# Tag Info

1

Perhaps it is the sinc function. The Wikipedia article states: It was introduced by Philip M. Woodward in his 1952 paper "Information theory and inverse probability in telecommunication", in which he said the function "occurs so often in Fourier analysis and its applications that it does seem to merit some notation of its own",[3] and his 1953 book ...

1

\begin{align*} Ai(v^2)&=\frac1\pi\int_0^\infty \cos\left(\frac13t^3+v^2t\right)\,dt\\&=\frac1{2\pi}\int_\mathbb R\cos\left(\frac13t^3+v^2t\right)\,dt\\&=\frac1{2\pi}\int_\mathbb R\cos\left(\frac13t^3+v^2t\right)+i \sin\left(\frac13t^3+v^2t\right)\,dt\\&=\frac1{2\pi i}\int_\mathbb R e^{i (\frac13t^3+v^2t)}\,d(it)\\&=\frac1{2\pi i}\int_Ie^{...

7

Before attacking the integral, I mention something about cubic theta function. The whole solution heavily exploits tools from modular forms. The "footnote" contains more information. The three cubic theta functions are defined by \begin{aligned} a(q) &= \sum_{m,n} q^{m^2+mn+n^2}\\ b(q) &= \sum_{m,n} \zeta_3^{m-n} q^{m^2+mn+n^2}\\ c(q) &= \sum_{... 3 For c=1, If we consider THE following integral I=\int_{0}^{\infty}x^{a}e^{x^{b}}e^{-\lambda(e^{x^{b}}-1)}dy then by letting y=x^{b} we get I=\frac{e^{\lambda}}{b^{a+1}}\int_{1}^{\infty} (log~y)^{\frac{a}{b}}e^{-\lambda y}dy if we consider \frac{a}{b} is an integer then I=\frac{e^{\lambda}}{b^{a+1}} (\frac{a}{b})! E_{0}^{\frac{a}{b}}(\lambda) ... 5 Not an answer, but an extended comment for now. This hypergeometric function is a special case, and some complicated quadratic and cubic transformations apply to it. See this like for reference: https://dlmf.nist.gov/15.8. The formulas 15.8.25 and 15.8.26 both apply here. However, the most interesting one is so called Ramanujan’s Cubic Transformation (15.... 1I=\int_0^\infty\operatorname{Li}_2^2\left(-\frac1{x^2}\right)\ dx\overset{IBP}{=}-4\int_0^\infty\operatorname{Li}_2\left(-\frac1{x^2}\right)\ln\left(1+\frac1{x^2}\right)\ dx$$By using$$\int_0^1\frac{x\ln^n(u)}{1-xu}\ du=(-1)^n n!\operatorname{Li}_{n+1}(x)$$setting n=1 and replacing x with -\frac1{x^2} we can write$$\operatorname{Li}_2\left(-\...

