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Generalized Laguerre polynomials have the generating function $$\sum_{n=0}^{\infty}\xi^n L_n^{(t)}(x)=(1-\xi)^{-t-1}e^{-\frac{\xi x}{1-\xi}}.$$ Multiplying this identity by $\xi^{t-1}$ and integrating w.r.t. $\xi$ from $0$ to $1$, one finds \begin{align}\sum_{n=0}^{\infty}\frac{L_n^{(t)}(x)}{t+n}&=\int_0^1\left(\frac{\xi}{1-\xi}\right)^{t-1} ...

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Some functions that have counter-intuitive properties on $\mathbb R$ that you should know : The Weierstrass function, continuous everywhere, differentiable nowhere function. The Cantor function, a uniformly continuous function that is not absolutely continuous. The Minkowski's question mark function, an increasing continuous function, non differentiable on ...

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Start by noting that $\lfloor\log_2(1-x)\rfloor = -(n+1)$ for all $x \in (1-2^{-n},1-2^{-(n+1)})$. Therefore, $\displaystyle\int_0^1 \dfrac{\,dx}{1-\lfloor \log_2(1-x)\rfloor} = \sum_{n = 0}^{\infty}\int_{1-2^{-n}}^{1-2^{-(n+1)}} \dfrac{\,dx}{1-\lfloor \log_2(1-x)\rfloor}$ $= \displaystyle\sum_{n = 0}^{\infty}\int_{1-2^{-n}}^{1-2^{-(n+1)}} ... 6 I dont think there is a list of functions that "every mathematician should know". There are simply to many functions, and besides the elementary functions (polynomials, rational functions, log, exp, power functions, trig and inverse trig functions) one must learn what is needed. In statistics, for instance, we use a lot special functions such as the gamma ... 3 Let us make the change of variables$t=\sinh \frac{x}{4}$. Since $$t^2+t^4=\frac{\sinh^2\frac{x}{2}}{4}=\frac{\cosh x-1}{8}$$ and$dt=\frac14 \cosh\frac{x}{4}dx, the initial integral can be rewritten as \begin{align} I(n)&=\frac{e^{\frac18}}{4^{n+1}}\int_0^{\infty}\cosh\frac{x}{4}\sinh^{2n}\frac{x}{2}e^{-\frac18\cosh x}dx=\\ ... 3 Try to substitute $$x=(b-a)t+a \;\text{ where }t\in [0,1]$$ Then use the beta function $$B(x,y) =\int_{0}^{1} t^{x-1}(1-t)^{y-1} dt =\cfrac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}$$ 3\newcommand{\+}{^{\dagger}} \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{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} ...

