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3

Ok, i will give it a shot: Writing $\int_{0}^{\infty}e^{-t(x+2)}=\frac{1}{x+2}$ and using $\Im(e^{ix})=\sin(x)$ we may reformulate the problem as follows: $$ I=\Im\left[\int_0^{\infty}dte^{-2 t}\underbrace{\int_0^{\infty}dxe^{i\pi x^2-tx}}_{J(t)}\right] $$ the inner intgral $J(t)$ is quite straightforward (and also well known because it is just the laplace ...


5

Integrating $f(z)=\dfrac{e^{iaz^2}}{z+b}$ along $[0,R]\cup Re^{i[0,\pi/2]}\cup i[R,0]$ gives \begin{align} \int^\infty_0\frac{\sin(ax^2)}{x+b}\ {\rm d}x &=\int^\infty_0\frac{b\cos(ay^2)-y\sin(ay^2)}{y^2+b^2}\ {\rm d}y \end{align} To compute the first integral, we consider the function $\displaystyle I(a)=\int^\infty_0\frac{e^{iay^2}}{y^2+b^2}\ {\rm d}y$ ...


1

The integral can be written: $$ \int^{\infty}_{0}\frac{x^n}{x^{m+n+1}}dx=\int^{\infty}_{0} x^nx^{-m-n-1} dx \\ =\int^{\infty}_{0} x^{-m-1} dx=\int^{\infty}_{0} \frac{1}{x^{m+1}} dx $$ which does not converge. You may be confusing the integral you have with the Beta function, which has the representation $$ ...


0

I would note that the j_0's are just sin y/y and so you can rewrite the numerators as proportional to exp[ibcos x] - exp[-ibcos x]. Then the whole integral will involve only complex exponentials and 1/(b cos x* b sin x). You can then pull out a and b such that you have factors of the form exp[i cos x]^b. A change of variable u = cos x will then convert ...


2

Too long for a comment: Here's a little intuitive tip: What do $~\dfrac{\sin t}t~$ and $~\dfrac{\cos t}{t^2}~$ both have in common? They are even functions. So, if you notice various series or integrals whose summand or integrand belongs to this category having a nice closed form, that should not surprise you. For instance, ...


4

"What remains" is actually the only nontrivial part of $\operatorname{Li}_2\left(e^{ix}\right)$, given by Clausen function. The fact that one can evaluate cosine series is related to the identity $$\operatorname{Li}_2\left(z\right)+\operatorname{Li}_2\left(z^{-1}\right)=-\frac{\ln^2\left(-z\right)}{2}-\frac{\pi^2}{6}.$$ Slightly rephrasing, we can rewrite ...


3

Here is a solution only using dilogarithm identities: $$ \operatorname{Li}_2(z) + \operatorname{Li}_2(1-z) = \zeta(2) - \log z \log(1-z). \tag{1} $$ $$ \operatorname{Li}_2(1-z) + \operatorname{Li}_2(1-\tfrac{1}{z}) = -\tfrac{1}{2}\log^2 z. \tag{2} $$ $$ \operatorname{Li}_2(z) + \operatorname{Li}_2(-z) = \tfrac{1}{2}\operatorname{Li}_2(z^2). \tag{3} ...


3

To complete the excellent answer of @JackD'Aurizio: \begin{align*} \Im\operatorname{Li}_2\left(1+i\sqrt3\right)=\frac{\pi \ln 2}{2}+\frac{5\sqrt3}{72}\left[\psi_1\left(\frac13\right)-\psi_1\left(\frac23\right)\right], \end{align*} where $\psi_1(z)$ denotes the trigamma function.


5

For first, we have: $$I=\Re\,\text{Li}_2(1\pm i\sqrt{3})=\Re\int_{1}^{i\sqrt{3}}\frac{\log t}{1-t}\,dt=\frac{\pi^2}{6}+\Re\int_{0}^{i\sqrt{3}}\frac{\log t}{1-t}\,dt\tag{1}$$ hence: $$ I = \frac{\pi^2}{6}-\int_{0}^{\sqrt{3}}\frac{\frac{\pi}{2}+t\log t}{1+t^2}\,dt\tag{2}=-\int_{0}^{\sqrt{3}}\frac{t\log t}{1+t^2}\,dt $$ but: $$\begin{eqnarray*} \int\frac{t\log ...


0

For this to have an answer, I slightly expand my comment: This ODE is a particular case of the confluent Heun equation $$\frac{{d}^{2}w}{{dz}^{2}}+\left(\frac{\gamma}{z}+\frac{\delta}{z-1}+\epsilon% \right)\frac{dw}{dz}+\frac{\alpha z-q}{z(z-1)}w=0,$$ with ...


