1)\left(\frac{1}{2n}-\frac{1}{2n+1}\right)=2\int_{0}^{1}...


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$\int x^x~dx=\int e^{x\ln x}~dx=\int\sum\limits_{n=0}^\infty\dfrac{x^n(\ln x)^n}{n!}dx$ For $\int x^n(\ln x)^n~dx$ , where $n$ is any non-negative integers, $\int x^n(\ln x)^n~dx$ $=\int(\ln x)^n~d\left(\dfrac{x^{n+1}}{n+1}\right)$ $=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\int\dfrac{x^{n+1}}{n+1}d((\ln x)^n)$ $=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\int\dfrac{nx^{n+1}(\...


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This is a very empirical approach of the problem since based on numerical simulations. Computing the values of $B_{1/2}(y+1,y)$ for $0\leq y\leq 5000$, what we can notice is that the logarithm varies as a linear function of $y$. For this range, I obtained $$\log\left(B_{1/2}(y+1,y)\right)\approx -1.38659 y-3.1602$$ which seems to be a very good fit (as ...


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I'll play around and see if anything interesting happens. $\begin{array}\\ B_{1/2}(y+1,y) &=\int_0^{1/2}x^y (1-x)^{y-1}dx\\ &\ge\int_0^{1/2}x^y (\frac12)^{y-1}dx\\ &=\dfrac1{2^{y-1}}\int_0^{1/2}x^ydx\\ &=\dfrac1{2^{y-1}}\dfrac{x^{y+1}}{y+1}\big|_0^{1/2}\\ &=\dfrac1{2^{2y}(y+1)}\\ \end{array} $ and $\begin{array}\\ B_{1/2}(y+1,y) &=\...


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This is straightforward steepest descent (or Laplace's method). Let $f(x):=\ln(x)+\ln(1-x),$ and $g(x)=(1-x)^{-1}$ so that the integral becomes: $$\int_0^{1/2}g(x)e^{yf(x)}dx.$$ On $[0,1/2]$, $f$ achieves its maximum at $x_0=1/2$. Furthermore, $f''(x)=-1/x^2-1/(1-x)^2$, and $f''(x_0)<0$. Then Laplace's method gives: $$\int_0^{1/2}g(x)e^{yf(x)}dx\sim \...


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While setting $t=1$ in the second-to-last equation gives a valid identity, it's not a very useful one. What you should instead do is expand the LHS of that equation in powers of $t$ just as you did already for the generating function. If you match this term-by-term to the RHS (i.e. require that the coefficient of $t^n$ be the same on both side) then you ...


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Here is another suggestion. I'm not sure if this would be "cleaner" than computing the derivative. For positive real $x$ you have: $$ \Gamma(x)=\lim_{n\to \infty}\dfrac{n!n^x}{x(x+1)\ldots(x+n)}. $$ Since the limits exist for $x$ and $1/x$ you can multiply the terms together and try to make an inductive argument that the terms are bounded by $1$ from below....


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If $$f(x)=\Gamma(x)\,\Gamma\left(\frac{1}{x}\right)$$ $$f'(x)=\frac{\Gamma \left(\frac{1}{x}\right) \Gamma (x) }{x^2}\left(x^2 \psi(x)-\psi \left(\frac{1}{x}\right)\right)$$ for which the only solution seems to be $x=1$. Working the second derivative (I shall not write it), you would find that $$f''(1)=\frac{\pi ^2}{3}-2 \gamma >0$$ confirming that we ...


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$$\sum_{n\geq 2}(-1)^n \,\text{Li}_2\!\left(\frac{2}{n}\right)=\sum_{n\geq 2}(-1)^n \sum_{m\geq 1}\frac{2^m}{n^m m^2}=\sum_{m\geq 1}\frac{2^m(\eta(m)-1)}{m^2}\\=\sum_{m\geq 1}\frac{2^m(\zeta(m)-1)-2\zeta(m)}{m^2}\\=\sum_{m\geq 2}\frac{(2^m-2)(\zeta(m)-1)}{m^2}+\frac{\pi^2}{3}-2$$ and: $$ \sum_{m\geq 2}\frac{\zeta(m)-1}{m}\,x^m = (1-\gamma)x+\log\Gamma(-x)$$ ...


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$$J_0 \left ( x \right )^2 = \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \frac{(-1)^{m+n}}{m!^2 n!^2} \frac{x^{2 (m+n)}}{2^{2 (m+n)}}$$ or, looking at coefficients of $x^{2 k}$, i.e., $m+n=k$: $$J_0 \left ( x \right )^2 = \sum_{k=0}^{\infty} \sum_{m=0}^k \frac{(-1)^k}{m!^2 (k-m)!^2} \frac{x^{2 k}}{2^{2 k}}$$ Now, $$\sum_{m=0}^k \frac{1}{m!^2 (k-m)!^2} = \...


