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We have $$I_{2n}=\frac{(2n)!!}{(2n+1)!!}\cdot \frac1{2n+2}\cdot \pi^{2n}.$$ To see this, we follow the suggestion by Terry Tao in the comments and apply the diagonalization of the integral operator with the kernel $1/(x+y)$ on $[0,1]$. Change the variable to $1/x\in [1,\infty)$ and use (1.18) here (this is Mehler integral operator, as I understand) to ...
There exist no closed-form expressions for arbitrary $d$ for the integral over the unitary group $\mathbb{U}(d)$ of a rational function of the matrix elements. There are asymptotic results for large $d$, see for example J. Math. Phys. 37, 4904 (1996). The leading order term for $\text{tr}\,C$ of order $d$ is $$\int\limits_{\mathbb{U}(d)}\dfrac{1}{\sum_{k,l}... 3 According to this answer,$$z_\infty:=\lim_n z_n=I:=\int_0^\infty F(s)G(s)\,ds,$$where$$F(s):=\prod_{k=1}^\infty\frac{1}{\sqrt{1+2s/k^3}},\quad G(s):=\sum_{k=1}^\infty\frac k{k^3+2s}.$$Mathematica can express F and G in terms of functions built-in in Mathematica (and these expressions should be rather straightforward to verify), and then the ... 2 Let's investigate$$ L(k,b) := \int_0^\infty \log(1+x) x^k e^{-bx}dx \tag1$$where k is a nonnegative integer, and b>0. I assume we already know$$ E(k,b) :=\int_0^\infty x^k e^{-bx} dx = \frac{k!}{b^{1+k}} \tag2$$We can evaluate$$ \widetilde{E}(n,b) := \int_0^\infty (1+x)^n e^{-bx} dx \tag3$$as a linear combination of E(k,b) for k=0,1,\dots,n.... 2 First integrate over \theta_1,\theta_2. Use the delta function representation (for k\in\mathbb{R})$$\int_{-\infty}^\infty e^{2\pi i k\theta}\,d\theta=\delta(k),$$to evaluate$$\int_{-\infty}^\infty \int_{-\infty}^\infty e^{2\pi i (v_1\cdot x)\theta_1+2\pi i(v_2\cdot x)\theta_2}\,d\theta_1 d\theta_2=\delta(v_1\cdot x)\delta(v_2\cdot x).$$Next for the ... 1 For completeness let R be the associated Riemann surface of \log and \pi : R \to \mathbb{C}^* (with \mathbb{C}^* := \mathbb{C} \setminus \{0\}) be the corresponding universal cover (see f.i. https://en.wikipedia.org/wiki/Complex_logarithm). Then \pi is surjective, continuous and each (Radon-) measure \mu on \mathbb{C}^* can be written in the ... 1 Breaking the integral into two terms, the first term is simply E_i\left[ \int_{t_i}^{t_{i+1}}\widehat{Z}_sds\right]. The second term is E_i \left[\int_{t_i}^{t_{i+1}} \overline{\widehat{Z}_i} ds\right]. The term in the expectation is \mathcal F_{t_i} measurable, and so the second term is just \int_{t_i}^{t_{i+1}} \overline{\widehat{Z}_i} ds. The ... 1 This is not properly an answer, after the comments of Tao it is difficult to give an answer. Only an explanation of my comment above. I still think that my series and the integral are equal. We can write the integral I in the form$$I=\frac{1}{2\pi}\int_{-\infty}^\infty\frac{1}{\zeta(1+it)\zeta(1-it)}\frac{dt}{t^2}.$$Hence I consider the function$$u(\...