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1

Okay, so I think I may have found the answer myself. So, really, the absolute value symbol is a trick. You can get rid of it by pulling out an $i$, and then you have $$\mathcal{I}:=\frac{1}{2\pi i}\int_a^b dx \frac{\log(x e^{-t_1})}{z-x}\frac{\sqrt{(z-a)(z-b)}}{\sqrt{(x-a)(x-b)}}$$ What you have to do is take a dumbbell contour around taking a clockwise &...

4

You can rewrite the integral as $$\int_0^{2\pi} \left(\sum_{j=0}^{n-1}\cos\Big(\sin t+\tfrac tn+2\pi \tfrac jn\Big)\right)\,dt.$$ But $\sum_{j=0}^{n-1}\cos\big(a+2\pi \tfrac jn\big)=0$ for all $a$. In particular, the equality holds if $\sin t$ is replaced by any $2\pi$-periodic function.

10

Diestel and Uhl commented on this in Measure theory and its applications:

1

You just need to use a higher precision, using exact numbers when possible, as shown in the Mathematica work below (with $k:=\kappa$ and $t:=\theta$). (However, this question is indeed better suited for Mathematica SE.)

3

It looks like you care only about the order of magnitude (i.e., an answer up to a constant factor), in which case it is fairly easy. First, ignore all coefficients. Setting them to $1$ just changes the answer at most constant number of times. Now, suppose we have the denominator of the form $\sum_{(\alpha,\beta)} x^\alpha r^\beta$ where $\alpha$ is a multi-...

2

The accuracy issue with the evaluation of the hypergeometric function can be avoided for integer $\kappa$, since then the full expression reduces to an error function (see my answer to your previous question). I tried this out for $\kappa=5$. Then $$\mathbb{E}\left[ a \mathcal{Q} \left( \sqrt{b } \gamma \right) \right]=\frac{a}{48 b^4 {\theta}^8} \left[\sqrt{... 4 Let's consider the second integral, which can be written in the following form:$$ I(p, q, i, j, k, l; a, b, c, d) := \int_0^\infty dt\, \exp(-p t^2) t^q j_i(a t) j_j(b t) j_k(c t) j_l(d t) $$where in your case, i = j = k = 1, l = 0, and q = -1. These kinds of integrals (as well their generalization to a product of arbitrarily many spherical Bessel ... 1 The best evaluation, a single sum, that I can derive is$$ \int_0^{\pi/2} \big( P_n^m(\cos{\theta}) \big)^2\ d\theta = \frac{\pi}{2} \frac{(2m)!}{m!^4} \Big( \frac{(n+m)!}{(n-m)!} \Big)^2 {}_4F_3 \bigl( \begin{smallmatrix} m+1/2, & m+1/2, & m-n, &m+n+1 \\ m+1 & m+1 & 2m+1 \end{smallmatrix} | 1\bigr) I derived it from the answer ... 3 Thanks to the comment by Johannes, the solution can indeed be obtained by using the following identities: $$P_\ell(z) = \frac{1}{2^\ell} \sum\limits_{k=0}^{\left\lfloor \frac{\ell}{2}\right\rfloor} (-1)^k \begin{pmatrix} \ell \\ k \end{pmatrix} \begin{pmatrix} 2\ell - 2k \\ \ell \end{pmatrix} z^{\ell - 2k} \tag{3} \label{3}$$ and: ... 3 To evaluate \begin{align} c_{n,m}&=\int_0^{1/2}\frac{x(x-1/2)}{\sin^2(2\pi x)}\, \sin(2\pi(2m+1)nx)\,dx\\ &=-\frac{1}{8}\int_0^{1}\frac{x(1-x)}{\sin^2(\pi x)}\, \sin(\pi(2m+1)nx)\,dx \end{align} we first notice by changing x\to 1-x, that the integral vanishes if n is even. In the following c_{2n+1,m} is calculated. We apply twice a result ... 2 Besides some condition ensuring that the map is well defined a simple and natural condition for the continuity is a Lipschitz condition with respect to the second variable, i.e.,|f(x,y)-f(x,z)|\le c|y-z|$$for some constant c independent of x,y,z. This is a standard assumption for (a version of) the Picard-Lindelöf theorem for ODEs. I repeated my ... 1 \newcommand\Ga\Gamma Without loss of generality a=1. Let then Q:=\mathcal Q, k:=\kappa>0, and t:=\theta\sqrt b>0, so that \sqrt b\,\gamma has the gamma distribution with parameters k,t. Let also c:=\Ga(k)t^k. Then, letting f denote the standard normal pdf, we have$$c\,EaQ(\sqrt b\,\gamma)=\int_0^\infty dy\,y^{k-1} e^{-y/t}Q(y) =\...

