# Tag Info

0

Rather than go over standard ground here is the reduce compter aided algebra answer; which certain looks correct. Let: $\mathrm{aa : =} \frac{x^m n!}{(- m + n) !m}$ Then hypergeometric summing with respect to m you get: $\mathrm{hypergeom} \left( \left\{ - n + 1 \hspace{0.17em} \mathrm{, \hspace{0.17em}} \hspace{0.17em} 1 \hspace{0.17em} \mathrm{, ... 1 We have the identity: $$\sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b}\tag{1}$$ hence by taking$a=\frac{N}{2}+1$and$b=\frac{N}{2}+\frac{1}{2}$it follows that: $$\psi\left(\frac{N}{2}+1\right)-\psi\left(\frac{N}{2}+\frac{1}{2}\right)=2\sum_{n\geq 0}\frac{1}{(2n+N+2)(2n+N+1)}\tag{2}$$ but$\sum_{m\geq 0}\frac{1}{(2m+1)(2m+2)}$is ... 1 Hint: Use Integration by parts, where you integrate$f(x)=e^{-x}$and differentiate$P_n(x)$, then use the Rodrigues' formula. 1 This can also be done via elliptic integrals and it is possible to get a closed form involving$\Gamma(1/4)$but the calculations are complicated. We need to use the power series for$\text{dn}(u, k)$and compare it with its Fourier series. Thus we have $$\text{dn}(u, k) = 1 - k^{2}\frac{u^{2}}{2!} + k^{2}(4 + k^{2})\frac{u^{4}}{4!} - \cdots\tag{1}$$ and the ... 2 It turns out to be an identity about Bernoulli numbers,$B_2=B_{14}-1$. They both are rational numbers, and their denominators have to be the same by the Von Staudt-Clausen theorem. Equality of numerators, too, happens more or less by chance. 2 Since you ask multiple questions, I suppose what follows isn't a complete answer. I learned this trick the other day, so I'm more than happy to share it here, If you want an integral that can generalize the recurrence, $$(1) \quad \mathfrak F(0)=1,\quad \mathfrak F(n+1)=\mathfrak F(n)\times F(n+1).\tag3$$ You can do so by assuming the formula is of the ... 2 The trick is to use $$t^2+b^2=(t+i b)(t-ib)$$ and use partial fraction decomposition $$\frac t{t^2+b^2}=\frac{1}{2 (t+i b)}+\frac{1}{2 (t-i b)}$$ So $$\int \frac{t\sin at}{t^2+b^2}dt=\frac{1} 2\int\frac{\sin at}{t+ib}dt+\frac{1} 2\int\frac{\sin at}{t-ib}dt$$ So, now, consider $$I=\int \frac{\sin at}{t+c}dt$$ Change variable$t=u-c$to get $$I=\int\frac{\sin ... 2 Why (3) is so different from (1) ? Because if we rewrite the double integrals as iterated ones, for (1) we get$$\int_0^1 \left(1-x_2\right)^{\alpha_3-1}\left({\color{blue}{\int_0^{x_2}x_1^{\alpha_1-1}\left(x_2-x_1\right)^{\alpha_2-1}dx_1}}\right)dx_2,$$and for (3) we find$$\int_0^1 x_1^{\alpha_1-1}\left({\color{red}{\int_0^{x_1} ... 2 Let me consider the two integrals $$I = \int \mathrm{e}^{-v}\int_0^{a - bv} \mathrm{e}^{-u^2} du\,dv$$ $$J= \int \int_0^{a + bv} \mathrm{e}^{-u^2} du\,dv$$ First $$\int_0^{a - bv} \mathrm{e}^{-u^2} du =\frac{\sqrt{\pi } }{2} \text{erf}(a-b v)$$ which makes $$I=\frac{ \sqrt{\pi }}{2} \int e^{-v} \text{erf}(a-b v)\,dv$$ This one can be integrated by parts ... 0 If you consider that the locus of the constraints$ \ M \ $is the intersection of an "upward-opening" circular paraboloid and an oblique plane, said plane being symmetrical about the plane$ \ y \ = \ x \ $and sloping "downward" in the "positive" direction, then the constraint curve is a tilted ellipse with its "high end" in the third quadrant of the$ \ ...