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I believe our definitions of the error function differ by a constant, but the following approach works anyway: $$\begin{eqnarray*}\sum_{k\geq 0}\frac{x^k}{k!!}&=&\sum_{n\geq 0}\frac{x^{2n}}{2^n n!}+\sum_{n\geq 0}\frac{x^{2n+1}}{(2n+1)!!}\\&=&e^{x^2/2}+\sum_{n\geq 0}\frac{2^n x^{2n+1}}{(2n+1)!}n!\\&=&e^{x^2/2}+\int_{0}^{+\infty}e^{-z}\...


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$\int_1^\infty e^{-at}(1-t^{-1})^b~dt$ $=\int_1^\infty t^{-b}(t-1)^be^{-at}~dt$ $=\int_0^\infty(t+1)^{-b}t^be^{-a(t+1)}~d(t+1)$ $=e^{-a}\int_0^\infty t^b(t+1)^{-b}e^{-at}~dt$ $=e^{-a}\Gamma(b+1)U(b+1,2,a)$ (according to http://en.wikipedia.org/wiki/Confluent_hypergeometric_function#Integral_representations) $=\dfrac{e^{-a}}{\Gamma(b)}G_{1,2}^{2,1}\left(...


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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}} \...


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I will here show how one could go about to derive the solution of your ODE without knowing a priori what the answer should be. Your ODE $(5)$ can after multiplication by $x^2$ be written $$x^2y^{\prime\prime}(x)+x(1-2a)y^{\prime}(x)+\left[\left(bcx^{c}\right)^2+a^2-p^2c^2\right]y(x)=0$$ This looks very close to the Bessel ODE so we will try to see if we ...


1

$\log \Gamma$ is a convex function, hence $\psi$ is an increasing function and $$\begin{eqnarray*}&&\int_{t}^{t+\frac{1}{2}}\psi(x)\,dx -\frac{\psi\left(t+\frac{1}{4}\right)+\psi\left(t+\frac{1}{2}\right)}{4}\\&=&\int_{t}^{t+\frac{1}{2}}\left(\psi(x)-\psi(t)\right)\,dx -\frac{\left(\psi\left(t+\frac{1}{4}\right)-\psi(t)\right)+\left(\psi\...


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We have that $$\log\left(\Gamma\left(x\right)\right)\sim x\log\left(x\right)-x-\frac{1}{2}\log\left(\frac{x}{2\pi}\right)+O\left(\frac{1}{x}\right) $$ and $$\psi\left(x\right)\sim\log\left(x\right)+O\left(\frac{1}{x}\right) $$ as $x\rightarrow\infty$ (see here and here) hence we have to evaluate $$\begin{align} &\frac{a+2}{4}\log\left(\frac{a+2}{4}\...


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I spent some time working on analogs of integration by parts that can be created using the quotient rule instead of the product rule, and versions based on more general versions of the product rule, but found that they were functionally no more or less useful than the well-known version of integration by parts. The straightforward examples you gave, (first ...


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[2016-06-07] Note: General solution added to provide a comparison with the example part. Here we show that according to OPs example \begin{align*} y(x)=x^{\frac{1}{2}}Z_{\frac{1}{3}}(2x^{\frac{3}{2}}) \end{align*} is a solution of the differential equation \begin{align*} y^{\prime\prime}+9xy=0\tag{1} \end{align*} In the following we use ...


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Another approach. Through the orthogonality of Legendre polynomials, if $$ f(x) = \sum_{k=0}^{3}c_k\cdot P_k(x)\tag{1} $$ we have: $$\forall k\in\{0,1,2,3\},\qquad c_k = \frac{2k+1}{2}\int_{-1}^{1}f(x)\,P_k(x)\,dx. \tag{2}$$


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These relations are easily proved if we use the nome $q = e^{2\pi i\tau}$ and use Ramanujan notation of $P(q), Q(q), R(q)$ also. We have \begin{align} \eta(\tau) &= \eta(q) = q^{1/24}\prod_{n = 1}^{\infty}(1 - q^{n})\tag{1}\\ E_{2}(\tau) &= P(q) = 1 - 24\sum_{n = 1}^{\infty}\frac{nq^{n}}{1 - q^{n}}\tag{2}\\ E_{4}(\tau) &= Q(q) = 1 + 240\sum_{n = ...


