# Tag Info

0

Maybe differentiate with respect to $a,b$ ? $$\frac{\partial}{\partial a} \frac{\partial}{\partial b}f(a,b,c) = \int_0^1 \frac{u^{c+2}}{(1- au)(1-bu)} \; du$$ Then partial fractions might work, you can plug in $u = 1/a, 1/b$ to get the coefficients $A,B$. $$\frac{1}{(1- au)(1-bu)} = \frac{A}{1-au} + \frac{B}{1-bu}$$ The result is an integral like, ...

2

First, we have \begin{align} \psi'(n) &=\sum_{k=0}^\infty\frac1{(k+n)^2}\\ &=\sum_{k=n}^\infty\frac1{k^2}\tag{1} \end{align} Then \begin{align} \sum_{n=1}^\infty\psi'(n)^2 &=\sum_{n=1}^\infty\sum_{j=n}^\infty\frac1{j^2}\sum_{k=n}^\infty\frac1{k^2}\tag{2}\\ ... 1 If you're looking for good numerical approximations, then for example\eqalign{- 0.000038839155&+ ( 2.472698342+ ( - 0.1185303776+ (\cr & 0.599575302+ ( - 4.237027543+ ( 9.643963778+ (\cr - & 12.22813840+ ( 8.085033824- 2.189508102\,x ) x ) x ) x ) x ) x ) x ) x} $$is an optimal degree 8 polynomial approximation on [0,1], with maximum ... 2 We have that the function:$$ B(\lambda)=\sum_{n=0}^{+\infty}\left(\frac{1}{(2n-1)4^n}\binom{2n}{n}\right)^2\lambda^{2n} = \frac{1}{\pi}\int_{0}^{\pi}\sqrt{1+\lambda^2+2\lambda\cos(2\theta)}\,d\theta $$satisfies the relation:$$B(\lambda) = 2\frac{1+\lambda}{\pi}\cdot E\left(\frac{4\lambda}{(1+\lambda)^2}\right)\tag{1}$$and the differential equation:$$ B ...

3

Notice $$k\frac{dE(k) }{dk} = k\int_0^{\pi/2} \frac{\partial}{\partial k}\sqrt{1-k^2\sin^2\theta} d\theta = \int_0^{\pi/2} \frac{-2k^2\sin^2\theta}{2\sqrt{1-k^2\sin^2\theta}} d\theta = E(k) - K(k)$$ This leads to $$\frac{K(k)}{E(k)} = 1 - k\frac{d}{dk}\log E(k) = 1 - x\frac{d}{dx}\log E(k)$$ and hence $$\int\frac{dx}{x}\frac{K(k)}{E(k)} = ... 0 I think that it is numerical error in the Mathematica implementation. I am using v7.0 and here is the output of command" Table[N[Re[MathieuC[MathieuCharacteristicA[2 - 10^(-n), -1], -1, 0]], 20], {n, 1, 50}]". When n\ge 46, the calculation breaks down. 0.53972224712427121309, 0.092750258079945471042, 0.0093744966834648209648, ... 1 First of all, this is an ODE, not a PDE, because you're differentiating with respect to just one independent variable. Second, I don't get g_1 as a solution. This is the most likely reason why the v(x) term doesn't drop out of the reduction of order. Reduction of order should always find a second solution if your first solution is correct and nonzero. ... 0 I know its an old post, but others stumbling upon this post might find it helpful. You can try the accurate approximate analytical expression for faster numerical evaluation such as this or answers in this post especially by Ron Gordon. 0 If C and c are constant:$$cf'(cx)=Cf'(x)f'(cx)=\frac Ccf'(x)=\frac{C^2}{c^2}f'(\frac xc)=\cdots=\frac{C^n}{c^n}|_{n\to\infty}f'(0)$$If c>1:$$f(cx)=\frac{C^n}{c^n}|_{n\to\infty}f'(0)x+Cf(x)=\left[\lim_{n\to\infty}\left(\frac Cc\right)^n\right]f'(0)\frac xc+\mathcal{Constant}$$If C>c: If f'(0)=0 , f(x)=\mathcal{Constant} If ... 3 Assuming that I properly understood the question, you can perform a Taylor expansion of the Lambert function at x=a. What is beautiful with this function is that all its derivatives express just involve the Lambert function itself. As a result,$$W(x)=W(a)+\frac{W(a) }{a (W(a)+1)}(x-a)-\frac{\left(W(a)^3+2 W(a)^2\right) }{2 a^2 ...