1

hint: $g(x,y) =x+y+x^2+y^2 = 12\implies xy + z^2 = xy + (12-x-y)^2= 3xy + 144+x^2+y^2-24(x+y)=3xy+144+12-25(x+y)=3xy-25(x+y)+156= f(x,y)$. Thus we have: $\nabla f = \lambda \nabla g\implies 3y-25 = \lambda(1+2x), 3x-25=\lambda(1+2y)\implies 3(y-x) = 2\lambda(x-y)\implies x = y$ or $\lambda = -\dfrac{3}{2}$. Can you continue from here?

4

We consider the Bessel functions \begin{align*} J_p(x)&=\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+1+p)}\left(\frac{x}{2}\right)^{2n+p}\\ J_{-p}(x)&=\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+1-p)}\left(\frac{x}{2}\right)^{2n-p}\\ \end{align*} Blue question: Due to the symmetry of $J_p(x)$ and $J_{-p}(x)$ the author seems ...

2

strictly monotone increasing. Any such function is continuous almost everywhere. But what do you mean by singularities? It could have jump discontinuities, in fact, it could have infinitely many. As pointed out by Merlinsbeard, it is in fact almost everywhere differentiable (which is even stronger)

6

Not sure if it is your book's problem or what, but the Weierstrass product formula is $$\frac1{\Gamma(z)} = ze^{\gamma z} \prod_{n=\color{red}1}^\infty \left(1+\frac zn\right)e^{-z/n}$$ (Note that $(1+\frac z0)e^{-z/0}$ is undefined, so the product cannot start from 0.) So after taking inverse square root your solution should be $$\left| \Gamma(\alpha + ... 6 this is \int_0^\infty e^{-bt}t^{(1-\alpha)-1}dt where b=-\log (1-a)>0. Substitute bt=u you will get Gamma function. 0 We can specialize:  _{u+1}F_{v}\left[\begin{array}{cc} -n & \alpha_{1},\ldots,\alpha_{u}\\ & \beta_{1},\ldots,\beta_{v} \end{array};z\right] = \frac{\left(\alpha_{1}\right)_{n}\cdots\left(\alpha_{u}\right)_{n}}{\left(\beta_{1}\right)_{n}\cdots\left(\beta_{v}\right)_{n}}\left(-z\right)^{n}\cdot_{v+1}F_{u}\left[\begin{array}{cc} -n & ... 3 Utterly impossible:$$y?1=y\cdot 1 = y,y' = \frac{d(y?1)}{dx}=\frac{dy}{dx}\cdot\frac{d1}{dx} = 0$$0 With the so-called logarithmic derivative, you can write$$(\log(yz))'=\frac{(yz)'}{yz}=\frac{y'}{y}+\frac{z'}{z}.$$This does not generalize to the second order. 0 It is unclear what you are asking. Why are you defining so many letters? In any case, Convolutions have a property you may be interested in: https://en.wikipedia.org/wiki/Convolution#Integration, but you may need some technical conditions on the function space for these to work. In particular, as the linked page mentions, you'll need Fubini's theorem or one ... 0 What you are looking for is called Weyl's limit point(LP)/limit circle(LC) classification. An end point is LC if both solutions are square integrable (w.r.t . the weight function) and LP otherwise. If both endpoints are LC then you need boundary conditions and you will have a complete set of eigenfunctions. Otherwise the spectrum might have a continuous ... 0 Observe that 2(n+1)(2\,n+1)!=(2\,n+2)!. Then$$ \sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{2(n+1)(2n+1)!}=\frac1x\sum_{n=0}^\infty \frac{(-1)^nx^{2n+2}}{(2\,n+2)!}. $$3 The story so far:$$ \lim_{n \rightarrow \infty} V_n(1) = 0. $$This has made a lot of people very angry and been widely regarded as a bad move. Induction, but increasing the dimension by 2 each time. You do need to know that \Gamma(1+n) = n! along with the special value$$ \Gamma(\frac{1}{2}) = \sqrt \pi $$The recursion is, in general,$$ z ...

