# Tag Info

## New answers tagged special-functions

3

Since, $m$ and $m^2$ have the same parity, $$\sum_{(m,n) \neq (0,0)} \frac{(-1)^{m+n}}{m^2 + n^2} = \sum_{(m,n) \neq (0,0)} \frac{(-1)^{m^2+n^2}}{m^2 + n^2}$$ It boils down to this question.

0

You have that $$f(k,x) = \frac{1}{\ln(k)} e^{x\ln(k)} (x\ln(k)) = y$$ Hence, the inverse of your function is $$\frac{1}{\ln(k)} W(y\ln(k))$$ Indeed, $$f\left(k,\frac{1}{\ln(k)} W(y\ln(k)) \right) = \frac{1}{\ln(k)}\exp \left( \ln(k)\frac{1}{\ln(k)} W(y\ln(k)) \right)\ln(k)\frac{1}{\ln(k)} W(y\ln(k))$$ = \frac{1}{\ln(k)}\exp \left( W(y\ln(k)) ... 1 The inverse of f(k,x) can be found in terms of the Lambert W. \begin{align*} y&=k^xx \\ y &= e^{x \ln k}x \\ y\ln k &= e^{x \ln k}x\ln k \\ W(y\ln k) &= x\ln k \\ x &= \frac{W(y\ln k)}{\ln k} \end{align*} 1 (A partial answer.) I tested your cfrac with the order 12 discussed by Naika (which in turn is a special case of a general cfrac by Ramanujan) and labeled as D_1(q) here,D_1(q)= \dfrac{q(1-q)} {1-q^3+\dfrac{q^3(1-q^2)(1-q^4)} {(1-q^3)(1+q^6)+\dfrac{q^3(1-q^8)(1-q^{10})} {(1-q^3)(1+q^{12})+\dfrac{q^3(1-q^{14})(1-q^{16})} {(1-q^3)(1+q^{18})+\ddots }}}} ...

27

You may write, for $N \geq 2$, \begin{align} e^{3/2}\prod_{n=2}^{N}e\left(1-\dfrac{1}{n^2}\right)^{n^2}&=e^{3/2}\times\prod_{n=2}^{N}e\times\prod_{n=2}^{N}\left(1-\dfrac{1}{n^2}\right)^{n^2}\\\\ &=e^{3/2}\times e^{N-1}\times\prod_{n=2}^{N}\dfrac{(n-1)^{n^2}}{n^{n^2}}\dfrac{(n+1)^{n^2}}{n^{n^2}}\\\\ &=e^{3/2}\times ... 11 One can write the log of the product as\sum_{n=2}^{\infty} \left [1+n^2 \log{\left (1-\frac1{n^2} \right )} \right ] $$Now,$$\log{\left (1-\frac1{n^2} \right )} = -\int_0^1 \frac{du}{n^2-u} $$So the sum is equal to$$-\int_0^1 du \, u \sum_{n=2}^{\infty} \frac1{n^2-u} \sum_{n=-\infty}^{\infty} \frac1{n^2-u} = -\frac{\pi \cot{\pi ...

2

Make your life easier writing$$B=\frac{2hf^3}{c^2\left(e^\frac{hf}{kT}-1\right)}=\frac{\alpha}{e^{\beta}-1}$$ using $\alpha=\frac{2hf^3}{c^2}$ and $\beta=\frac{hf}{kT}$. So $$\frac{dB}{dT}=\frac{dB}{d\beta}\times\frac{d\beta}{dT}$$ Now $$\frac{dB}{d\beta}=-\frac{\alpha e^{\beta }}{\left(e^{\beta }-1\right)^2}$$ $$\frac{d\beta}{dT}=-\frac{hf}{kT^2}$$ So ...

