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1

Here are some useful components to developing a solution to the proposed integral. \begin{align}\tag{1} \int_{0}^{\infty} Ei^{4}(-x) \, dx &= -4 \, \int_{0}^{\infty} e^{-x} \, Ei^{3}(-x) \, dx \\ \int_{0}^{\infty} e^{-x} \, Ei^{3}(-x) \, dx &= - \ln^{2}2 + \int_{0}^{\infty} x \, e^{-x} \, Ei^{2}(-x) \, dx - 2 \, \int_{0}^{\infty} e^{-x} \, Ei(-x) \, ...


2

Let $a = \displaystyle{ \frac{11 + \sqrt{106}}{2}}$ and $b = \displaystyle{ \frac{21 + 2 \sqrt{106}}{2}}.$ Then $$x = (a + \sqrt{a^2 - 1}) (b + \sqrt{b^2 + 1}).$$ As requested, this exhibits $x$ as a product of two quartic units. (For the purists, note that $a$ and $b$ are only half algebraic integers, but the expressions above are genuinely units. I wrote ...


0

Note that the derivative of the Hurwitz Zeta Function can be calculated as following (similiar to calculating the derivative of the Riemann Zeta): \begin{align} \frac{d}{ds} \zeta(s,q) &= \frac{d}{ds} \sum_{n=0}^{\infty} (n+q)^{-s} \\ &= \sum_{n=0}^{\infty} \frac{d}{ds} (n+q)^{-s} \\ & = \sum_{n=0}^{\infty} - \ln(n+q) (n+q)^{-s} \\ &= ...


1

In each interval $\left(\pi n - \frac{\pi}{2}, \pi n + \frac{\pi}{2}\right)$ there is exactly one solution $x_n$ (i.e. $\tan x_n = \ln x_n$), and, when $n$ is large, it appears that $x_n$ is approximately $\pi n + \frac{\pi}{2}$. Let's show this. Since $\tan$ is $\pi$-periodic we have $$\tan\left(\pi n + \frac{\pi}{2} - x_n\right) = ...


1

Numerically, I get $$ S_1+S_2 = 0.14836252987273216621 $$ which agrees with $$ \frac{\pi^6}{6480} $$ Also numerically, $$ S_1 = 0.074181264936366083104 \\ S_2 = 0.074181264936366083104 $$ are seemingly equal.


8

This is a very intuitive approach based on the fact that I assume to be in the exam room with no computer and even no calculator. If there is a simple root $\sqrt{64x+5}$ should reduce to a whole number and $x=\frac1{16}$ is obvious (since $9$ is the closest square to $5$). From here, we can go backward (verify that every time we get another square) and ...


4

\begin{align*} \sqrt {x + \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}}} - \sqrt x & = 1\\ \sqrt {x + \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}}} & = 1+\sqrt{x}\\ x + \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}} & = 1+x+2\sqrt{x}\\ \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}} & = 1+2\sqrt{x}\\ 4x + \sqrt {16x + \sqrt {64x + 5}} & = ...


3

With the extra $\sqrt x$ there, you just square both side and keep doing it, miraculously some terms cancel out :)


13

$$\sqrt {x + \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}}} = 1+ \sqrt x$$ Squaring $$ \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}} = 1+ 2\sqrt x$$ Squaring $$ \sqrt {16x + \sqrt {64x + 5}} = 1+ 4\sqrt x$$ Squaring $$ \sqrt {64x + 5} = 1+ 8\sqrt x$$ Squaring $$ 5 = 1+ 16\sqrt x$$


6

You get rid of the square roots by successive squarings and changes of side. $$\sqrt {x + \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}}} = 1,$$ $$\sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}} = -x+1,$$ $$\sqrt {16x + \sqrt {64x + 5}} = x^2-6x+1,$$ $$\sqrt {64x + 5} = x^4-12x^3+38x^2-28x+1,$$ $$0=x^8-24x^7+220x^6-968x^5+2118x^4-2152x^3+860x^2-120x-4.$$ Using a ...


4

The best solution I can imagine is to square repeatedly and then use numerical methods to find roots of the resulting polynomial and then checking for extranneous answers from our squaring.


2

(a) Fucntion $y$ describes strainght line iff coefficient of $x^2$ is $0$, so $$1-2\cos \theta=0\implies\cos\theta=\frac12$$ From this we get $$\sin \theta =\pm\sqrt{1-\cos^2\theta}=\pm\frac{\sqrt3}2$$ So, equations of these lines are $$y=\pm\frac{\sqrt3}2x+\frac12$$ (b) Let $y=0$. Discriminant of quadratic equation for $x$ is ...


