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43 votes
Accepted

Is there a closed form for $\int_0^\infty\frac{\tanh^3(x)}{x^2}dx$?

Following the suggestion I made in a comment, the integral can be rewritten as the contour integral $$ I_{3,2} = \frac{1}{2\pi i} \oint \frac{\operatorname{tanh}^3 z}{z^2} \log(-z) \, dz , $$ where ...
Igor Khavkine's user avatar
39 votes
Accepted

Why these surprising proportionalities of integrals involving odd zeta values?

For $n\geq 1$ and $m\geq 0$, an application of integration by parts ($u=\log^n(1-x)$, $dv=\log^m(x)\,dx/x$) followed by the substitution $x\mapsto 1-x$ shows that $$ \frac{I_{n,m}}{I_{m+1,n-1}}=\frac{...
Julian Rosen's user avatar
  • 8,941
36 votes
Accepted

How do i solve this : $\displaystyle \ f'=e^{{f}^{-1}}$?

There is no such function. Since $f$ would have to map $\mathbb R$ onto $\mathbb R$ for the equation to make sense at all $x\in\mathbb R$, it follows that $f^{-1}(x)\to -\infty$ also as $x\to -\infty$,...
Christian Remling's user avatar
30 votes

When can an invertible function be inverted in closed form?

I recommend the following paper: MR1501299 Ritt, J. F. Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925), no. 1, 68–90. (freely available on the web). It indeed gives a short ...
Alexandre Eremenko's user avatar
28 votes
Accepted

What is the value of this double sum in closed form?

Consider the integral $$I=\int_0^1\int_0^1\frac{zdzdt}{(1-zt)(1-z(1-t))}=\sum_{k,j\geqslant 0} \int_0^1\int_0^1 z^{k+j+1}t^k(1-t)^jdzdt=\\ \sum_{k,j\geqslant 0} \frac1{k+j+2}\cdot \frac{k!j!}{(k+j+1)!}...
Fedor Petrov's user avatar
20 votes

Is there a closed form for $\int_0^\infty\frac{\tanh^3(x)}{x^2}dx$?

Rewrite the integrand and apply Taylor expansion to $\frac1{(1+e^{-2x})^3}$ so that $$\frac{\tanh^3x}{x^2}=\sum_{j\geq0}(-1)^j\binom{j+2}2 \frac{(1-e^{-2x})^3}{x^2}\binom{j+2}2e^{-2jx}.$$ Integrate ...
T. Amdeberhan's user avatar
18 votes
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On $\zeta(7)$ as the integration of the product of an indefinite integral due to Lobachevskii by a power of the inverse Gudermannian function

First, $\mathrm{gd}^{-1}(z) = \ln \frac{1 + \tan(\frac{z}{2})}{1 + \tan(\frac{z}{2})}$. So we substitute $t = \tan(\frac{z}{2})$, obtaining: $$\mathcal{G}_n = \int\limits_{0}^{1}2\cdot\ln^{n + 1}\...
Rybin Dmitry's user avatar
16 votes
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How to prove that $\int _0^\infty\frac{\text{arcsinh}^nx}{x^m}dx$ is a rational combination of zeta values?

Let me first establish the case $m=2$. The general case is established in the "Added" section below. By making the substitution $x=\sinh t$ we get \begin{align*} I(n,2)&=\int_0^\infty t^n\frac{\...
GH from MO's user avatar
  • 99.4k
16 votes

Is there real or complex analytic function whose positive real zeros are the primes?

