148

The answer is 'no'. Making the substitution $$ x = \frac{(t-1)(t-5)(t^2+2t+5)}{16t^2}, $$ one finds $$ {\textstyle\sqrt{x+\sqrt{x+\sqrt{x+1}}}\,\mathrm{d}x} = \frac{(t^2-2t+5)(t^2-5)\sqrt{t^4{-}2t^2{-}40t+25}\ \mathrm{d}t}{32t^4}. $$ Denote the right hand side of the above equation by $\beta$. Now, setting $$ Q(t) = \frac{(25-25t-14t^2-3t^3+t^4)}{96t^3}, ...


98

I'm adding a separate answer for the general question that the OP asked, which settles the question in the negative for all $n>2$ (and gives an alternate proof for $n=3$ to the one I gave above). Recall that the OP defined a sequence of algebraic functions $f_n$ by the rule $f_0(x) = 1$, $f_1(x) = \sqrt{x+1}$, and $f_{n+1}(x) = \sqrt{x + f_n(x)}$ for all ...


71

I have proved this equality by means of Cauchy’s Theorem applied to an adequate function. Since my solution is too long to post it here, I posted it in arXiv, you can get it at http://arxiv.org/abs/1402.3830 The function $$G(z)=\frac{\log(1+(1+i)\,f(z)\,)}z$$ where $$f(x)=\frac{\operatorname{arctanh}(x)-\arctan(x)}{\pi}$$ extended analytically.


65

Tonight I read here [the answer by esg to another your question] that $\frac1{2\pi}\int_{-\pi}^\pi e^{-ik t}(1+e^{it})^ndt=\binom{n}{k}$, which is, well, obvious at least when both $n$ and $k$ are positive integers: just expand the binomial $(1+e^{it})^n$ and integrate. Denoting $\alpha=k/n$ we may rewrite this as $\frac1{2\pi}\int_{-\pi}^\pi (f(t))^n dt=\...


48

Finding an anti-derivative of $x\tan x$ amounts to finding an anti-derivative of $f=\frac{x}{e^x+1}$. Consider the field $K=\mathbb C(x,e^x)$. Note that $K$ is closed under taking derivatives. If $f$ is elementary integrable, then Liouville's Theorem gives elements $u_i\in K$, $\gamma_i\in\mathbb C$, $v\in K$ with \begin{equation} \frac{x}{e^x+1}=\sum\...


37

Gradshteyn and Ryzhik gives the recursion relation: (see also volume 1 of Integrals and Series by Prudnikov, Brychkov, and Marichev, page 387) so if $m$ (your $q$) and $p$ are both even or both odd, with $p\ge m$ to assure convergence of the integral, we arrive at integrals of the form $$\int_0^\infty \frac{\sin^{2n+1}x}{x}\,dx = \frac{(2n-1)!!}{(2n)!!}\...


37

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 $$ \frac{I_{n,m}}{I_{m+1,n-1}}=\frac{n}{m+1}. $$ All of your examples are special cases of this identity.


36

We have \begin{align} & 2\int_0^{\pi/2}\frac{\sin x}{1+\sqrt{\sin 2x}} \, dx=\int_0^{\pi/2}\frac{\sin x+\cos x}{1+\sqrt{\sin 2x}} \, dx=\frac12\int_0^\pi\frac{\sqrt{1+\sin y}}{1+\sqrt{\sin y}} \, dy \\[6pt] = {} &\int_0^{\pi/2}\frac{\sqrt{1+\sin y}}{1+\sqrt{\sin y}} \, dy =\int_0^1\frac{\sqrt{1+t}}{(1+\sqrt{t})\sqrt{1-t^2}} \, dt=\int_0^1\frac{dt}{(...


34

Actually, I now think that the easiest method is to do this: Write $k=\sin z$, so that $|k|<1$, and make the substitution $x = \arcsin(k\sin\theta)$, where $0\le \theta\le \frac\pi2$. The integral becomes $$ \int_0^{\pi/2} \frac{\arcsin(k\sin\theta)}{k\sin\theta \,\,(1-k^2\sin^2\theta)^{(1/2)}}\ \mathrm{d}\theta = \int_0^{\pi/2} \sum_{n=0}^{\infty} c_n\...


30

It has nothing to do with the conflict with Borel which developed later, and one can find a pretty explicit answer in the aforementioned letters of Lebesgue to Borel. (These letters were first published in 1991 in Cahiers du séminaire d’histoire des mathématiques; selected letters with updated commentaries were also published later by Bru and Dugac in an ...