2

If we ignore all "coincidences", including associativity and commutativity where it applies, we can enumerate all valid exprsssions by arranging the $n$ numbers in one of $n!$ orders insertingof parentheses among them in C(n-1)$$=\frac{(2n-2)!}{(n-1)!n!} ways inserting operators among the subexpressions in 4^{n-1} ways That gives us a total of ... 2 The \log function in the order 2 Clausen function is the natural logarithm.$$\mathrm{Cl}_2(\phi) = -\int\limits_{0}^{\phi} \log_e |2 \sin \frac{x}{2}| \mathrm{d}xLogarithms in calculus expressions are always base e, rather than base 10, unless explicitly stated otherwise. The subscript is often omitted by lazy authors because this ... 2 No. See the Jacobi theta function. 2 Deriving from integral approach is easier: \begin{align} \int(1+x^n)^{-\frac{1}{m}}~dx &=\int_0^x(1+t^n)^{-\frac{1}{m}}~dt+C\\ &=\int_0^{x^n}(1+t)^{-\frac{1}{m}}~d(t^\frac{1}{n})+C\\ &=\dfrac{1}{n}\int_0^{x^n}t^{\frac{1}{n}-1}(1+t)^{-\frac{1}{m}}~dt+C\\ &=\dfrac{1}{n}\int_0^1(x^nt)^{\frac{1}{n}-1}(1+x^nt)^{-\frac{1}{m}}~d(x^nt)+C\\ ... 2 Anticlimax ahead:\sum_{k=0}^\infty \frac{1}{4^k(k!)^2} < \sum_{k=0}^\infty \frac{1}{4^k\,k!}$$follows since 4^k(k!)^2 \geqslant 4^k\, k! for all k and the inequality is strict for k\geqslant 2. For the inequality$$\sum_{k=0}^\infty \frac{1}{5^k\,k!} < \sum_{k=0}^\infty \frac{1}{4^k(k!)^2},$$we note that the terms for k = 0 are equal ... 2 Not a full answer, but a partial one: Notice that the powers of two in the denominator are increasing in periodic repetitions of 3, 2, 1. That is, in this case the exponents on the 2 are 2, 5, 7, 8, 11, 13, 14, 17, 19, 20.... This is sequence A047268 on the OEIS website, defined as the sequence of numbers congruent to {1, 3, 5} mod(6). I fiddled around ... 2 This is the same as the integral with 1-x replaced by x, and this answer really solves the x version of the integral. In the interval (2^{-(n+1)},2^{-n}) this function takes the value {1}{1+n+1} = \frac{1}{n+2}. So this integral is:$$\sum_{n=0}^\infty \frac{1}{2^{n+1}} \frac{1}{n+2}$$The general sum:$$\sum_{n=0}^\infty \frac {x^{n+1}}{n+2}$$... 2 One function I would love to mention: The Greatest integer function or the Floor function. f:\mathbb{R}\rightarrow \mathbb{R} defined by f(x)=[x] where [x] denotes the greatest integer less than or equal to x 2 I shall describe an expansion for this integral \mathcal{I}(a,b,c) in powers of c^{-2} . To do so I will make a few changes of parameters first. Observe that the substitution x=bk yields$$ \mathcal{I}(a,b,c) =\int_0^\infty dx \,e^{-a x^2/b^2} J_0(x)\frac{x^3}{b^{4} c^{2}+x^4}.$$Defining t=b^2/4a, \epsilon=1/b^4 c^2, and I(\alpha,\epsilon) = b^4 ... 2 A reference: Chaudhry, M.A. et al. Asymptotics and closed form of a generalized incomplete gamma function. 2 A closed form for the integral is the Incomplete Beta function :$$\int{ \frac{1}{\sqrt{x+1}}{x^n}} dx = (-1)^{n+1}B_{-x}(n+1 , 1/2)+constant$$The inverse function needs numerical calculus to be evaluated : http://www.dtic.mil/dtic/tr/fulltext/u2/a467901.pdf 2 \newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} ... 2 To compute the second integral, let r=\sqrt{\alpha^2+\beta^2} and observe that we may write (\alpha,\beta)=(r \cos\phi,r\sin\phi) for some particular \phi. The sum-to-products formula then gives$$\alpha \cos\theta+\beta\sin\theta = r \cos\phi\cos\theta+r\sin \phi \sin\theta = r\cos(\theta-\phi).$$So the integral becomes$$\int_0^{2\pi}\,d\theta ... 2 This identity follows from the distribution relation on the Wikipedia page: $$\displaystyle\sum_{m=0}^{n-1}\zeta\left(z,a+\frac{m}{n}\right)=n^z\zeta\left(z,na\right).\tag{1}$$ It suffices to set in (1)a=\frac{q}{2}$,$n=2$and differentiate it once with respect to$z$. 2 Continuing from O.L.'s answer, the following is an evaluation of $$\int_{0}^{\infty} \frac{\sin 2x}{x} \text{Ci}(x) \ dx .$$ First notice that making the substution$ \displaystyle u = \frac{t}{x}$, $$\text{Ci}(x) = - \int_{x}^{\infty} \frac{\cos t}{t} \ dt = - \int_{1}^{\infty} \frac{\cos xu}{u} \ du.$$ Then $$\int_{0}^{\infty} \frac{\sin 2x}{x} ... 2 For the integral$$I:=\int_0^z {\frac {\exp {(w\cdot \frac{a^2-1}{2a^2})}}{\sqrt {(z-w) w}} dw} $$by completing the square in the denumenator and simplify we get$$\sqrt{\frac{1}{4} z-\frac{1}{4} z+zw-w^2}=\frac{1}{2} z\sqrt {1-(1-\frac{2w}{z})^2}$$Now using the substitution$$\sin\theta = 1-\frac {2w}{z} $$Therefore I reduces to$$ I=\int_{\frac ... 1 Defining$I$as the definite integral, $$I:= \int\limits_{0}^{1}\left[\frac{\zeta{(2)}-2\log^2{2}}{1-x}-\frac{1}{x(1-x)}\left(2\operatorname{Li}_2{\left(\frac{1-\sqrt{1-x}}{2}\right)}-\log^2{\left(\frac{1+\sqrt{1-x}}{2}\right)}\right)\right]\mathrm{d}x,$$ prove: $$I=\frac52\zeta{(3)}-2\zeta{(2)}\log{2}-\frac43\log^3{2}.$$ Substituting ... 1 No you can't do this: if you substitute your$A$and$B$into the RHS, then bring it all to a common denominator, you will find that you do not get the LHS. Actually if you think about it this is fairly obvious: the numerator on the RHS after combining terms will be a constant times$x$plus a constant, and this is not what the numerator looks like on the ... 1 Define$f(x):= \ln \left(K_0(\sqrt{x})\right)$and compute (e.g. with the help of a CAS) $$f'(x)=-\frac{1}{2}\frac{K_1(\sqrt{x})}{\sqrt{x}K_0(\sqrt{x})}$$ $$f''(x) = \frac{1}{4} \frac{\sqrt{x}K_0(\sqrt{x})^2+2K_0(\sqrt{x})K_1(\sqrt{x})-\sqrt{x}K_1(\sqrt{x})^2} {x^{3/2}K_0(\sqrt{x})^2}$$ From http://dlmf.nist.gov/10.37 we have$K_1(x)>K_0(x)>0\;$for ... 1 I've recently come across this integral, and I think that your solution is wrong. If I am right, it holds that$f(z)=-\sqrt{\pi} e^{z^2} \text{erf}(z)$. To see this, note that $$f(z)= -\frac{i}{\sqrt{\pi}} e^{z^2} \int_{-\infty}^\infty \frac{e^{-t^2-2izt}dt}{t}=-\frac{i}{\sqrt{\pi}} e^{z^2} I(z)$$ via substitution. Using the fact that$\int_{-\infty}^\infty ...