H. Laurent introduced the function $$f(z)=\sum_{n=1}^\infty(\sin\pi z)^2\left(\frac{1}{n^2\sin\frac{\pi z}{n}}-\frac{1}{\pi n (n-z)}\right)^2$$ whose real roots are the positive primes and have no ...
juan's user avatar
  • 6,976
15 votes

Find a formula for the recurrent sequence $q_{n+1}=q_n(q_n+1)+1$

I'm not quite sure what you mean by an "analytic formula." As Fedor Petrov indicated, there is unlikely to be a closed formula. However, there is a convergent power series. More precisely, consider ...
Joe Silverman's user avatar
14 votes

Find a formula for the recurrent sequence $q_{n+1}=q_n(q_n+1)+1$

This is a second answer with a somewhat different viewpoint from my other answer. It is an expansion, in some sense, of Gerry Myerson's answer. There is a general theory for estimating these sorts of ...
Joe Silverman's user avatar
13 votes
Accepted

Invertibility of specific function

The answer to the question of whether the inverse has a closed form depends of course on one's definition of "closed form." One plausible definition is that a closed-form function is a ...
13 votes
Accepted

Closed form for ₄F₃(n,n,n,2n;1+n,1+n,1+n;−1)

Fiddling with Maple, I get: if $n$ is a positive real number, then $$ {{_4\mathrm F_3}(n,n,n,2\,n;\,n+1,n+1,n+1;\,-1)}={\frac {{n}^{2} \sqrt {\pi}\,\Gamma(n+1)\,\psi^{(1)}(n)}{{4}^{n}\,\Gamma \left( n ...
Gerald Edgar's user avatar
  • 40.3k
13 votes
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On 12 cfracs: for Catalan's $K$, Gieseking's $\kappa$, and $\pi^2$, $\pi^3$, plus three for $\zeta(3)$ using Zagier's "six sporadic sequences"

We have $$C_2(-17,-6,-72)=-(5/8)L(\chi_{-3},2)$$ and $$C_2(10,3,-9)=(1/2)L(\chi_{-3},2)$$ so both are proportional to what you call Gieseking's constant but which is simply the value at 2 of the L ...
Henri Cohen's user avatar
  • 11.8k
12 votes
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Closed Form for $\sum\limits_{n=-2a}^\infty(n+a){2a\choose-n}^4,~a\not\in\mathbb Z$

Some experimentation suggests that the second series is given by $$ S_4^-(a) = \sum_{n=-2a}^\infty (n+a) \binom{2a}{-n}^4 = \frac{1}{4\cos(2\pi a) \,\Gamma(2a+1)^2\,\Gamma(-4a)}, $$ which agrees with ...
Timothy Budd's user avatar
  • 3,555
11 votes

When can an invertible function be inverted in closed form?

Closed-form functions need the definition which set of functions is allowed to represent the function. Take e.g. the algebraic definition of the Elementary functions by Liouville and Ritt (Wikipedia: ...
IV_'s user avatar
  • 1,063
11 votes
Accepted

On Zagier's missing continued fraction with multiple limits?

Set $Q=(1/2)L(\chi_{-3},2)$ (related to your Gieseking constant) and $P=2\pi^2/81$. The limits are almost certainly (not proved), \begin{align} \lim_{m\to\infty}C_2(6m+0) &= -Q\\ \lim_{m\to\infty}...
Henri Cohen's user avatar
  • 11.8k
10 votes

Find a formula for the recurrent sequence $q_{n+1}=q_n(q_n+1)+1$

The sequence (at any rate, the case $q_0=1$) has been studied, and references are given at OEIS. The closest thing to a formula given there is $a(n) = [c^{2^n}]$ for $n > 0$, where $c = 1....
Gerry Myerson's user avatar
10 votes
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Is this closed-form summation a special case of known Lerch zeta function formulas?

This is the Fourier series for the RHS, as a function of $\alpha\in (0,2\pi)$, $$ f(\alpha)=\frac{2\pi i}{1-e^{-2\pi iz}}\, e^{-iz\alpha} . $$ The series representation follows by computing the ...
Christian Remling's user avatar
9 votes

Are these two new ways of representing odd zeta values as integrals known?