2

Let the integral in question be \begin{align}\tag{1} I_{a} = \int_{0}^{\infty} e^{-st} \, \frac{\ln(1+a t)}{1 + t} \, dt. \end{align} For the case of $a=1$ the following is obtained. By utilizing \begin{align} \int_{0}^{\infty} e^{-s t} \, \ln^{2} t \, dt = \frac{1}{s} \, ( \zeta(2) + (\gamma + \ln s)^{2} ) \end{align} for the change $t \to t+1$ leads to ...

2

Integrating by parts and using the known series results, we get that $$\int_0^1 \frac{\text{Li}_2 \left(-\frac{1}{1-z}\right)-\text{Li}_2 \left(-\frac{1}{1+z}\right)}{z}dz$$ $$=\int_0^1 \frac{\log (z+2) \log (z)}{z+1}dz+\underbrace{\int_0^1\frac{\log (2-z) \log (z)}{1-z} dz}_{\large \sum _{k=1}^{\infty } \frac{(-1)^k H_k}{k^2}=-5/8 \zeta ... 1$$B(f,T)=\frac{2hf^3}{c^2}\frac{1}{e^\frac{hf}{kT}-1}\frac{\partial B}{\partial T}=\frac{2hf^3}{c^2}\frac{\partial}{\partial T}\left(\frac{1}{e^\frac{hf}{kT}-1}\right)==-\frac{2hf^3}{c^2}\left(\frac{1}{e^\frac{hf}{kT}-1}\right)^2\frac{\partial}{\partial T}\left(e^\frac{hf}{kT}-1\right)=$$... 2 My suggestion is to look at \log B. Take the derivate of the logged function wrt to T and multiply by B(T). 1 This is conditional on Dickson's conjecture, but may be of interest. You may choose any admissible prime k-tuple (b_1=0,b_2,b_3,\ldots,b_k). By definition the b_i avoid some residue modulo every prime, and hence so do \{(2+b_i)n+1\}. Then it is a consequence of Dickson's conjecture that q_i=(2+b_i)n+1 are simultaneously prime for 1\le i \le k for ... 0 There is Legendre's formula which counts the number of positive integers less than or equal to a number n which are not divisible by any of the first k primes:$$\begin{align} &\phi(n,k)=\lfloor n \rfloor-\sum_{p_i\le k}\left\lfloor \dfrac{ n }{(p_i)}\right\rfloor+\sum_{p_i<p_j\le k}\left\lfloor\dfrac{ ...