2

Notice that $x-x^2/2\leq \ln(1+x) \leq x$ for $x\geq 0$t hen : $$\frac{xk }{n^2} \geq \ln\left(1+\frac{ xk}{n^2}\right) \geq \frac{x k}{n^2} -\frac{x^2 k^2}{n^4}$$ Sum form $k=1$ to $n$ : $$\frac{(n+1)x}{2n} \geq \sum_{k=1}^n \ln\left(1+\frac{ xk}{n^2}\right) \geq \frac{(n+1)x}{2n} - \frac{(n+1)(2n+1)x}{6n^3}$$ Then the limit of the middle sum is $x/2$ ...

0

A possible candidate is $$y=-\sin\left[x-\left(\frac\pi 2-\alpha\right) y\right]$$ This may not quite meet all the specifications set out in your original question but it does have the following characteristics: has values of $0$ at $n\pi$ minimum points have been displaced from $\;(2n+\frac 12)\pi\;$ to $\; 2n\pi+\alpha\;$ maximum points have been ...

4

Here I use the Daniel's method: Take the logarithm of $\prod _{k=1}^n (1+\dfrac{kx}{n^2} ),$ then we have $$\ln\left(\prod_{k=1}^n\ \left(1+\dfrac{kx}{n^2 }\right) \right)=\sum_{k=1}^n\ln\left(1+\dfrac{kx}{n^2 }\right).$$ By the Taylor series expansion of logarithms, $$\ln\left(1+\dfrac{kx}{n^2} \right)=\sum_{m=1}^\infty\dfrac{(-1)^{m-1}}{m}\left(\dfrac{ ... 6 We have:$$ I = \frac{1}{4}\int_{0}^{1}z^{-2/3}(1-z)^{-2/3}\log z \,dz$$hence:$$ I =\frac{1}{4}\left. \frac{d}{d\alpha}\left(\int_{0}^{1}z^{-2/3+\alpha}(1-z)^{-2/3}\,dz\right) \right|_{\alpha=0}=\frac{\Gamma(1/3)}{4}\left.\frac{d}{d\alpha}\left(\frac{\Gamma(1/3+\alpha)}{\Gamma(2/3+\alpha)}\right)\right|_{\alpha=0}$$and by using the identity \Gamma' = ... 1 I have not been able to get rid of hypergeometric functions and I apologize for that. What a CAS found is that$$\int\frac{\ln{x}\,\mathrm{d}x}{\sqrt[3]{x(1-x^2)^2}}=\frac{\left(x \left(x^2-1\right)^2\right)^{2/3}}{\left(1-x^2\right)^{4/3}}A(x)$$where$$A(x)=\frac{3}{2} \, _2F_1\left(\frac{1}{3},\frac{2}{3};\frac{4}{3};x^2\right) \log (x)-\frac{9}{4} \, ...

0

I have figured out the answer. All you need to do is draw an ellipse that is tangent to the lines $y = \pm 1$ at the points of intersection between those lines and a line of slope $\alpha$. The ellipse also needs to pass through $(1,0)$, $(-1,0)$ and $(0,\sin(\alpha))$. In doing this I figured out the equation needs to be: $$f(\theta) = ... 0 This should do it with the right values of a and b:$$ x\mapsto a\sin x + b \sin (2x) In order to get extreme values at the prescribed points, we need the derivative to be 0 at those points. So we need \begin{align} -1 & = a\sin\alpha + b \sin(2\alpha) \\ 0 & = a\cos\alpha+2b\cos(2\alpha) \end{align} In matrix form \begin{bmatrix} ...

0

Legendre polynomials are solutions of the Legendre differential equation $$[(1-x^2)y']' + n(n+1)y = 0$$ Applying the series method by substituting $y=\sum_{j=0}^n a_jx^j$, one finds a recurrence relation between the coefficients $a_j$ that lead to the elementary solutions $P_n(x)$ that are polynomials of degree $n$ which contain only odd-numbered or only ...

0

$\int_0^t\dfrac{e^{-a^2z}}{\sqrt{z}(z+v)}dz$ $=\int_0^1\dfrac{e^{-a^2tz}}{\sqrt{tz}(tz+v)}d(tz)$ $=\dfrac{\sqrt{t}}{v}\int_0^1\dfrac{e^{-a^2tz}}{\sqrt{z}\left(1+\dfrac{tz}{v}\right)}dz$ $=\dfrac{2\sqrt{t}}{v}\Phi_1\left(\dfrac{1}{2},1,\dfrac{3}{2};-\dfrac{t}{v},-a^2t\right)$ (according to About the confluent versions of Appell Hypergeometric Function and ...