0

Using Landen's identity $$\operatorname{Li}_2(x)+\operatorname{Li}_2\left(\frac{x}{x-1}\right)=-\frac12\ln^2(1-x)$$ set $x=\sqrt{2}-1$ we get $$\operatorname{Li}_2(\sqrt{2}-1)+\operatorname{Li}_2\left(\frac{\sqrt{2}-1}{\sqrt{2}-2}\right)=-\frac12\ln^2(2-\sqrt{2})\tag{1}$$ we have \operatorname{Li}_2\left(\frac{\sqrt{2}-1}{\sqrt{2}-2}*\color{red}{\frac{... 1 \begin{align} \int\frac{\operatorname{Li}_2(x)^2}{x}\ dx&\overset{IBP}{=}\operatorname{Li}_3(x)\operatorname{Li}_2(x)+\int\frac{\operatorname{Li}_3(x)\ln(1-x)}{x}\ dx\\ &=\operatorname{Li}_3(x)\operatorname{Li}_2(x)+\sum_{n=1}^\infty\frac1{n^3}\int x^{n-1}\ln(1-x)\ dx \end{align} This is the simplest form I can get. 1 Using the generalized integral expression of the polylogrithmic function which can be found in the book (Almost) Impossible Integrals, Sums and series page 4.\int_0^1\frac{x\ln^n(u)}{1-xu}\ du=(-1)^n n!\operatorname{Li}_{n+1}(x)$$and by setting n=1 and replacing x with x^2 we get$$\operatorname{Li}_{2}(x^2)=-\int_0^1\frac{x^2\ln u}{1-x^2u}\ du$$... 0 Since$$\frac{d}{dy}\operatorname{Li}_2(1-y)=\frac{\ln y}{1-y}Then \begin{align} \operatorname{Li}_2(1-y)|_0^x&=\int_0^x\frac{\ln y}{1-y}\ dy \quad \text{apply integration by parts}\\ \operatorname{Li}_2(1-x)-\zeta(2)&=-\ln(1-y)\ln y|_0^x+\int_0^x\frac{\ln(1-y)}{y}\ dy\\ &=-\ln(1-x)\ln x-\operatorname{Li}_2(y)|_0^x\\ &=-\ln(1-x)\ln x-\... 2 \begin{align} I&=\int_0^1\operatorname{Li}_3(1-x^2)\ dx\overset{IBP}{=}2\int_0^1\frac{x^2\operatorname{Li}_2(1-x^2)}{1-x^2}\ dx\\ &=2\int_0^1\left(\frac1{1-x^2}-1\right)\operatorname{Li}_2(1-x^2)\ dx\\ &=2\int_0^1\frac{\operatorname{Li}_2(1-x^2)}{1-x^2}\ dx-2\int_0^1\operatorname{Li}_2(1-x^2)\ dx\tag{1} \end{align} By the OP, the second integral ... 5 This is related to the q-Pochhammer Symbol,f_n(x):=\prod_{i=1}^nq^ix(1-q^ix)=q^{n(n+1)/2}x^n(qx;q)_{n}.1 Let f(z) be the integrand. By the previous question you linked to, f(z) is equal to \frac{x+\sqrt{x(x+4)}}{2} on the positive real line, and is holomorphic on \mathbb{C} \backslash [-4,0]. So for positive real x we can rewrite f(x) as follows: \begin{align*} f(x)&=\frac{x+\sqrt{x(x+4)}}{2}\\ &=\frac{1}{2}x+\frac{1}{2}x\sqrt{1+\frac{4}{x}}... 3 I am not an expert on continued fractions, so take this with a grain of salt. But here's my take. We don't know what the continued fraction is, except that we know it is a function of z. Call it y(z): \displaystyle y(z) = z + \frac{z}{z+{\frac{z}{etc.}}} But now notice that the denominator in the fraction is also y(z)! And so, \displaystyle y(... 2 This is not an answer, but some information in attempt to clear the air. Using the first definition for the limit function, we have:f(z)=(1-z)^2F_2(\alpha;1,1;\alpha,\alpha;z,z)$$Using the integral definition for this Appell function we obtain:$$f(z)=(1-z)^2(\alpha-1)^2 \int_0^1 \int_0^1 \frac{(1-u)^{\alpha-2}(1-v)^{\alpha-2} du dv}{(1-z (u+v))^\...

0

We want to solve the following integral: $$I(x) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} \frac{1}{jt(1+jt)} e^{\frac{-jNat}{1+jt}} e^{jxt} dt$$ To solve, we first recall a property of the Fourier transform. Namely: $$F^{-1}\big\{ \frac{G(t)}{jt} \big\} = \int_{-\infty}^{x} g(\tau) d\tau$$ where $g(x)$ is the inverse Fourier transform of $G(t)$, and ...

0

You are right, the series is a little messed up, but there's no reason to try guessing the proper form. We can always use the general way to get the hypergeometric form of a series. Consider: $$f(z)=z \Gamma(3)\left[\frac{1}{\Gamma(3)}+\frac{(\gamma)_1}{\Gamma(4)}\frac{z}{1!}+\frac{(\gamma)_{2}}{\Gamma(5)}\frac{z^2}{2!}+\frac{(\gamma)_{3}}{\Gamma(6)}\frac{... 1 Here is some references mainly focused on special functions, I think it will help you a lot. 1.~~ "Special functions and their applications" by N. N. Lebedev, Richard A. Silverman 2.~~ "Special Functions for Scientists and Engineers" by W. Bell 3.~~ "Special Functions: An Introduction to the Classical Functions of Mathematical Physics" by Nico M. ... 2 The last formula is valid as the limit for m approaching an integral value. For your values of the parameters, the function becomes a finite sum:$$U(n, n + m + 1, z) = \frac {m! \hspace {1px} z^{-n}} {(n - 1)!} \sum_{k = 0}^m \frac {(n + k - 1)!} {(m - k)!} \frac {z^{-k}} {k!}.$$There are some contrived ways to write a power function as a G-function; ... 1 A highly thorough and accessible reference on the Mathieu equation and the parameters that characterize its stability/instability curves can be found in Chapter 6 of Rand's Lecture Notes on Nonlinear Vibrations. Hope this helps! 2 Reduction to c=1 is just "time scaling":$$\mathcal{L}^{-1}\left\{\frac{1}{s^b-1}\right\}=f(x)\implies\mathcal{L}^{-1}\left\{\frac{1}{s^b-c}\right\}=c^{1/b-1}f(c^{1/b}x).$$For the former, there's also a "series path": around s=+\infty,$$\frac{1}{s^b-1}=\sum_{n=1}^{\infty}s^{-nb}\implies\mathcal{L}^{-1}\left\{\frac{1}{s^b-1}\right\}=\sum_{n=1}^{\infty}\...