9

I changed my evaluation slightly, and I was able to get the result in a very simple form. First notice that $$\int_{-\infty}^{\infty} \text{Ei}^{2}(-|x|) e^{ikx} \, dx = 2 \int_{0}^{\infty} \text{Ei}^{2}(-x) \cos(kx) \, dx.$$ Then integrating by parts, and assuming for now that $k >0$, \begin{align}\int_{-\infty}^{\infty} \text{Ei}^{2}(-|x|) ... 1 Here's another variation of the theme based upon Stirling Numbers. Starting from \begin{align*} f_n(x)&=\int_1^x\binom{t-1}{n}dt =\frac{1}{n!}\int_1^x{(t-1)}_ndt =\frac{1}{n!}\int_0^{x-1}{(u)}_ndu \end{align*} we can use the Stirling Numbers of the first kind s(n,k) which can be defined for n\geq 0 and 0\leq k \leq n by ... 0 Some notes: \begin{align} {}_{2}F_{1}(i, 1; 1+i; x) = \sum_{n=0}^{\infty} \frac{(i)_{n} \, x^{n}}{(1+i)_{n}}. \end{align} Now, \begin{align} \frac{(i)_{n}}{(1+i)_{n}} = \frac{\Gamma(i+1) \, \Gamma(n+i)}{\Gamma(i) \, \Gamma(n+i+1)} = \frac{i}{n+i} = \frac{1}{1-i n} = \frac{1+i n}{n^{2}+1} \end{align} which leads to \begin{align} {}_{2}F_{1}(i, 1; 1+i; x) = ... 3 Got a reference on MO; see http://mathoverflow.net/questions/210144/number-of-primes-one-larger-than-divisors-of-a-fixed-number-which-is-lcm-of-1-2#comment520981_210144 and http://www.math.drexel.edu/~eschmutz/PAPERS/lambda.pdf In Theorem 1 on the first page of the Erdos-Pomerance-Schmutz article, they announce the existence of a constant c and a ... 2 There is indeed a relation with the Bernoulli numbers of second kind. If I'm right, this is the result : Theorem : generating function of f_n(x)\sum_{n=0}^{+\infty} f_n(x)z^n = \dfrac{(1+z)^{x-1}-1}{\log (1+z)}$$Corrolary :$$f_n(x) = \sum_{k=0}^n {{x-1}\choose{k+1}} \dfrac{b_{n-k}}{(n-k)!}$$I'll present here mainly the "formal" steps ... 3 Take Cauchy's differentiation formula:$$ f^{(n)}(a) = \frac{n!}{2\pi i} \oint_\gamma \frac{f(z)}{(z-a)^{n+1}}\, dz $$and plug a holomorphic f such that f^{(n)}(a)=1. For example, f(z)=\exp(z), a=0, and \gamma the unit circle:$$ \frac{1}{n!} = \frac{1}{2\pi i} \oint_\gamma \frac{e^z}{z^{n+1}}\, dz $$Does that count? 3 You can write the inverse Laplace transform of 1/s^{n+1}, evaluated at t=1, as 1/n!. The integral is$$ \int_{c-i\infty}^{c+i\infty}\frac{1}{2\pi is^{n+1}}e^s\,ds=\frac{1}{n!}, $$for suitable real c. 8 Following @RobertIsrael, we have$$\frac{1}{2\pi}\int_0^{2\pi}e^{e^{i\phi}}e^{-in\phi}d\phi=\oint_{|z|=1}e^{z}z^{-n}\frac{dz}{iz}\tag 1$$We note that the integrand on the right-hand side of (1) has a pole of order n+1 at z=0. The residue is given by$$\text{Res}\left(-i\frac{e^z}{z^{n+1}},z=0\right)=\frac{1}{n!}\lim_{z\to ...

11

$$\dfrac{1}{2\pi} \int_0^{2\pi} e^{e^{i\theta}} e^{-in\theta}\; d\theta$$

2

This is related. The function $$f(a,b,x)=\frac{\sin[b\sqrt{a^{2}+x^{2}}]}{\sqrt{a^{2}+x^{2}}}$$ has Fourier transform $$\hat{f}(a,b,s)= \left\{\begin{array}{cc} \frac{\pi}{2}J_{0}[a\sqrt{b^{2}-s^{2}}], & 0 < s < b \\ 0 & b < s < \infty \end{array}\right.$$ The derivative of ...

5

This time I let the target Carmichael number be the least common multiple of the numbers from $1$ to $w.$ This is more efficient in terms of the number of divisors. The Superior Highly Composite Numbers and the Colossally Abundant Numbers share the main property of this LCM, which is that the exponent of some prime $p$ is proportional to $1/ \log p.$ As a ...

1

It took a while, but I wrote something in C++ with GMP to find the biggest possible $n$ that has a fixed Carmichael number, furthermore I took the Carmichael numbers to be $w!$ for $4 \leq w \leq 12,$ which takes a fair amount of time as it is. The quantity $\log f(n) / \log n$ evidently gets arbitrarily close to $1$ this way. ...