1

For just one example in probability: consider a football match (soccer for North Americans). A fairly reasonable model is that each team, independently, scores goals according to a Poisson process. The expression for the probability that the score is tied after 90 minutes of play involves the modified Bessel function $I_0$.

15

Your integral has an elementary closed form, that was correctly stated by Cleo in her answer without proof: $$S=\int_0^1\left({_2F_1}\!\left(-\tfrac14,\tfrac54;\,1;\,\tfrac{x}2\right)\right)^2dx=\frac{8\sqrt2+4\ln\left(\sqrt2-1\right)}{3\pi}.\tag1$$ Proof: Using DLMF 14.3.6 we can express the hypergeometric function in the integrand as the Legendre ...

3

First, consider the $n=3$ case to have a point of reference. Since we integrate over all directions on the sphere, we may take $\vec{y}$ to define the vertical axis i.e. $\vec{x}\cdot\vec{y}=y \cos\theta$ where $\theta$ is the azimuthal angle. Then the integral in spherical coordinates (normalized by the surface area) is ...

8

Step 1. If $$I_1(n)=\sum_{1\leq k\leq\sqrt{n}}\left(\Gamma\left(\frac{k}{n}\right)\right)^{-k}$$ Then $\lim\limits_{n\to\infty}I_1(n)=0$. Proof. Indeed, since $\Gamma$ is decreasing on $(0,1]$ we have $$I_1(n)\leq\sum_{1\leq ... 0 The first and third identity are easy to prove. Recall that the modular discriminant \Delta = \Delta(\omega_1, \omega_2) is defined by$$\Delta = g_2^3 - 27g_3^2,$$where g_2 = g_2(\omega_1, \omega_2) = 60G_4(\omega_1, \omega_2) and g_3 = g_3(\omega_1, \omega_2) = 140G_6(\omega_1, \omega_2) are the Weierstrassian invariants. Moreover, we define the ... 1 After two small corrections in your equation, the simplification becomes easy : 0 Here are several equivalent definitions:$$\begin{align} f_a(x) &= \tanh(a\tanh^{-1}x) \\ &= \frac{(1+x)^a-(1-x)^a}{(1-x)^a+(1+x)^a} \\ &= \frac{\binom a1x+\binom a3x^3+\binom a5x^5+\cdots}{\binom a0+\binom a2x^2+\binom a4x^4+\cdots}. \end{align}$$Near \pm1, f behaves like a shifted, scaled copy of x^a. Also, f'(0)=a. 7$$\begin{eqnarray*}\psi(1+a)-\psi(1+b)&=&\sum_{n=1}^{\infty}\left(\frac{a}{n(n+a)}-\frac{b}{n(n+b)}\right)=\sum_{n=1}^{\infty}\frac{a(n+b)-b(n+a)}{n(n+a)(n+b)}\\&=&(a-b)\sum_{n=1}^{\infty}\frac{1}{(n+a)(n+b)}.\end{eqnarray*}4 I will use only the standard reflection and duplication identities: \begin{align} &\psi_1(z)+\psi_1(1-z)=\frac{\pi^2}{\sin^2\pi z},\\ &\psi_1(z)+\psi_1\bigl(z+\text{\frac12}\bigr)=4\psi_1(2z). \end{align} Use the first of them to replace \psi_1\bigl(\frac1{10}\bigr) by  ... 4 Consider the multiplication formula for the polygamma function \psi_n(mz)=\frac{1}{m^{n+1}}\sum_{k=0}^{m-1}\psi_n\left(z+\frac km\right)\quad;\quad\text{for}\ n\ge1\tag1 $$and its reflection formula$$ \psi_n(1-z)+(-1)^{n+1}\psi_n(1-z)=(-1)^n\pi\frac{d^n}{dz^n}\cot\pi z.\tag2 $$Using (1) by setting n=1,\ m=2, and z=\dfrac1{10}, we obtain ... 1 Another approach using contour integration that avoids having to deform the contour around branch cuts is to consider$$ \displaystyle f(z) = \frac{\log(z+ e^{i \pi /4})}{(1+z^{2})^{2}}$$and integrate around a contour that consists of the real axis and the upper half of |z|=R. Then letting R \to \infty,$$ \begin{align} &\int_{-\infty}^{0} ...