1

The solution given by user1337 in the comments is correct. To derive it rewrite your equation $$\frac{1-e^{X}}{X} = K$$ to $$\frac{1-KX}{K}e^{\frac{1-XK}{K}} = \frac{e^{\frac{1}{K}}}{K}$$ This is now on the form $W(Z) e^{W(Z)} = Z$ with $W(Z) = \frac{1-XK}{K}$ and $Z = \frac{e^{\frac{1}{K}}}{K}$.

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We have $$\tag+\frac{1-e^X}X = K \iff 1-KX = e^X \iff (1-KX)\cdot e^{-X} = 1$$ We want to have something of the form $Ye^Y$ to apply $W$, hence we write $$e^{-X} = e^{-\frac{KX}K} = \frac{e^{\frac 1K - \frac{KX}K}}{e^{1/K}} = \frac{e^{\frac{1-KX}K}}{e^{1/K}}$$ So, we divide $(+)$ by $K$ to get $$\frac{1-KX}{K} \cdot e^{\frac{1-KX}K} = \frac{e^{1/K}}K ... 1 I bet that the answers to the first and third question (about convergence and zeroes) are affirmative, since the function$$ f(z) = \left(1-\frac{x^2}{4}+\frac{x^4}{64}\right)\mathbb{1}_{[0,1]}(x)+\sqrt{\frac{2}{\pi x}}\cos(x-\pi/4)\mathbb{1}_{[1,+\infty)}(x), by following Abramowitz and Stegun, is a very good approximation for $J_0(x)$, but I do not think ...

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