1

Assuming the variable in your equation is $x$ and $\theta$ is a fixed real number (which the statement of your problem fails to specify), then you're looking to get only first or zeroth powers of $x$, as in : $y = ax + b, (a,b) \in \mathbb{R}^2$ where $ax$ is the first-power term and $b = bx^0$ the zeroth-power term. This means that terms of higher power ...


0

This is a bit old, but anyway. As for the question for the LHS being $\leq 1$ when $b =1 $, I think one can prove directly that \begin{equation} \int_0^1\frac{\sin x}{x}dx \leq 1. \quad(*) \end{equation} Integrating by parts $$\int_\epsilon^1\frac{\sin x}{x}dx = -\sin\epsilon\ln\epsilon - \int^1_\epsilon\ln x\cos x dx.$$ Then use that $$- \int^1_\epsilon\ln ...


2

I'd say it's quite plausible to find the roots! :) Rewrite as $\cos x=4^{x-x^{\cos x}}$. For $x\ge1$, $x-x^{\cos x}\ge0$ is true because $x^{\cos x}\le x^1$, thus $4^{x-x^{\cos x}}\ge4^0=1$, with equality if and only if $\cos x=1\iff x=2\pi k$ for $k\in\mathbb N$. For $x<0$, the problem is nonsense. For $x=0$, equality holds. For $0<x<1$, we have ...


1

There exists an addition formula for Bessel functions due to Graf. In the case you are interested in this addition formula takes the form $$H^{(1)}_{\nu}\left(R_1+R_2\right)=\sum_{k\in\mathbb{Z}}H^{(1)}_{\nu-k}\left(R_1\right)J_k\left(R_2\right),$$ where it is assumed that $|R_2|<|R_1|$. I'm not sure whether this can be called a simplification, but there ...


3

There may be some cases where complex variables, the residue theorem and the residue at infinity are helpful. Suppose your OGF is $f(z)$ and the desired EGF is $g(w).$ Then we have $$g(w) = \sum_{n\ge 0} \frac{w^n}{n!} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} f(z) \; dz.$$ This will simplify together with some conditions on convergence ...


1

You can try the following $$B(x,y)=\int_0^1 t^{x-1}(1-t)^{y-1}dt$$ Substitution now: $$u:=\frac t{1-t}=\frac1{1-t}-1\implies du=\frac{dt}{(1-t)^2}$$ so we get $$B(x,y)=\int_0^\infty \frac{u^{x-1}}{(u+1)^{x-1}}\frac1{(u+1)^{y-1}} (u+1)^2du=\int_0^\infty\frac{u^{x-1}}{(u+1)^{x+y}}du$$ Now, we have $$\Gamma(x):=\int_0^\infty ...


1

The integral $$ \int_{0}^{+\infty} x\, J_n(x)\,dx $$ is not converging for any $n\in\mathbb{N}$, since $J_n(x)$ decays like $\frac{1}{\sqrt{x}}$ as $x\to +\infty$. On the other hand, for any $n\in\mathbb{N}$ we have: $$ \int_{0}^{+\infty} J_n(x)\,dx = 1 $$ since: $$ \mathcal{L}(J_n(x)) = \frac{1}{\sqrt{1+s^2}\left(s+\sqrt{1+s^2}\right)^n}.$$


2

Ok let's see: There is a famous integral representation for the Hurwitz function due to Hermite which reads as follows: $$ \zeta(s,a)= \frac{a^{-s}}{2}+\frac{a^{1-s}}{1-s}+2\int_0^{\infty}dt \frac{\sin(s\arctan(t/a))}{(e^{2 \pi t}-1)(t^2+a^2)^{s/2}} \quad (1) $$ You may check for yourself that it is allowed to switch differentiation w.r.t to $s$ with ...


0

Suppose $(x_n)$ be a sequence in $f^-$$^1$$[F]$ converging to x∈$R$ since $f$ is continuous on $R$ that means $f$ is continuous on each point of $R$ By $Sequential-Criterian$ of continuity $f(x_n)$ converges to $f(x)$. $F$ is closed and $f(x_n)$∈$F$ implies that $f(x)$∈$F$. Finally we have $x$∈$f^-$$^1$$[F]$, Hence $f^-$$^1$$[F]$ is closed.


3

If you search for Lambert function in Wikipedia (http://en.wikipedia.org/wiki/Lambert_W_function), you could find interesting approximations for large values of the argument such as $$W(x)\approx L_1-L_2+\frac{L_2}{L_1}+\frac{L_2(L_2-2)}{2L_1^2}+\frac{L_2(2L_2^2-9L_2+6)}{6L_1^3}+\cdots$$ where $L_1=\log(x)$ and $L_2=\log(L_1)$ (this is for the pricipal ...