1

This probably isn't the answer you're hoping for, but there isn't going to be an analytic expression for the root. Bessel functions themselves (as in alone, not combined) have non-analytic roots that must be solved numerically, so combining these transcendental functions and looking for a root is surely only going to be solvable with further numerical ...

1

All sinc function integrals of the type $$I_n=\int_{0}^{\infty}\frac{\sin^n{(x)}}{x^n}\mathrm{d}x$$ can be expressed using the following general form: $$\displaystyle \int_{0}^{\infty}\frac{\sin^a{(x)}}{x^b}\mathrm{d}x=\frac{(-1)^{\lfloor(a-b)/2 \rfloor} \cdot \pi^ {1-c}}{ 2^{a-c}(b-1)!}\sum_{k=0}^{\lfloor a/2-c \rfloor } (-1)^k { a \choose ... 2 I have not found a source, but I think I reconstructed it correctly:$$\psi^{(n)}\!\left(\tfrac34\right)-\psi^{(n)}\!\left(\tfrac14\right)=(-1)^n\,4^{n+1}\,n!\,\,\Im\operatorname{Li}_{n+1}(i),\ n\in\mathbb N.$$In the source where I saw it, the imaginary part was probably written as a difference of two polylog terms. It can be proved using: ... 2 Let f_0\colon [m,1] be any continuous (and maybe monotonic) function with f(m)=\frac1k, f(1)=1. Then we can define f(x)=k^{-r}f_0(m^{-r}x) where r=\lfloor\log_mx\rfloor  and obtain a continuous solution on (0,1] and every continuos solution on (0,1] can be obtained this way. If k>1, this solution extends to [0,1]. If k<1, there ... 1 Here's a crude but rigorous bound, while we wait for the question to be settled definitively. Let x\geq 0. By definition (and half-angle formulas),$$I_0(x) = \frac{1}{\pi} \int_0^\pi e^{x \cos \theta} d\theta = \frac{2 e^x}{\pi} \int_0^{\pi/2} e^{-2 x \sin^2 \alpha} d \alpha,$$where we do the change of variables \alpha = \theta/2. Now, for ... 3 First thing set y=1/t and n=\lfloor 1/x\rfloor (for simplicity), then$$\begin{aligned}f(x)= \int_0^x \left\{ \frac{1}{t} \right\}\,\mathrm{d}t &= \int_{1/x}^{\infty} \frac{\{y\}}{y^2} \ \mathrm{d}y = \left(\sum_{k=n+1}^{\infty} \int_k^{k+1} \frac{x-k}{x^2} \ \mathrm{d}y\right) +\int_{1/x}^{n+1} \frac{y-n}{y^2} \end{aligned}$$Integrate simply to ... 1 Hint$$\frac{d}{dx}\Big(e^{-x} \sqrt{x} I_0(x)\Big)=\frac{e^{-x} ((1-2 x) I_0(x)+2 x I_1(x))}{2 \sqrt{x}}$$Then, appeared Jack D'Aurizio's answer to which nothing needs to be added ! Just to put some approximation beside Taylor and Newton, the function$$f(x)=(1-2 x) I_0(x)+2 x I_1(x)$$is very well represented by a Pade approximant built at x=0. For ... 0 By just considering the derivative of \sqrt{x}\,I_0(x)\,e^{-x} and setting it equal to zero we have that the maximum is attained in the only point for which:$$f(x)=(1-2x)\,I_0(x)+2x\, I_1(x) = 0.\tag{1}$$Such function is convex and decreasing over (0,x_0=2.3555\ldots), concave and decreasing over (x_0,+\infty). Since f(0)=1 and f(x_0)<0, we ... 4 Here's another approach for evaluating the one without the x in front. First notice that it's equivalent to showing that$$\int_{0}^{\pi /6} \log^{2}(2 \sin x) \ dx = \frac{7 \pi^{3}}{216}. $$Using the principal branch of the logarithm,$$ \begin{align} \log(1-e^{2ix}) &= \log (e^{-ix}-e^{ix}) + \log(e^{ix}) \\ &= \log(-2i \sin x) + ix \\ ...

1

For real $x$, the gamma function provides a continuous (indeed, even smooth) function that gives the factorials at the positive integers. However, you can choose points in between the positive integers and define your function to be whatever value you want there, and still find a smooth function that fits those values plus gives factorials at the positive ...