1

Not a complete solution, merely reducing the number of parameters in the integrand by one. \begin{align*} \frac{1}{2\pi\mathrm{i}}&\int_{\gamma - \mathrm{i}\infty}^{\gamma + \mathrm{i}\infty} \frac{1}{s^b-c} \mathrm{e}^{st} \,\mathrm{d}s \\ &= \frac{1}{2\pi\mathrm{i}} \frac{1}{c} \int_{\gamma - \mathrm{i}\infty}^{\gamma + \mathrm{i}\infty} \frac{1}...

4

With \begin{equation*} I_n = \int_{0}^{1}\left(\dfrac{x-1}{\ln(x)}\right)^n\, dx \end{equation*} and the substitution $x=e^{-y}$ we get that \begin{equation*} I_n= \int_{0}^{\infty}\dfrac{f(y)}{y^n}\, dy \end{equation*} where \begin{equation*} f(y) = \left(1-e^{-y}\right)^ne^{-y} =\sum_{k=0}^{n}\binom{n}{k}(-1)^{k}e^{-(k+1)y}. \end{equation*} Then $f(0)=... 8 The general formula for$I_n$is $$I_n = \frac 1 {(n-1)!}\sum_{k=1}^n {n \choose k} (-1)^{n-k} (k+1)^{n-1} \ln (k+1)$$ and for$J_n(t), we have $$J_n(t) = \frac 1 {(n-1)!}\sum_{k=1}^n {n \choose k} (-1)^{n-k} (kt+1)^{n-1} \ln (kt+1).$$ I've checked that this formula returns correct values \begin{align*} I_1 =& \ln 2\\ I_2 =& - 4\ln 2 + 3\ln 3\\ ... 1 Basically, your mistake is the choice ofC$. The integrals over$\Sigma_{-,\varepsilon}$and$\Sigma_{+,\varepsilon}$both diverge, and this breaks the rest of the above. It's important to understand the "how and when" of$\int_C\Gamma(-s)F(s)\,ds$. Surely, if$F(s)$is regular at$s\in\mathbb{Z}_{\geqslant 0}$, and$C_n(is simple positively oriented and) ... 2 A third answer attempt for the lack of space in the two previous ones. I hope the community forgives me this one time. Let's try dealing with the following integral, since it all comes down to it anyway: $$R(z)=z \int_0^1 u \psi (z u) du$$ I want to use the exhaustion formula referenced in this question: $$\int_0^1f(x)\,dx=-\sum_{n=1}^\infty\sum_{m=1}^{2^... 2 Adding another answer with a different attempt, this time using the series. From one of the linked questions we find out the Taylor series representation:$$\log \Gamma(z)=\sum_{k=2}^{\infty} \frac{\zeta(k)}{k} (1-z)^{k} +\gamma (1-z)I(z)=\sum_{k=2}^{\infty} \frac{\zeta(k)}{k} \frac{1- (1-z)^{k+1}}{k+1} +\frac{\gamma}{2} z (2-z)$$Comparing with the ... 2 This answer is meant to connect the ones given by @Harry Peter and @Start wearing purple, clarifying a few questions emerged in the comments. The integral of interest can be evaluated in the way pointed out by @Harry Peter, without forgetting to set some conditions on the parameters. First of all, for \left|z\right|<1 we can use the power series ... 0 As Robert Israel's answer says, we can reduce this to proving the inequality$$ x( I_1(x)^4 - I_0(x)^4) + 2 I_0(x)^3 I_1(x) < 0 ; $$a more useful form for our purposes is$$ 1 - \left( \frac{I_1(x)}{I_0(x)} \right)^4 - \frac{2}{x} \frac{I_1(x)}{I_0(x)} > 0 . $$It suffices to find a simple upper bound for the ratio, then verify it satisfies the ... 8 Let's use integration by parts:$$I(z)=\int_0^z\ln\Gamma(t)~dt=z \ln\Gamma(z)-\int_0^z t \psi(t) dt\psi(t)=\log t-\frac{1}{2t}-2 \int_0^\infty \frac{udu}{(u^2+t^2)(e^{2 \pi u}-1)}\int_0^z t \log t dt=\frac{z^2}{4} (2 \log z-1)\frac{1}{2}\int_0^z dt=\frac{z}{2}2 \int_0^z \frac{t dt}{u^2+t^2}=\log \left(1+ \frac{z^2}{u^2} \right)$$... 1 To recap my findings from @martycohen's guidance, I got to this result for the inverse Laplace transform I need:$$ \mathcal{L}^{-1}\Big\{\frac{1}{s(s-a)}e^{b/s}\Big\}(t) = \frac{e^{at}}{a}\sum_{k=1}^\infty \frac{(b/a)^k}{k!}\frac{\gamma(k+1,at)}{\Gamma(k+1)}.$$The book "An Introduction to the Classical Functions of Mathematical Physics" by Temme (1996) ... 4 In general, assuming convergence, we have$$ \sum_{n=1}^\infty \sin\left(\frac{\pi}{6}\,n\right)\, a_{n} = \sum_{n=1}^\infty \sin\left(\frac{\pi}{3}\,n\right)\, a_{2n} + \sum_{n=1}^\infty \sin\left(\frac{\pi}{2}\,n\right)\, \left(\frac12 a_{n}+\frac32 a_{3n} \right)$$To see this, we partition the naturals into evens (2n) and odds (6n+1,6n+3,6n+5). ... 0 Since the Kummer equation is a Sturm–Liouville equation, the zeros of F(A;B;z) are simple, so since F is real for real z if the parameters are real and F(A;B;0)=1, F takes negative values if and only if it has a zero. So we need to determine when F(a;b;-x) has at least one positive zero. We have the relation F(A;B;z) = e^z F(B-A;B;-z), and so ... 4 Using the power series of e^x, we get$$ -\int_0^x\frac{1-e^{-t}}t\,\mathrm{d}t=\sum_{n=1}^\infty\frac1n\frac{(-x)^n}{n!}\tag1 $$Since$$\newcommand{\Ei}{\operatorname{Ei}} \Ei(-x)=-\int_x^\infty\frac{e^{-t}}t\,\mathrm{d}t\tag2 $$we have$$ \begin{align} \sum_{n=1}^\infty\frac1n\frac{(-x)^n}{n!} &=\color{#C00}{-\int_0^x\frac{1-e^{-t}}t\,\mathrm{d}t}+\... 3 As I showed in my answer to your other question Closed form for an infinite series involving lower incomplete gamma functions,Q(u, v)+Q(v, u) =e^{u+v}+I_0(2\sqrt{uv}) $. I also suggested that you research the Marcum Q-function. 4$Q(t) = \frac{e^{at}}{a}\Big[e^{b/a}-e^{-at}\sum_{k=0}^\infty \sum_{l=0}^k \frac{(at)^l(b/a)^k}{k!l!}\Big]. $I'll blindly try to reverse the order of summation and see what happens.$\begin{array}\\ S(u, v) &=\sum_{k=0}^\infty \sum_{l=0}^k \frac{u^lv^k}{k!l!}\\ &=\sum_{l=0}^\infty\sum_{k=l}^\infty \frac{u^lv^k}{k!l!}\\ &=\sum_{l=0}^\infty\...