Using the reflection formula followed by the recurrence formula and the Beta integral representation (DLMF) \begin{align} \frac{x(1-x)}{\sin \pi x}&=\frac{1}{\pi}x(1-x)\Gamma(x)\Gamma(1-x)\\ &...
Paul Enta's user avatar
  • 781
9 votes

counting points on unit sphere mod p

Since this old question has resurfaced, let me sketch two ways to prove the stated formulæ using algebraic geometry: The first way is fairly elementary. Let us stick for definiteness with the number ...
Gro-Tsen's user avatar
  • 30.2k
9 votes

Could there be a special-function counterexample to Schanuel's conjecture?

Yes, some special cases of the hypergeometric function give roots of polynomial equations whose Galois groups are not solvable. The simplest examples are the solutions of $y(1-y)^t = x$ for rational $...
Noam D. Elkies's user avatar
9 votes

Find a formula for the recurrent sequence $q_{n+1}=q_n(q_n+1)+1$

If we denote $A_n=q_n+1/2$, then $$A_n=A_{n-1}^2+5/4$$ with $A_0=q_0+1/2\ge 3/2$ by $q_0\in\mathbb{N}$. Further, $$\log A_n=2\log A_{n-1}+\log\left(1+\frac{5}{4A_{n-1}^2}\right),$$ namely $$\frac{1}{...
Zhou's user avatar
  • 967
9 votes

Why mpmath computes $\sum_{n=2}^\infty (-1)^n\log(n)=\log\left(\frac1 2 \sqrt{2} \sqrt{\pi}\right)$

A summation method for this... $$ F(s) = -\sum_{n=2}^\infty \frac{(-1)^n}{n^s} = -(2^{1-s}-1)\zeta(s) \qquad\text{for $s>0$} $$ Differentiate: $$ \sum_{n=2}^\infty \frac{(-1)^n\log n}{n^s} = F'(s) ...
Gerald Edgar's user avatar
  • 40.3k
9 votes
Accepted

Partition numbers as the specific sums of the A161511

If $2m=2^{j_1}+2^{j_2}+\ldots+2^{j_k}$ for $0<j_1<j_2<\ldots<j_k$, then $\ell(m)=j_k-1$, $a(m)=j_1+(j_2-1)+(j_3-2)+\ldots+(j_k-(k-1))$, as the $i$-th (from the left) 1 in the binary ...
Fedor Petrov's user avatar
8 votes
Accepted

Does this system have a closed-form solution? $x_j^2 = \sum_{i=1}^n B_{ij} x_i$

Sorry, there cannot be a simple solution. For example, taking $n=3$ and $B_{ij} = (i+j-1)/6$, we compute numerically (by iterating the contraction mapping as you suggest) $$ (x_1,x_2,x_3) = (1....
Noam D. Elkies's user avatar
8 votes

On the search for an explicit form of a particular integral

Let $$ I_n=\int_0^1 \frac{x^n x^x (1-x)^{1-x} \sin (\pi x)}{\pi e}\mathrm dx $$ and $$ b_0=-1,\quad b_n=\frac{-1}{n}\sum_{k=1}^n\frac{b_{n-k}}{k+1} $$ Then $$ \boxed{I_n\equiv b_{n+2}} $$ so indeed ...
AccidentalFourierTransform's user avatar
8 votes
Accepted

Difficult trigonometric integral

Here is an outline of the approach I have taken to solve this integral. First rewrite the integral $(1)$ in Cartesian variables: $$I=\int_{-\infty}^{\infty} \mathrm{d}^3v~ \frac{3x^2y^2v_z^2}{y^...
Dinesh Shankar's user avatar
8 votes

how to solve $\sum_{i=0}^n (x-\mu_i)e^{-(x-\mu_i)^2} = 0$

There always exists at least one solution: If $x >\mu_i$ for all $i$, then each of the terms in the sum is positive and if $x < \mu_i$ for all $i$, then each of the terms in the sum is negative. ...
Robert Bryant's user avatar
8 votes

Invertibility of specific function

The change of variable given above by Pietro Majer shows that this is equivalent to Kepler's equation Wikepedia on Kepler's Equation which is believed not to have any closed form solution (let alone ...
Rob Corless's user avatar

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