2

I will sketch the very elegant argument given by Keith Ball in An Elementary Introduction to Modern Convex Geometry. Let we consider: $$I(n) = \int_{\mathbb{R}^n}\exp\left(-\sum_{k=1}^{n}x_k^2\right)\,d\mu. \tag{1}$$ By Fubini's theorem, it is just $\Gamma\left(\frac{1}{2}\right)^n = \pi^{n/2}$. Let: $$S_R = \int_{x_1^2+\ldots+x_n^2=R^2}1\,d\mu. ... 1 Let I(s) be the integral given by$$I(s)=\int_0^\infty t^{s-1}\sin(t)\,dt \tag 1$$for \text{Re}(s)<1. We can use Euler's Formula to write (1) as$$I(s)=\frac1{2i}\left(\int_0^\infty t^{s-1}e^{it}\,dt-\int_0^\infty t^{s-1}e^{-it}\,dt\right) \tag 2$$Now, moving to the complex plane, we analyze the closed contour integral(s) ... 4 For \text{Re}(z)\in(-1,1), we may use the Laplace transform to get:$$ \int_{0}^{+\infty}\sin(t)\,t^{z-1}\,dt = \frac{1}{\Gamma(1-z)}\int_{0}^{+\infty}\frac{ds}{s^z(1+s^2)}\tag{1}$$since \mathcal{L}(\sin t)=\frac{1}{1+s^2} and \mathcal{L}^{-1}\left(t^{z-1}\right)=\frac{1}{s^z \Gamma(1-z)}. With the substitution \frac{1}{1+s^2}=u, the RHS of (1) ... 1 Never mind, thanks to the comment by J. M. I found the source of this expression. The series are connected to the exponential integral:$$\text{Ei}(t)=-\int_{-t}^{\infty} \frac{e^{-p}}{p} dp=\gamma+\log |t|+\sum_{n=1}^{\infty} \frac{t^n}{n!n}$$The continued fraction turns out to be a particular case of incomplete Gamma function:$$\Gamma ...

1

This answer has more explanatory character since most of the calculcations we need were already done by OP or provided in the answer of @user5713492. For convenience only we introduce the operator notation \begin{align*} D_x:=\frac{d}{dx} \end{align*} First step: We focus at the $\color{red}{red}$ part of OPs calculation and obtain ...

13

Let we put everything together. $$I = \int_{0}^{1}\sin(\pi x)\log\Gamma(x)\,dx = \int_{0}^{1}\sin(\pi z)\log\Gamma(1-z)\,dz \tag{1}$$ leads to: $$I = \frac{1}{2}\int_{0}^{1}\sin(\pi x)\log\left(\Gamma(x)\,\Gamma(1-x)\right)\,dx \tag{2}$$ but $\Gamma(x)\,\Gamma(1-x) = \frac{\pi}{\sin(\pi x)}$, hence: $$I = \frac{\log ... 18 Let u=\cos(x), then$$ \begin{align} \int_0^\pi\sin(x)\log(\sin(x))\,\mathrm{d}x &=\frac12\int_{-1}^1\log\left(1-u^2\right)\,\mathrm{d}u\\ &=\frac12\left(\int_{-1}^1\log(1-u)\,\mathrm{d}u+\int_{-1}^1\log(1+u)\,\mathrm{d}u\right)\\ &=\int_0^2\log(v)\,\mathrm{d}v\\[3pt] &=\left.v\log(v)-v\right]_0^2\\[9pt] &=2\log(2)-2\tag{1} \end{align} ...