1

If $y(x)$ is a solution of the first equation, let $$ z \left( x \right) =y \left( x \right) {{\rm e}^{\beta\,x/2}}{x}^{ \gamma/2} \left( x-1 \right) ^{\delta/2}$$ Then if I'm not mistaken, $z$ satisfies the second differential equation with $$ A = -\dfrac{\beta^2}{4},\ B = \dfrac{2q + (\delta-\beta)\gamma}{2},\ C = \dfrac{2\alpha\beta - ...


3

To explain how an algebraic equation for $Z:=\operatorname{cn} \left(\frac{2K}{3}|m\right)$ can be obtained, notice that The usual parity properties imply that $$\operatorname{cn} \left(\frac{4K}{3}\biggl|\,m\right)=\operatorname{cn} \left(2K-\frac{2K}{3}\biggl|\,m\right)=-\operatorname{cn} \left(-\frac{2K}{3}\biggl|\,m\right)=-\operatorname{cn} ...


3

Assuming $1/2$ is $k$, as in Maple ... Maple JacobiCN(2/3*EllipticK(1/2),1/2) evaluates to $0.473058826656122429170671314726$. From ISC we find that this is a solution of $$ Z^4-2Z^3-6Z+3=0 $$ which may be written $$ {\frac { \left( 1+2\,\sqrt [3]{6} \right) ^{3/4}+\sqrt [4]{1+2\, \sqrt [3]{6}}-\sqrt {-2\,\sqrt [3]{6}\sqrt {1+2\,\sqrt [3]{6}}+2\, \sqrt ...


5

The Weierstraß $\wp$-function has a pole of order $2$ at all lattice points. Hence $\wp^{(m)}$ has a pole of order $m+2$ at all lattice points. If $p_0,p_1,\dotsc, p_{k-1}$ are polynomials, the function $$h(z) = \sum_{m = 0}^{k-1} p_m(z)\wp^{(m)}(z)$$ has at most a pole of order $k+1$ at the lattice points, but if $p_k$ is a nonzero polynomial, then ...


1

There is no closed form of this integral that can be readily determined. It can be determined that $\textrm{sech}^{2}(x)$ has coefficients related to Bernoulli numbers and are similar in form to those of $\tanh(x)$ and has the form $$\textrm{sech}^{2}(x) = \sum_{n=0}^{\infty} \frac{(-1)^{n} \, \theta_{n}}{n!} \, x^{2n}.$$ Now the integral in question is $$ ...


4

Assume $a$ is a real number such that $a^{s-1}\neq1$. If you set $$ z(s)=\zeta (s) \left(1-\frac{1}{a^{s-1}}\right), $$ then, using the functional equation for $\zeta$, you obtain $$ \begin{align} z(1-s)&=\zeta (1-s) \left(1-\frac{1}{a^{-s}}\right)\\\\ &=2\frac{\Gamma(s)}{(2\pi)^s} \cos\left(\frac{\pi}{2}s\right) \zeta(s)\times ...


4

Expanding my comment: the substitution $1-x^2\mapsto x$, followed by expanding the trilogarithm and keeping in mind Legrende's duplication formula $B(n+1,\frac12)=2^{2n+1}B(n+1,n+1)$, we arrive at $$I=\sum_{n=1}^{\infty} ...


2

If we define the ramp function $r$ as $$ r(t)= \begin{cases} t, & t\ge 0\\\\ 0, & t<0 \end{cases}$$ then the function $f$ plotted in the post can be represented as $$f(t)=r(t)-r(t-2)$$ Note that if one introduces (i.e., adds) a step function, the resulting plot would exhibit a jump discontinuity. Inasmuch as the plot exhibits no jump, then ...


5

The answer is yes. Given the nome $q = \exp(i\pi\tau)$, elliptic lambda function $\lambda(\tau)$, Dedekind eta function $\eta(\tau)$, Jacobi theta functions $\vartheta_n(0,q)$, and Ramanujan's octic cfrac, the following relations are known, $$\begin{aligned} u(\tau) & = \big(\lambda(\tau)\big)^{1/8} = \frac{\sqrt{2}\, \eta(\tfrac{\tau}{2})\, ...


3

To prove the third identity, it suffices to take the limit $z\to -\frac12$ of the Ramanujan's cubic transformation. It yields $$_2F_1\left(\frac13,\frac23;1;-\frac18\right)=\lim_{z\to-\frac12^+} \frac{_2F_1\left(\frac13,\frac23;1;1-\left(\frac{1-z}{1+2z}\right)^3\right)}{1+2z}=\frac23\,C_{\mathrm{B4CC}}^2,\tag{1}$$ where the limit is evaluated using the ...