1

In matrix form, $$\begin{bmatrix} p_0\\ p_1\\ p_2\\ p_3\end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ -\frac{1}{2} & 0 & \frac{3}{2} & 0\\ 0 & -\frac{3}{2} & 0 & \frac{5}{2}\end{bmatrix} \begin{bmatrix} 1\\ x\\ x^2\\ x^3\end{bmatrix}$$ We want to find a weight vector $\mathrm{w}$ such that $$...


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Continuing from your steps, we compare the coefficients of each term on the L.H.S and R.H.S. $$2x^3-x^2-3x+2=\dfrac A2(5x^3−3x) + \dfrac B2(3x^2−1) + Cx +D$$ So, $$ 2x^3-x^2-3x+2=\dfrac{5}{2}Ax^3+\dfrac{3}{2}Bx^2+(C-\dfrac{3}{2}A)x+(D-\dfrac{B}{2})$$ For the $x^3$ term, we have $\dfrac{5}{2}A=2 \Rightarrow A=\dfrac{4}{5}$ For the $x^2$ term, we have $\...


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First get the $x^3$ term right. That will set A. That is $4/5$. Then do the quadratic to set B, and so on. Edit: the coefficient of $x^3$ on both sides are 2 and $5/2*A $. Set these equal to solve for A.


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Your function has roots for $t=(2n+\tfrac{1}{2})i, \; n \in \mathbb{Z}>0$. These are the well-known trivial zeroes of the $\zeta$ function, because $\frac{1}{2}+(2n+\tfrac{1}{2})i\times i= -2n.\;$ The trivial zeroes are a consequense of the $\sin$ term in the reflection formula. PS: I guess you are interested in the zeroes on the critical line $\tfrac{1}{...


1

$$\begin{eqnarray*} \int_{x}^{+\infty}\frac{\Gamma(a,t)}{t}\,dt = \int_{1}^{+\infty}\frac{\Gamma(a,sx)}{s}\,ds&=&\int_{1}^{+\infty}\int_{sx}^{+\infty}\frac{1}{s}u^{a-1}e^{-u}\,du\,ds\\&=&x^a\iint_{(1,+\infty)^2}(sv)^{a-1}e^{-svx}\,dv\,ds\\&=&x^a\int_{1}^{+\infty}\int_{1}^{p}\frac{1}{s} p^{a-1}e^{-px}\,ds\,dp\\&=&x^a\int_{1}^{+\...


2

Let $q = e^{-\pi x}$ and then $$\eta(ix) = q^{1/12}\prod_{n = 1}^{\infty}(1 - q^{2n}) = 2^{-1/3}\sqrt{\frac{2K}{\pi}}(kk')^{1/6}$$ and $x = K'/K$. Then $$\eta^{4}(ix) = 2^{-4/3}\left(\frac{2K}{\pi}\right)^{2}(kk')^{2/3}$$ Now we can see that $x = 0$ then $k = 1$ and $x \to \infty$ implies $k \to 0$ and $$\frac{dx}{dk} = -\frac{\pi}{2kk'^{2}K^{2}}$$ and hence ...


1

About the first request, it is the Legendre duplication formula (you can find a proof here) and we get $$\frac{1}{\Gamma\left(n+1/2\right)}=\frac{n!4^{n}}{\left(2n\right)!\Gamma\left(1/2\right)}\tag{1} .$$ About the second request, we have, using $(1)$ $$J_{-1/2}\left(x\right)=\left(\frac{x}{2}\right)^{-1/2}\sum_{n\geq0}\frac{\left(-1\right)^{n}}{n!\Gamma\...


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This answer is too long for a comment and refers to BLAZE's comment above as to how to compute $$\frac{d}{dx}\frac{dY}{dX}$$ Setting the Scene $y(x)=x^{\alpha}Y(X),\quad X=\beta x^{\gamma}$ Now it is obvious that \begin{align} \frac{dy}{dx} &= \alpha x^{\alpha - 1}Y+x^{\alpha}\frac{dY}{dx} \\ &= \alpha x^{\alpha-1}Y+x^{\alpha}\frac{dY}{...


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Here is a possible argument but I would like to see a better one. If $\nu=n\in\mathbb{Z}$ we could show that $J_{\nu}(x) = (-1)^n J_{-\nu}(x)$. We know that $J_{\nu}(x)$ and $J_{-\nu}(x)$ are solutions, of the Bessel equation so any linear combination is a solution as well. We consider the difference: \begin{eqnarray*} J_{-\nu}(x) - (-1)^{\nu} J_{\nu}(...



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