3

Let's focus on the case $x \in (0,1)$, the difficult part. Conjecture 1 Let $x \in (0,1)$ then $$\int_0^x \left\{ \frac{1}{t} \right\}\,\mathrm{d}t = 1 - \gamma + H_{\{1/x\}} - x\lfloor1/x\rfloor + \log x$$ (where $\{x\} = x - \lfloor x\rfloor$ is the fractional part of a number, and $H_n = \sum_{k=1}^n 1/k$ are the harmonic numbers.) Please, ...

2

One interesting thing about $I_1$ is that it satisfies a PDE. Specifically, it satisfies a version of Schroedinger's equation:: $$\partial_a^2 I_1 =-\int_0^{2\pi} \cos^2 \theta\;e^{i a\cos[\theta]+i b\cos^2[\theta])}d\theta=i\partial_b I_1.$$ Similar remarks apply to $I_2$. Since the $a=0$ and $b=0$ cases are both reducible to a zeroth-order Bessel function, ...

1

If $x$ is the variable of differentiation, $$\frac{\partial}{\partial x} \left[ t^x \right] \ne x t^{x-1}.$$ That is why you're getting a different result. Instead you should have $$\frac{\partial}{\partial x}\left[ t^x \right] = t^x \log t.$$ And the extra $\log t$ is what introduces the digamma function.

5

It is hypergeometric function. We usually write it as $_2F_1[a,b;c;z]$, not $_2F_1[a,b,c,z]$. For $|z|\lt 1$, the hypergeometric function is defined as $$_2F_1[a,b;c;z]=\sum_{n=0}^{\infty}\frac{(a)_n(b)_n}{(c)_n}\cdot\frac{z^n}{n!}$$ where $$(q)_n=\begin{cases}q(q+1)\cdots (q+n-1)&\text{if}\ n\gt 0\\1&\text{if}\ n=0\end{cases}$$ Here, $(q)_n$ is ...

10

The conjecture is true, as are the other cases reported in the comments where $f(z) := {}_2F_1 \left( \frac13, \frac13; \frac56; z \right)$ takes algebraic values for special rational values of $z$. There are a few others obtained from the symmetry $z \leftrightarrow 1-z$ (these ${}_2F_1$ parameters correspond to a hyperbolic triangle group with index ...

10

Continuing from Olivier Oloa's answer, \begin{align} \sum_{k=1}^{\infty} \big(\psi^{(1)} (k) \big)^{2} &= \int_{0}^{1} \int_{0}^{1} \frac{\ln u \ln v}{(1-uv)(1-u)(1-v)} \ du \ dv \\ &= \int_{0}^{1} \frac{\ln v}{(1-v)^{2}} \int_{0}^{1} \left(\frac{\ln u}{1-u} - \frac{v \ln u}{1-vu} \right) \ du \ dv \\ &= \int_{0}^{1} \frac{\ln v}{(1-v)^{2}} ... 1 I played a little with your conditions and this is what I came up with: Let \alpha \in ]0,1[, then: f_\alpha(x)=\frac{\left|\frac{1-\frac{2}{\pi}\arccos\left(\alpha\sin\left(\frac{\pi}{2}x\right)\right)}{1-\frac{2}{\pi}\arccos\left(\alpha\right)} ...

4

I think I have found something for you. See http://authors.library.caltech.edu/43489/ for the Integral Tables of the Bateman Project. In Volume 1, p.310 (PDF p. 322) Formula (23), which I hope should be applicable for your case. The last formula of david-h's computation is an example of a Mellin integral transformation. $\int_0^\infty (1+x)^\nu (1+\alpha ... 3 An equivalent form of the integral as a result of the inversion substitution$x=\frac{1}{u}$followed by the shift substitution$u=t+1is: \begin{align} I&=\int_{0}^{1}x^{-2/3}\,(1-x)^{-2/3}\,(1+8x)^{-1/3}\,\ln{(1+8x)}\,\mathrm{d}x\\ ... 6 The function f(a) has the following integral representation in terms of the Struve function \operatorname{H}_{\nu}{(z)}:\begin{align} f(a) &={_2F_3}{\left(1,1;\frac32,1-a,2+a;-\pi^2\right)}\\ &=(1+a)\Gamma{(1-a)}\int_{0}^{1}\left[\pi x(1-x^2)\right]^{a}\sqrt{x}\operatorname{H}_{-\frac12-a}{(2\pi\,x)}\,\mathrm{d}x. \end{align}\$ Proof: The ...

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