5

To prove L.G. result, one just needs to apply twice integration by parts, then prove through its favourite technique (for instance, differentiation under the integral sign and computation of a few derivatives of a Beta function) that: $$I_0=\int_{0}^{1}x^2 \log(x)\,\frac{dx}{1-x^2}=1-\frac{\pi^2}{8}, $$ $$I_{-}=\int_{0}^{1}x ...


2

$$-\frac72\zeta\left(3\right)+\pi^2\left(\ln 2-1\right)+8$$


1

$\int_{1}^{x} [u]([u]+1) f(u) du = 1 \cdot 2 \cdot \int_{1}^{2} f(u) du + 2 \cdot 3 \int_{2}^{3} f(u) du + \dots + [x]\cdot ([x]+1) \int_{[x]}^x f(u) du = 2 \cdot (\int_{1}^{2} f(u) du + \int_{2}^{3} f(u) du + \int_{3}^{4} f(u) du + \dots + \int_{[x]}^{x} f(u) du) + \\ 4 \cdot (\int_{2}^{3} f(u) du + \int_{3}^{4} f(u) du + \int_{4}^{5} f(u) du + \dots + ...


0

I ended up taking a different approach to show it only for half-integer $\nu$, which is 90% of the cases I care about. I think it should be possible to extend that to all $\nu > \tfrac12$, but I haven't gotten that to work yet. Define the Matérn kernel by $$k_\nu(x) = \frac{1}{\Gamma(\nu) 2^{\nu - 1}} \left( \sqrt{2 \nu} x \right)^\nu K_\nu\left( \sqrt{2 ...


0

From Antonio DJC's comment i explicit compute \begin{align} &\theta_1(q, y)= -i y^{\frac{1}{2}}q^{\frac{1}{8}} \prod_{n=1}^\infty \left(1-q^n\right)\left(1-y q^n\right) \left(1-y^{-1}q^{n-1} \right) \\ &\phantom{\Theta(q,y)}= -i y^{\frac{1}{2}}q^{\frac{1}{8}}(1-y^{-1}) \prod_{n=1}^\infty \left(1-q^n\right)\left(1-y q^n\right) \left(1-y^{-1}q^{n} ...


1

Rewrite the product in your first equation as $-iq^{1/8}y^{1/2}(1-y^{-1})\prod_{k=1}^{\infty}(1-q^k)(1-yq^k)(1-y^{-1}q^k)$ (just tweak the third part of the product), divide out $(1-y^{-1})$ from the equation, then use L'Hôpital's rule (with respect to $z$ as it approaches $0$) to calculate the limit of the left-hand theta expression; the result you want ...


1

$$K_\nu(z)=\frac{2^\nu\,\Gamma\left(\nu+\frac{1}{2}\right)}{z^\nu\sqrt{\pi}}\int_{0}^{+\infty}\frac{\cos(zt)}{(t^2+1)^{\nu+\frac{1}{2}}}\,dt $$ so we have to compare $$ I_2=\int_{0}^{+\infty}\int_{0}^{+\infty}\frac{\cos(zt)\cos(zs)}{\left((t-s)^2+(1+ts)^2\right)^{\nu+\frac{1}{2}}}\,ds\,dt$$ and $$ I_1 = ...


1

I've found a generalization of your conjecture: If $k>0$ real number, then $${\large\int}_{-1}^1\frac{dx}{\sqrt[3]{k\pm\tfrac{4\sqrt5k}{9}\,x}\ \left(1-x^2\right)^{\small2/3}}\stackrel{?}{=} \frac{1}{\sqrt[3]{k}}\cdot\frac{3^{\small13/6}}{2^{\small4/3}5^{\small5/6}\pi}\Gamma^{3}\left(\tfrac{1}{3}\right).$$ For $k=9$ with $+$ sign it gives ...


2

Looking here, there is an interesting expansion for large arguments $$E_1(x)=\frac{e^{-x}}{x}\sum_{k=0}^\infty \frac {k!}{(-x)^k}$$ For $x=25$, using $n$ terms, we have $$S_1=5.332970444146184 \times 10^{-13}$$ $$S_2=5.350747012293338 \times 10^{-13}$$ $$S_3=5.348613824115680 \times 10^{-13}$$ $$S_4=5.348955134224105 \times 10^{-13}$$ ...