1

Mathematica gives an answer in terms of a hypergeometric function: $$\frac{a 2^{-\frac{\kappa}{2}-\frac{3}{2}} b^{-\frac{\kappa}{2}} {\theta}^{-k} \, _2F_2\left(\frac{\kappa}{2}+\frac{1}{2},\frac{\kappa}{2};\frac{1}{2},\frac{\kappa}{2}+1;\frac{1}{2 b {\theta}^2}\right)}{\sqrt{\pi } \Gamma \left(\frac{\kappa}{2}+1\right)}-\frac{a 2^{-\frac{\kappa}{2}-2} \... 0 An solution for this integral can be found at Mathmatica.SE, which is reproduced next. After applying the change of variable technique with x=2^r-1 we get$$f=\frac{e^{-\frac{\sqrt{r}}{b}} r^{\frac{d}{2}-1} \log _2(r+1)}{2 \left(b^d \Gamma (d)\right)} \text{Integrate}[f,\{r,0,\infty \},\text{Assumptions}\to d\in \mathbb{R}\land b\in \mathbb{R}\land d&...

6

As you've pointed out, the only parameter that matters here is the angle $\theta$ between $x$ and $y$. To see how, consider instead the Gaussian integral: $$I(x,y)=\frac{1}{(2\pi)^{(d+1)/2}}\int_{u\in\mathbb{R}^{d+1}}\max(0,x^Tu)\cdot\max(0,y^Tu)\exp(-\frac{1}{2}u\cdot u)du$$ The integral you are interested in is obtained by changing to $(d+1)$ dimensional ...

2

I'll suggest here another possible characterisation, expanding on a suggestion of the OP in one of the comments. Again, this is an assertion that certain known properties of line integration characterise it uniquely; this doesn't provide a "new" construction of line integration. Unlike my previous answer, here all the action takes place on the one ...

10

I don't know if it's exactly what you're looking for, but line integration is the unique way to assign a real number $I(\omega,c)\in\mathbb{R}$ to every pair of a smooth $1$-form $\omega$ on a smooth manifold $M$ with boundary and smooth path $c\colon[0,1]\to M$ such that: (adjunction) if $f\colon M\to N$ is a smooth map of smooth manifolds with boundary, $\... 4 A nice version of the manifold version of Fubini's theorem is in Differentialgeometrie und Fasserbündel by Rolf Sulanke and Peter Wintgen: Let$\phi \in C^1(M,N)$, where$M,N$are smooth manifolds of dimensions$m,n$, respectively, with$m \ge n$. Let$\omega \in \Omega^{m-n}(M)$and$\eta \in \Omega^n(N)$, and let$f : M \to {\bf R}$be measurable (meaning, ... 1 I think this is $$-\int_{0+\epsilon}^{1-\epsilon} {\frac {2\;{\rm W} \left(-\frac12\,\ln \left( 1-{x }^{2} \right) \right)}{\ln \left( 1-{x}^{2} \right) }}\,{\rm d}x$$ not elementary. 0 Hint: you may try$n=2and follow in general this :\begin{align} \int_0^\infty \left(\frac{\log(x)}{e^x}\right)^n\,dx&=\int_0^\infty e^{-nx}\log^n(x)\,dx\\\\ &=\frac1n\sum_{k=0}^n\binom{n}{k}\log^{n-k}(n)\int_0^\infty e^{-x}\log^n(x)\,dx\\\\ &=\frac1n\sum_{k=0}^n\binom{n}{k}\log^{n-k}(n) \left.\left(\frac{d^n \Gamma(x+1)}{dx^n}\right)\right|_{x=... 16 Let $$f(a):=\int_0^\infty x^{a-1}e^{-2x}\,dx.$$ Then $$f''(1)=\int_0^\infty \ln^2x\,e^{-2x}\,dx,$$ which is the integral in question. On the other hand,f(a)=2^{-a}\,\Gamma(a)$, and hence the integral in question is $$f''(1)=\frac{\ln^2 2}2 - \Gamma'(1)\ln2 + \Gamma''(1)/2 =\frac{\pi^2}{12}+ \frac{(\gamma +\ln2)^2}2=1.6293\dots,$$ where$\gamma=0.57721\dots\$...

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