5

If you want a symmetrical generalization for $n$th roots then define $$C_{k,n}(z)=\frac{1}{n}\sum_{\zeta^n=1}\zeta^k e^{\zeta z}.$$ Then $C_{0,2}(z)=\cosh z$ and $C_{1,2}(z)=\sinh z$. Note that $C_{k,n}(z)=C_{0,n}^{(k)}(z)$. In particular, for $n=3$, if $\omega$ is the cube root of unity in the upper half plane then $$\begin{array}{lll} C_{0,3}(z) & ... 1 One thing you definitely want to do is simply find the sequence of integers for which your quantity increases, and factor those. This is the entire story for a number of optimization problems that go back to Ramanujan. Those are all multiplicative functions. The bad news for you is that the Carmichael function is not multiplicative. Please take a look at ... 6 Since we know the value of \eta(3i), the point is just to compute the value of the product:$$ \prod_{n\geq 0}(1+e^{-6\pi n})=\exp\sum_{n\geq 0}\log\left(1+e^{-6\pi n}\right)=\exp\sum_{n\geq 0}\int_{n}^{n+1}\frac{6\pi n}{1+e^{6\pi s}}\,ds$$where:$$\sum_{n\geq 0}\int_{n}^{n+1}\frac{6\pi n}{1+e^{6\pi s}}\,ds = \int_{0}^{+\infty}\frac{6\pi s\,}{1+e^{6\pi ...

5

After persevering with a Mathematica session, I found that $F(6i)$ is the root of $96$-deg eqn (no wonder it was hard to find!) but could be prettified as, $$\eta(6i) = \frac{1}{2\cdot 6^{3/8}} \left(\frac{5-\sqrt{3}}{2}-\frac{3^{3/4}}{\sqrt{2}}\right)^{1/6}\,\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$$ However, the second question is still open.

0

By exploiting: $$\sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b} \tag{1}$$ we have that our limit equals: $$\begin{eqnarray*} \lim_{x\to 0^+}\left(\frac{1}{x}+\sum_{n\geq 0}\left(\frac{1}{n+x}-\frac{1}{n+\frac{x}{2}}\right)\right)&=&\lim_{x\to 0^+}\sum_{n\geq 1}\left(\frac{1}{n+x}-\frac{2}{2n+x}\right)\\&=&-\lim_{x\to ... 2 The following is not a direct answer to your question, but is rather too long for a comment. This is basically Ramanujan's approximation$$\pi \approx \frac{24}{\sqrt{n}}\log(2^{1/4}g_{n}) = \frac{6}{\sqrt{n}}\log (2g_{n}^{4}) = \frac{6}{\sqrt{n}}\log (2u)\tag{1}$$where g_{n} is Ramanujan's class invariant given by$$g_{n} = 2^{-1/4}e^{\pi\sqrt{n}/24}(1 ...

2

1. $f(x)=-x\ln{|x|}$. Verify: $f(0^{+})=f(0^{-})=0$, so $\lim_{x\to0}f(x)=0$; $f^{'}(x)=-\ln{|x|}-1$, so $\space f^{'}(0^{+})=+\infty$. 2. $f(x)=\frac{\arctan{x}}{\sqrt{|x|}}$.Verify: $f(0^{+})=f(0^{-})=0$ (L'Hôpital's rule: $\lim_{x\to0^{+}}f(x)=lim_{x\to0^{+}}\frac{\frac{1}{1+x^2}}{\frac{1}{2\sqrt{x}}}=im_{x\to0^{+}}\frac{2\sqrt{x}}{1+x^2}=0$) ...

0

From you question, it is unclear exactly what shape you are looking for, and there are many functions that could describe the behaviour you're after. However, two possible options could be the negative exponential and a negative gompertz function. Possible forms of these could be: Negative exponential $y(x)=e^{−ax+ln(1-b)}+b,$ where $b=0.2$ and a is a ...

0

How about something like $$y=1-0.8\,\frac{e^{ax}-1}{e^{1095a}-1}?$$ The sign of $a$ determines concavity or convexity; $a=-0.005$ gives a nice graph.

1

I have been presented with a similar problem, that is why I am answering in this section. Joriki, do you use the p-test to test for convergence? If you have two functions $f(x)$ and $g(x)$ with $f(x)\geq g(x)\geq0$, then if the integral of $g(x)$ diverges, so should the integral over $f(x)$. And if the integral over $f(x)$ converges, then this should also ...