3

I give the solution according to Lucian's comment. We start things off by using the $\zeta$ Riemann's function, defined as: $$\zeta(s)= \sum_{n=1}^{\infty}\frac{1}{n^s}, \; \mathfrak{Re}(s)>1$$ which converges absolutely since all terms are positive. That the series converges if $\mathfrak{Re}(s)>1$ is an immediate consequence of the integral test. ...


3

One method is to recognize that $x^2 y''$ is frequently seen in many differential equations whereas the other terms are not. With this then one can consider a function of the type $y(x) = f(x) g(x)$. \begin{align} y(x) &= f g \\ y' &= f g' + f' g \\ y'' &= f g'' + 2 f' g' + f'' \end{align} Now, \begin{align} 0 &= x^2y''+(2x^2+x)y'+(2x^2+x)y ...


0

You can write the two hypergeometric functions as integrals (zero to one or zero to infinity). This integrals are integrals of powers of linear functions of the variables i.e. powers of polynomials. Then integrate to find another integral of polynomials. This will be another hypergeometric function. Another way is to expand in series both hypergeometric ...


4

Hint The problem is better conditioned if, instead, you look for the roots of $$h(x)=\log(x)\cos(x)-\sin(x)=0$$ and the convergence of Newton method is much faster. For example, using $x_0=4$, the iterates are : $4.09701$, $4.09546$ which is the solution for six significant digits. Using $x_0=7$, the iterates are : $7.42088$, $7.39041$, $7.39037$. Using ...


3

There are some significant problems with the content of that page. First of all, the $L$-series is usually defined as an Euler product, but the product appearing at the bottom of that page is not correct (when $p$ is good, the Euler factor should be $(1-a(p)p^{-s}+p^{1-2s})^{-1}$. With this definition, then $$L(E,s) = \sum_{n\geq 1} \frac{a(n)}{n^{s}}$$ ...


1

Suppose we are interested in the value of $$S_b(n) = \sum_{k=0}^n \frac{(-1)^{k+1}}{(n-k+1)!(k+1)!} \frac{\Gamma(b+n+k+2)}{\Gamma(b+k+1)}.$$ This is $$\sum_{k=0}^n \frac{(-1)^{k+1}}{(n-k+1)!(k+1)!} (n+1)! {b+n+k+1\choose n+1} \\ = \frac{1}{n+2} \sum_{k=0}^n \frac{(-1)^{k+1} (n+2)!}{(n-k+1)!(k+1)!} {b+n+k+1\choose n+1} \\ = \frac{1}{n+2} \sum_{k=0}^n ...


1

$$ \int e^{ax} dx = \frac{1}{a} e^{ax} + c $$ Take $\left .\frac{d}{da}\right |_{a=1}$ on both sides $n$ times, and algebra to get rid of $(-1)^n$, you'll have an integral equal to $n!$. This is an intuitive way to get the Gamma function. You've shown that for integers it holds from this simple derivation. Mathematicians then went through a great deal ...


2

Hint. This works very directly. Suppose you have a sequence $(x_n)$ in $f^{-1}[F]$ converging to $x \in \mathbf R$. As $f$ is continuous (on the whole of $\mathbf R$!) we know that $f(x_n)$ converges to? As $F$ is closed and $f(x_n) \in F$, this implies $\ldots$.


3

I have come up with following expression using books: 1) Integrals and Series - Special Functions, and 2) Table Of Integrals, Series And Products. This looks different from what you exactly need, but with a simple transformation, you can find solution for when $b=c$. Further, if you deeply go through the literature, you can also find more general solution ...


5

$$ x^2=e^x\implies x/2=\pm\tfrac12e^{x/2}\implies-x/2\,e^{-x/2}=\pm\tfrac12 $$ Therefore, $$ x=-2\mathrm{W}\!\left(\pm\tfrac12\right) $$ Since $\mathrm{W}(x)$ is real only for $x\ge-\frac1e$, we only have one real solution: $$ x=-2\mathrm{W}\!\left(\tfrac12\right)=-0.70346742249839165205 $$


5

The Lambert W function is the inverse of $xe^x$. We want to find the inverse of $e^x x^{-2}$, dividing by $x^2$. Now: \begin{align} y &= x^{-2} e^x \\ y^{-0.5} &= x e^{-0.5x} &&\vee y^{-0.5} = -x e^{-0.5x} \\ -0.5y^{-0.5} &= -0.5x e^{-0.5x} &&\vee 0.5y^{-0.5} = -0.5x e^{-0.5x} \\ W(-0.5y^{-0.5}) &= -0.5x &&\vee ...