0

Through the recurrence relation we have: $$ \frac{H_{2r-1}}{x} = H_{2r-2}-(2r-2)\frac{H_{2r-3}}{x}\tag{1} $$ so: $$ \frac{H_{2r-1}}{x} = H_{2r-2}-(2r-2)H_{2r-4}+(2r-2)(2r-4)\frac{H_{2r-5}}{x}\tag{2} $$ and by induction (since $\frac{H_1}{x}=H_0$): $$ \frac{H_{2r-2}}{x}=\sum_{k=0}^{r}(-1)^k H_{2r-2k}\prod_{j=1}^{k}(2r-2j) \tag{3}$$ that is easy to rearrange ...


2

Generally, power series representation is not a good choice for larger input, even if the radius of convergence if infinite. In this case, you need about 90 leading terms just simply to achieve the desired order of magnitude (and more terms for the accurate digits). Indeed, Mathematica 10.2 confirms that if we let $$ S_n = -\gamma - \log 25 - ...


2

Without seeing your code it is difficult. But I guess it is catastrophic cancellation. $E_1(25) \approx 5.3488997553\times 10^{-13}$ and the single terms are much larger, e.g. the for $n=10$ the value is $2628070.75729$ and for $n=24$ the term is $238585088.1445781!$ You will get similar problems is you compute $e^{-x}$ or $\sin(x)$ for $x=25$ using the ...


1

If the parameters $\{b_k\}$ of $_4F_3$ were a little more generic ($b_k\notin \mathbb{Z}$ and $b_j-b_k\notin\mathbb{Z}$) the proof would be very simple: it would suffice to check that the appropriate generalized hypergeometric equation is satisfied and $t\to 0 $ behaviors of both sides match. In our non-generic case the equation is easily verified but the ...


1

Indeed you can express associated Legendre functions in terms of hypergeometric functions. See, for example, Abramowitz / Stegun, section 8.1, or the Digital Library of Mathematical Functions, section 14.3: http://people.math.sfu.ca/~cbm/aands/page_332.htm http://people.math.sfu.ca/~cbm/aands/page_333.htm http://dlmf.nist.gov/14.3 In your particular ...


1

Regarding the empty product you're right. In case the lower index of a product is greater than the upper index, the product is equal to one (analogously to the empty sum which is zero in such cases). We can read e.g. in H. Bateman, Higher Transcendental Function Volume 1, section 5.3, p.207 Definition of $G$-Function: ... where an empty product ...


1

Let $f(z)$ be the ordinary generating function and $\mathcal{L}$ be the Laplace transform. Then the EGF $g(z)$ has the form $$ g(z)=\mathcal{L}^{-1}\left(\frac{1}{z} f(z)\Big |_{z=\frac{1}{z}}\right) $$


7

We can do the first two cases by integrating by parts and using the fact $$\text{Ei}(x) \sim \ln (x) + \gamma + \mathcal{O}(x) \ \text{as} \ x \to 0^{+}.$$ For the first one, we have $$ \begin{align} \int_{0}^{1} \text{Ei}(x) \, dx &= x \text{Ei}(x) \Big|^{1}_{0^{+}} - \int_{0}^{1}e^{x} \, dx \\ &= \text{Ei}(1)- e +1. \end{align}$$ And for the ...


2

Notice $\tau(p)=2$ and so for sufficiently large primes, say $p> p_r$ we have $\dfrac{2}{p^a}$ is going to be less than $\frac{1}{r}$. for any positive real $r$ Also notice $\tau(p^a)=a+1$ and so for sufficiently large $a$ we have $\dfrac{a+1}{p^{a\delta}}<\frac{1}{r}$. If we set $r=1$: This means the value $\frac{\tau(n)}{n^a}$ reaches a maximum, ...


1

Look at this part of the formula given in Lemma 1: $$ \begin{multline} -\frac{2cp^{\alpha+1}}{\sqrt{\pi}}\Gamma(-\alpha-1)G\left(\frac{1}{2},\frac{3}{2},\frac{3+\alpha}{2},1+\frac{\alpha}{2}; -\frac{c^2p^2}{4}\right) \\ + \frac{1}{c^\alpha\sqrt{\pi}\alpha}\Gamma\left(\frac{\alpha+1}{2}\right) ...


1

The very first conclusion is, strictly speaking, incorrect. The asymptotics of the Hankel function is $$H_0^{(1)}(z\to 0)\sim \frac{2i}{\pi}\ln z+O(1).$$ This looks similar to your third formula, but in fact that equation should contain an additional term proportional to $\ln k$ which diverges as $k\to 0$. This reflects the logarithmic divergence you get in ...