1

Closed-forms (in terms of other constants) of the Barnes G-function $G(z)$ for fractional $z = \frac{p}{q}$ with $0<p<q$ are known only (for now) for SEVEN special values. Given the Clausen function $\operatorname{Cl}_2(z)$ and, \begin{aligned} A \;&= \text{Glaisher–Kinkelin constant}\\ \operatorname{Cl}_2\left(\frac\pi2\right) &=\text{... 1 A solution using Abel's summation as suggested by Cornel. Let \ \displaystyle S\  denote \ \displaystyle \sum_{k=1}^\infty\frac{H_k^{(2)}}{(2k+1)^2}\  and by using Abel's summation: \displaystyle\sum_{k=1}^n a_k b_k=A_nb_{n+1}+\sum_{k=1}^{n}A_k\left(b_k-b_{k+1}\right)\  where \ \displaystyle A_n=\sum_{i=1}^n a_i\  and letting let \ \displaystyle ... 0 This is not an answer, but may be helpful as a start. First of all, we might as well define x = 2 \sqrt{\lambda}. Then with g(x) = I_0(x)/I_1(x), you want x (g(x) - 1/g(x)) to be monotone (decreasing) for x > 0. This turns out to be equivalent to - {{ I}_{0}\left(x\right)} ^{4}x+ {{ I}_{1 }\left(x\right)} ^{4}x+2\, {{ I}_{0}\left(x\right)}...

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