10

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 ...


7

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(\...


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