5

\eqalign{ & \int_{x\, = \,0}^\pi {\sin (x)\ln (\sin (x))dx} = - {1 \over 2}\int_{x\, = \,0}^\pi {\ln (1 - \cos ^{\,2} (x))d\cos (x)} = \cr & = - {1 \over 2}\int_{t\, = \,1}^{ - 1} {\left( {\ln (1 - t) + \ln (1 + t)} \right)dt} = \; \cdots \cr}

2

We are trying to evaluate $$\int_{-1}^1\frac{d^{\ell+m-n}}{dx^{\ell+m-n}}(x^2-1)^{\ell}\frac{d^{\ell-m+n}}{dx^{\ell-m+n}}(x^2-1)^{\ell}dx$$ For $0\le n<m$. Let $$u=\frac{d^{\ell-m+n}}{dx^{\ell-m+n}}(x^2-1)^{\ell}$$ Then $$du=\frac{d^{\ell+1-m+n}}{dx^{\ell+1-m+n}}(x^2-1)^{\ell}dx$$ And $$dv=\frac{d^{\ell+m-n}}{dx^{\ell+m-n}}(x^2-1)^{\ell}dx$$ So ...

2

Hint: The ingredients are $$B\left(a,b\right)=\frac{\Gamma\left(a\right)\Gamma\left(b\right)}{\Gamma\left(a+b\right)}$$the link between Gamma and Polygamma functions ...

1

The polylogarithm function is defined as $$\textrm{Li}_{s}\left(z\right)=\sum_{k\geq1}\frac{z^{k}}{k^{s}}$$ for all complex $s$ and for $\left|z\right|<1$. So observe that $$\frac{\partial}{\partial s}\left(\textrm{Li}_{s}\left(\frac{1}{e}\right)\right)=-\sum_{k\geq1}\frac{\log\left(k\right)}{e^{k}k^{s}}$$ and so $$\frac{\partial}{\partial ... 2 Hint: \int_0^\infty\dfrac{e^{-x^2}}{\sqrt{t^2+x}}~dx =2\int_0^\infty e^{-x^2}~d\left(\sqrt{t^2+x}\right) =2\int_t^\infty e^{-(x^2-t^2)^2}~dx =2\int_t^\infty e^{-x^4+2t^2x^2-t^4}~dx Similar to Evaluating \int_{1}^{\infty}\exp(-(x(2n-x)/b)^2)\,\mathrm dx 2$$ \begin{align} 2\int_{0}^{\infty} \frac{e^x-x-1}{x(e^{2x}-1)} \, dx &= 2 \int_{0}^{\infty} \frac{1}{x(e^{2x}-1)} \sum_{n=2}^{\infty} \frac{x^{n}}{n!}\\ & = 2 \sum_{n=2}^{\infty}\frac{1}{n!} \int_{0}^{\infty} \frac{x^{n-1}}{e^{2x}-1} \, dx \\ &= 2 \sum_{n=2}^{\infty}\frac{1}{n!2^{n}} \int_{0}^{\infty} \frac{u^{n-1}}{e^{u}-1} \, du \\ &= 2 ...

7

Hint. One may set $$f(s):=2\int_0^\infty \frac{e^{sx}-sx-1}{x(e^{2x}-1)}dx, \quad 0<s<2. \tag1$$ In order to get rid of the factor $x$ in the denominator, we may differentiate under the integral sign getting $$f'(s)=2\int_0^\infty \frac{e^{sx}-1}{e^{2x}-1}dx, \quad 0<s<2. \tag2$$ Then expanding the latter integrand in $e^{-kx}$ terms and ...

2

By Frullani's theorem we have: $$\int_{0}^{+\infty}\frac{e^x-1-x}{x}\,e^{-2m x}\,dx = -\frac{1}{2m}+\log\left(\frac{2m}{2m-1}\right)\tag{1}$$ hence it is straightforward to prove the claim by summing $(1)$ over $m\geq 1$, then exploiting: $$\gamma = \sum_{n\geq 1}\left[\frac{1}{n}-\log\left(1+\frac{1}{n}\right)\right] \tag{2}$$ that is just the usual ...

1

I think you have misunderstood the nature of the solutions to Bessel's equation: the differential equation $$y'' + \frac{1}{x}y' + \left(\lambda^2-\frac{n^2}{x^2}\right) y = 0$$ (Bessel's equation, with eigenvalue $\lambda^2$, $\lambda>0$) has a regular singular point at $x=0$, an irregular singular point at $\infty$, and all other points are regular ...