28

Consider $$f(x,y) = \cases{ \text{sgn}(x) \dfrac{x^2-y^2}{x^2} & for $0 < |y| < |x|$ \cr 0 & otherwise }$$ Then $\displaystyle \int_{-1}^1 f(x,y)\ dy = 4 x/3$ for $-1 \le x \le 1$, so $\displaystyle \frac{\partial}{\partial x} \int_{-1}^1 f(x,y)\ dy = 4/3$, but $\dfrac{\partial f}{\partial x} (0,y) = 0$ so $\displaystyle \int_{-1}^1 \frac{\...


28

Following the suggestion I made in a comment, the integral can be rewritten as the contour integral $$ I_{3,2} = \frac{1}{2\pi i} \oint \frac{\operatorname{tahn}^3 z}{z^2} \log(-z) \, dz , $$ where the clockwise contour tightly encircles the positive real axis, which coincides with the branch cut of the logarithm. The reason that this integral is ...


27

The Deutsche Mathematiker-Vereinigung has a biographical entry: Heinrich Hake was born August 4, 1891 in Düsseldorf. He attended high school in Attendorn (Westfalen), studied in Göttingen and Bonn, where he completed his studies in 1914. After military service he conducted Ph.D. research with prof. Hans Hahn in Bonn, where he defended his dissertation on 2 ...


26

Property (2) gives $T(1) = T(1)+a$ for any real $a$, which is not solvable in any real algebra (or vector space) $A$. Property (3) leads to a similar issue as it implies that $T(1) = aT(1)$ for all $a>0$. Note that many common ways of evaluating divergent sums and integrals (e.g. zeta function regularisation) do not actually obey (2) or (3). For ...


26

Following Gerald Edgar, $$W_3(1)\equiv\mathbb{E}[|Y|]=\int dx\int dy \,\sqrt{x^2+y^2}\,p(x,y)=2\pi\int_0^\infty r^2p(r)dr$$ with $p(x,y)dxdy=p(r)\,rdrd\phi$ the rotationally invariant distribution of the complex variable $Y=x+iy=r\cos\phi+ir\sin\phi$. From here $$\mathbb{E}[|X|]=\int dx\int dy \,|x|\,p(x,y)=\int_0^\infty \int_0^{2\pi} |\cos\phi|r^2 p(r)\,...


25

Let's record what is possible, so far. The case $n=2$, from the comments: $$\int_0^1\frac{dy}{x(x+y)}=\frac1{x^2}\int_0^1\frac{dy}{1+\frac{y}x}=\int_0^1\sum_{k\geq0}\frac{(-1)^ky^k}{x^{k+2}}dy=\sum_{k\geq0}\frac{(-1)^k}{k+1}\frac1{x^{k+2}}.$$ Hence, $$\int_1^{\infty}\sum_{k\geq0}\frac{(-1)^k}{k+1}\frac{dx}{x^{k+2}}=\sum_{k\geq0}\frac{(-1)^k}{(k+1)^2}=\frac{\...


24

Following the hint by Noam D. Elkies, we just need to show that the remainder $$R_n:=\int_{-\infty}^0{n!\over\Gamma(n+1-x)\Gamma(1+x)}dx+ \int_n^{+\infty} {n!\over\Gamma(n+1-x)\Gamma(1+x)}dx $$ satisfies $$0\le R_n<1.$$ The integrand writes $${n!\over\Gamma(n+1-x)\Gamma(1+x)}={n! \over (n-x)(n-1-x)\dots(1-x)\Gamma(1-x)x\Gamma(x)}={n!\over \pi}{ \sin \pi ...


24

Comment This is related to some papers (e.g. this) by P.M. Borwein et. al. on short random walks in the plane. Let $X_1, X_2, \dots$ be i.i.d. random variables, uniformly distributed on the unit circle $|z|=1$ in the complex plane. Then $$ X_1+X_2+X_3 $$ is a random variable in the plane, and your integral is $$ (2\pi)^3\;\mathbb{E}\Big[\big|\mathrm{Re}(...


23

Here is a proof which doesn't use the identity $\int_{-\infty}^\infty {n \choose x}\,dx= 2^n$: Using the representation ${ n \choose x}=\frac{1}{2\pi}\int_{-\pi}^\pi e^{-ixt}\left(1+e^{it}\right)^n\,dt$ (valid for $n,x\in\mathbb{R}, n>-1$) we find \begin{align*} \int_0^n {n \choose x}\,dx&=\frac{1}{2\pi} \int_0^n\int_{-\pi}^\pi e^{-ixt}\left(1+e^{it}\...