3

I'm going to use the following 3 identities along with the known value $\text{Li}_{2} \left(\frac{1}{2} \right) = \frac{\pi^{2}}{12} - \frac{1}{2} \log^{2}(2)$: $$\text{Li}_{2}(1-z) = - \text{Li}_{2} \left(1- \frac{1}{z} \right) - \frac{1}{2} \log^{2} (z) , \quad z \notin (-\infty,0] \tag{1}$$ $$\text{Li}_{2}(z) = - \text{Li}_{2} \left(\frac{1}{z} \right) ... 1 With the help of Mathematica I got:$$\mathcal{L}\left(\frac{J_1(R\sqrt{x})}{x^{3/2}}\right)=\frac{|R|}{2}\left(1-2\gamma-\frac{4s}{R^2}\left(1-e^{-\frac{R^2}{4s}}\right)+\log\frac{4}{R^2}-\Gamma\left(0,\frac{R^2}{4s}\right)\right)$$and by expanding the RHS as a series it is not difficult to check it matches your series. 0 Substitute z = e^{u}, e^{u/2}du = \frac{dz}{\sqrt z}.$$ \int \sqrt{z\frac{z + z^{-1}}{2} - z\cos v} \frac{dz}{z} = \frac{1}{\sqrt{2}}\int \frac{\sqrt{z^2 + 1 - 2z\cos v}}{z} dz = (9) = \\ = \frac{1}{\sqrt{2}}\sqrt{z^2 + 1 - 2z\cos v} - \frac{\cos v}{\sqrt{2}} \int\frac{dz}{\sqrt{z^2 + 1 - 2z\cos v}} + \frac{1}{\sqrt{2}} \int \frac{dz}{z\sqrt{z^2 + 1 - ...

7

At Vladimir Reshetnikov's request I'm going to show how to find an antiderivative for $$\frac{\log(1+2x) \log(1-x)}{1+x}$$ using the identity $$2 \log(x) \log(y) = \log^{2}(x) + \log^{2}(y) - \log^{2} \left(\frac{x}{y} \right)$$ where $x$ and $y$ are positive real values. \begin{align} &\int \frac{\log(1+2x) \log(1-x)}{1+x} \, dx \\ &= ... 1 Note that by Rodrigues' formula, P_{n+1}=\frac{1}{2^{n+1}(n+1)!}D^{n+1}[(x^{2}-1)^{n+1}]. Multiplying by (n+1), we have: \begin{align*} &{\left(n+1\right)P_{n+1}(x)}={\left(n+1\right)\frac{1}{2^{n+1}(n+1)!}D^{n+1}[(x^{2}-1)^{n+1}]}\\ =\;&{\left(n+1\right)\frac{1}{2^{n}n!}D^{n}[x(x^{2}-1)^{n}]}\\ \end{align*} Using the Leibniz rule: \begin{align*} ... 1 Using Rodrigues' formula \displaystyle P_n(x)=\frac{1}{2^n n!}D^n\left(x^2-1\right)^n, we can write \begin{align*} &\color{blue}{\left(n+1\right)P_{n+1}(x)}-\color{green}{\left(2n+1\right)xP_n(x)}+\color{red}{nP_{n-1}(x)}=\\ ... 1 Based on John Barber's comment, I plotted the fractal he mentioned using the domain coloring below: The above image assigns a color to every point in the complex plane (only [-10,10] of each axis is shown). Plotting \log(z)/\sqrt z gives: And iterating 100 times gives: There appears to be 5 colors: yellow, dark yellow, brown, dark brown, and red ... 1 In reference to Kirill's answer, I will show that indeed I(3) = \int_{1}^{\infty} \int_{1}^{\infty} \int_{1}^{\infty} \frac{1}{xyz} \frac{dx \, dy \, dz}{x+y+z-2} = \frac{7}{2}\zeta(3).$$I will make the same change of variables I made in my answer to your other more recent question.$$ \begin{align} I(3) &= \int_{1}^{\infty} \int_{1}^{\infty} ...

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