3

While the other response is a good exercise in exploiting recurrence relations, a much more direct route is provided by the technique of summing under the integral which converts the series into an integral of the generating function: $$\begin{align} f{\left(x\right)} &=\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}P_{n}{\left(x\right)}\\ ...


3

Let: $$f(x) = \sum _{n=0}^{\infty} \frac{x^{n+1} \, P_{n}(x)}{n+1}$$ As you said: $$f^{'}(x) = \sum_{n=0}^{\infty} x^{n} \, P_{n}(x) + \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1} \, P_{n}^{'}(x)$$ And using the generating function you showed, we get: $$f^{'}(x) = \frac{1}{\sqrt{1-x^{2}}} + \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1 } \, P_{n}^{'}(x)$$ Now, note ...


0

From the definition : $f_0(x,y)=x+y$ $f_n(x,0)=x$ $f_1(x,y+1)=2.f_1(x,y)+y+1$, so $$f_1(x,y+1)+(y+1+2)=2(f_1(x,y)+y+2)=2^{y+1}(x+2)$$ That implies $$f_1(x,y)=2^y(x+2)-y-2$$ $$f_2(x,y+1)=2^{f_2(x,y)+y+1}(f_2(x,y)+2)-(f_2(x,y)+y+1)-2$$ You can then compute using definitions : $f_2(7,1)=2294$ $f_2(3,2)=5742397643169488579854258$ $f_2(0,3)=88080360$ ...


1

Use Cauchy-Schwarz: $w^T x_i \leq ||w|| \cdot ||x_i||$ First, take the max over the right-hand side: $w^Tx_i \leq ||w|| \max_i ||x_i||$. Since this holds for all $i$ on the left-hand side, take $\max_i$ on that side, as well. So: $$ \max_i w^T x_i \leq ||w|| \max_i ||x_i|| $$ Then let $z = f(x_1, \ldots, x_N) = \frac{\max{||x_i||}}{||w||} w$.


1

Here's a hopefully-straightforward way of seeing how other folks' solutions come about.. For simplicity I'll eliminate the 'awkward' cases where any of $x,y,\alpha$ are zero or negative, and consider just $x, y, \alpha\gt0$. We can start by executing a simple change of variables: let $z=\frac xy$ and $w=xy$. It's easy to see that we don't lose any data by ...


2

Functions defined on $\dot{\mathbb R}^n:={\mathbb R}^n\setminus\{{\bf 0}\}$ and satisfying an identity of the form $$f(\lambda{\bf x})=\lambda^d\>f({\bf x})\qquad\bigl(\lambda\in\dot{\mathbb R}, \ {\bf x}\in\dot{\mathbb R}^n\bigr)$$ (or $=|\lambda|^d\>f({\bf x})$, depending on the context) for some constant $d\in{\mathbb R}$ are called homogeneous of ...


6

The point $(0,0)$ is going to be a nuisance to you in two ways. One, if you allow $\alpha = 0$ in your equation, then for all $x, y$, we have $f(x,y) = f(\alpha x, \alpha y) = f(0,0)$, so your function is constant. Okay, fine, so let's say that $\alpha \not= 0$, or $\alpha > 0$. But now even if you want $f$ to just be continuous at $(0,0)$, you've still ...


3

What comes to mind for me is the map which takes points in $\mathbb R^2$ and associates them with lines through the origin - which are the points of real projective space $\mathbb {RP}$. Points in real projective space are identified by coordinates similarly to that in $\mathbb R$, but one less coordinate is required. In this case, lines can be specified by ...


1

I used my q-continued fraction see Ramanujan theta function and its continued fraction ,which to my surprise is a q-analogoue of Gauss's well known continued fraction for pi.Let $$\psi^2(q^2) = \cfrac{1}{1-q+\cfrac{q(1-q)^2}{1-q^3+\cfrac{q^2(1-q^2)^2}{1-q^5+\cfrac{q^3(1-q^3)^2}{1-q^7+\ddots}}}}$$ And then multiplying both sides by $(1-q)$ and letting ...


3

Clearly we know that $$\vartheta_{2}(q) = 2q^{1/4}\psi(q^{2})$$ so we have $$\psi^{2}(q^{2})(1 - q) = \frac{q^{1/2}}{4}\vartheta_{2}^{2}(q)(1 - q) \to \frac{1}{4}\cdot \pi = \frac{\pi}{4}$$ from this answer to your previous question. For the sake of clarity, I have considered only real variable $q$ so that from equation $q = e^{2\pi i \tau}$ the variable ...



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