22

It's not quite $\pi/4$ . . . Using the same formula $\prod_{m=1}^\infty \cos(x/2^m) = \frac{\sin x}{x} = \text{sinc}\,x$, we write the integrand as $$ \prod_{n=1}^\infty \text{sinc}\,\frac{2z}{2n-1}, $$ and then the integrals $$ I_N := \int_0^\infty dz \prod_{n=1}^N \text{sinc}\,\frac{2z}{2n-1} $$ of the partial products are $1/2$ of the notorious Borwein ...


21

I'm posting an answer just to inform that the question has received an answer by Nick Strehlke on MSE. \begin{align*} \int_0^\infty {\sin^p x\over x^q}\,dx & = \left\{\begin{array}{ll} \displaystyle{(-1)^{(p+q)/2}\pi\over 2^{p+1}(q-1)!}\sum_{k = 0}^p(-1)^k{p\choose k} |p - 2k|^{q-1} & \text{$p,q$ even,} \\[2em] \displaystyle {(-1)^{(p+q)/2-1}\pi\...


21

This relation seems to be part of the story of Ramanujan's mock theta function. The identity you discuss is a specialization of transformation properties of a mock theta function evaluated at $\tau =i$. See this paper of Zwegers for proofs of the general transformation identity (and he gives other references to earlier work of Watson): in particular, ...


21

The following proposition answers OP's question regarding the upper bound of $$\tau(C) \Doteq f(C)/\lambda^2(C).$$ Let $B_n$ be the closed Euclidean unit ball of $\mathbb{R^n}$ centred at $0$, that is $$B_n = \{ (x_1,\dots, x_n) \in \mathbb{R^n} \, \vert \,\, x_1^2 + \cdots + x_n^2 \le 1\},$$ and let $\tau_n = \tau(B_n)$. Proposition. The Euclidean ...


20

From a geometric measure theory perspective, it is standard to define Radon measures $\mu$ to be Borel regular measures that give finite measure to any compact set. Of course, their connection with linear functionals is very important, but in all the references I know, they start with a notion of a Radon measure and then prove representation theorems that ...


20

We can think of $I_{n}$ as being a classical partition function for $n$ beads on a circle which cannot pass through each other, with logarithmic interaction potential between each bead and its next-to-nearest neighbors on either side. For $I_{2n}$ the beads fall into two ``colors" which do not have logarithmic interactions with each other; while for $I_{...


20

Following Robert Israel's answer, we also scale everything to $[-1,1]$ (thus multiplying the result by 2). As he mentions, the optimal polynomial is always a square of some other polynomial, $p_{2n}=p_{2n+1}=q_n^2$, and $q_n$ is either even or odd (see Lemma below). So we are left to find the minimal $L_2[-1,1]$-norm of an odd/even polynomial $q_n$ such that ...


20

The main applications of Lebesgue integral to concrete problems of analysis found before Poincare's death are the Riesz-Fischer theorem (1907) and Fatou's work (1906). All this is somewhat remote from the main interests of Poincare. Applications of measure theory to mechanics (ergodic theory) were found later, after his death. You cannot expect even the ...


20

Fix a compact group $G$ and consider its category of Banach representations: the objects are (complex) Banach spaces $X$ endowed with a $G$-action by automorphims (not necessarily isometries) such that the action maps $G\times X\to X$ are jointly continuous and the morphisms are bounded maps $X\to Y$ commuting with the $G$-actions. Denote the trivial ...


19

We can start with a substitution $x=2y$ $$\int_0^{\infty}dx\prod_{n=1}^{\infty}\text{sinc}\left(\frac{x}{2n-1}\right)=2\int_0^{\infty}dy\prod_{n=1}^{\infty}\text{sinc}\left(\frac{2y}{2n-1}\right).$$ Now the double angle formula $$\text{sinc}(2a)= \text{sinc}\left(a\right)\cos(a)$$ can be iterated to arrive to the familiar formula for $\text{sinc}$ $$\text{...


18

Yes, this is true, it follows from formulas in Higher-Dimensional Box Integrals, by Jonathan M. Borwein, O-Yeat Chan, and R. E. Crandall. \begin{align} &\text{define}\;\;C_{m}(s)=\int_{[0,1]^m}(1+|\vec{r}|^2)^{s/2}\,d\vec{r},\;\;\text{we need}\;\;C_{2n-1}(-2n).\\ &\text{the box integral is}\;\;B_m(s)=\int_{[0,1]^m}|\vec{r}|^s\,d\vec{r},\\ &\...


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