2

We have defined the function $\mathcal{I}:[-1,\infty)^{2}\rightarrow\mathbb{R}$ via the integral representation, $$\mathcal{I}{\left(a,b\right)}:=\int_{0}^{1}\frac{\ln{\left(1+ax\right)}\ln{\left(1+bx\right)}}{x}\,\mathrm{d}x;~~~\small{\left(a,b\right)\in[-1,\infty)^{2}}.\tag{1}$$ The goal is to find a single closed-form expression for the integral ...

1

Let $x=a+t$, where $a$ is an integer, and $0\le t\lt 1$. Suppose that $\frac{k-1}{n}\le t\lt \frac{k}{n}$, where $1\le k\le n$. Then $\lfloor nx\rfloor=na+n-k$. Now compute the left-hand side. The terms $\left\lfloor a+t+\frac{i}{n}\right\rfloor$ are equal to $a$ if $0\le i\lt k$, and are equal to $a+1$ for $k\le i\le n-1$. So there are $k$ values of $i$ ...

1

When $n=0$ this sum can be expressed in terms of theta function. Since $H_n(t+\alpha)$ is a polynomial in $t+\alpha$, and as a consequence in $t$, this sum can be represented as a sum of derivatives of theta functions. For $\beta=1$ a simple expression can be obtained. Consider the case $n\ge 1$. Using the integral representation of Hermite polynomials $$... 1 A way to interpret the delta function in this context is through an integral. Start with the example of the Fourier transform composed with its inverse.$$ f = \frac{1}{2\pi}\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}f(t)e^{-ist}dt\right)e^{isx}ds. $$This may be correctly written as$$ ...

0

I am going to play with it and see if anything happens. Turns out that this can be represented in terms of two simpler integrals. I reached a point where I can't go further. I'll enter what I have done in hopes that it might be useful to someone else. First I'll make it easier to type, changing \int_{\pi/2}^{\pi}\int_{\pi/2}^{\pi}J_{0}\left(\alpha ... 0 It can be computed by using the complex error function (aka the Faddeeva function): $$1-{\rm{erf}}(z)=e^{-z^2}w(iz)$$ Matlab and C packages for the Faddeeva function are available in the Matlab Central. 2 \begin{align*} C(x+n,y+n) &= \int_{-\infty}^\infty \frac{dt}{(1+t^2)^n(1+it)^x(1-it)^y} \\ &= \int_{-\infty}^\infty \frac{du/\sqrt{n}}{(1+u^2/n)^n(1+iu/\sqrt{n})^x(1-iu/\sqrt{n})^y} \qquad (\textrm{let }u=t/\sqrt{n}) \\ &= \int_{-\infty}^\infty \frac{1}{\sqrt{n}} e^{-u^2} + O(1/n) \qquad (\textrm{standard limit for }e) \\ &= ... 1 $$y(t) = L^{-1} [ F(s) e^{-s} - F(s) e^{-2s}] = (t-1)u(t-1) - f(t-2)u(t-2)$$u(t-t_0)$is the unit step function. It has the value$0$for all values of$t<t_0$and has the value$1$for all values of$t>t_0$. Using this definition you can see that for$0<t<1$all unit step functions vanish. Resulting in$y(t)=0$. For$1<t<2$only ... 2 For$t<1$we have$u(t-1)=0$and for$t<2$we have$u(t-2)=0$. So for$t<1$we have$f(t-1)u(t-1)=0$and$f(t-2)u(t-2)=0$and then$y(t)=0$. For$t> 1$we have$u(t-1)=1$and for$t<2$we have$u(t-2)=0$. So for$1<t<2$we have$f(t-1)u(t-1)=f(t-1)$and$f(t-2)u(t-2)=0$and then$\$y(t)=f(t-1)={1 \over 2} - e^{-(t-1)} + {1 \over 2} ...

Top 50 recent answers are included