7

Write the equation as $x^2e^{-x} = 1$. Then $x^2e^{-x} = 4\left(\left(-\frac{x}{2}\right)e^{-\frac{x}{2}}\right)^2$.


0

Hint: The incomplete beta function in the integrand has the integral representation $$\operatorname{B}_{x}{\left(a-1,0\right)}:=\int_{0}^{x}\frac{t^{a-2}}{1-t}\,\mathrm{d}t;~~~\small{\Re{\left(a\right)}>1}.$$ Thus, for $\Re{\left(a\right)}>1$, integrating by parts yields: $$\begin{align} \int ...


0

Jn(x+iy)=Jn(x) + Jn(iy)=Jn(x) + In(y)(i)^n where In is the modified Bessel function. you got to have a modified Bessel function.


0

Does anyone know how to evaluate the following integral ? No. No one knows how to evaluate that integral. The proof is by reduction to the absurd: If anyone would have known how to express that integral in closed form, then the much simpler case with $q=0,~\alpha=1,$ and only one $\Gamma(s)$ in the denominator would also have been known to possess a ...


1

The problematic component of the proposed integral is the term $e^{ab x^2}$. This can be eliminated by expanding this exponential into the corresponding power series for which the integral in question becomes \begin{align} I &= \sum_{n=0}^{\infty} \frac{(ab)^{n}}{n!} \, \int_{0}^{1} \, e^{-ax} \, x^{2n + \mu-1} \, (1-x)^{\nu - 1} \, dx \\ &= ...


0

Let's denote two most general linearly independent solutions of Mathieu equation with characteristic $a$ as $\operatorname{C}(a,q,x)$ and $\operatorname{S}(a,q,x)$. Now if we want the solution to be $2\pi n$-periodic, where $n\in\mathbb{N}$, we'll have to consider special forms of these solutions. Even solution will require characteristic $a_r(q)$ where ...


1

You can define not ony two couples of independant solutions (C,S) and (ce,se), but as many couples as you want : $$F=\alpha C + \beta S $$ $$G=\mu C + \nu S $$ where $\alpha, \beta, \mu, \nu$ are any constants, with condition $\alpha \nu - \beta \mu \neq 0$ Then $F$ and $G$ is a couple of independant solutions. But $C$, $S$, $F$, $G$ are not four ...


0

So we have to prove that $j:\mathbb{H} \longrightarrow \mathbb{C}$ is surjective. By the given q series of $j(\tau) $it is clear that $j(\tau)$ $$ j(\tau ) = \frac{1}{q} +744+ 196884q+ 21493760q^2.+ c_2q^3......\qquad where \quad q = e^{2\pi i \tau}$$only goes to infinty when q=0 or the imaginary part of $\tau$ goes to $\infty$. and the pole is a order of ...


0

I'm not entirely sure how useful this partial answer is for you, but we could probably use the ideas from this paper by Flajolet and Prodinger on generalizing Dobiński's formula to complex arguments. To wit, since $$\mathscr{B}_k(z)\exp\,z=\sum_{n=0}^\infty \frac{n^k z^n}{n!}$$ where $\mathscr{B}_k(z)$ is the Bell polynomial, and we have the exponential ...


2

Surprisingly, there are formulae, though a bit complicated (as they involve the Meijer $G$-function). In this paper, Brychkov and Geddes display some formulae for $\mathbf{H}_0^\ast(z)$ and $\mathbf{L}_0^\ast(z)$, where we borrow the "Petiau notation" $$f_\nu^\ast(z)=\left.\frac{\mathrm d}{\mathrm du}f_u(z)\right|_{u=\nu}$$ from a previous paper by ...


3

Your proof is perfectly fine. It seems to me that this approach is what the asker had in mind.


1

Set $x=6\pi$ and $t=\frac{\pi}{2}$ and the two formulas don't agree. you might want $$ \sin(-x+t)u(t)$$


2

Something is wrong. $$\sin(-x+t)u(-x+t)=\begin{cases}0&\text{if $t\le x$}\\\sin(-x+t)&\text{if $t>x$}\end{cases} $$ For your given function, you'd need someting like $f(x,t)=\sin(-x+t)u(t)$ and be careful what happens at $t=0$ (there are different specifications of $u$ in the wild, sometimes $u(0)=\frac12$, sometimes $u(0)=1$, ...)



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