Some integrals and sums(series)

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Some integrals and sums(series)

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Source: Inspired from silver08

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#1 h Dec 26, 2024, 2:51 PM • 1 Y

Y by centslordm

I saw Silver's post, so I thought I'd share some integrals and sums as well.

It's Christmas!!! (or boxing day.)

$\begin{align*} 1. & \quad \int_{0}^{1} \frac{K(-x) - E(-x)}{x \sqrt{x+1}} \ln\left(\frac{1-x}{1+x}\right) \, dx = \frac{\pi - 4 \ln(2)}{4\sqrt{\pi}} \cdot \Gamma^2\left(\frac{1}{4}\right) \\ & \text{where:} \\ & \quad E(x) = \int_{0}^{1} \frac{\sqrt{1 - t^2 x}}{\sqrt{1 - t^2}} \, dt, \quad K(x) = \int_{0}^{1} \frac{1}{\sqrt{1 - t^2} \sqrt{1 - t^2 x}} \, dt. \end{align*}$
$\begin{align*} 2. & \quad I = \int_{0}^{\infty} \frac{1}{1+x} \ln\left(\prod_{k=1}^{\infty}\left(1 + e^{-(2k+1)\sqrt{x}}\right) \prod_{k=1}^{\infty}\left(1 + e^{-(2k+1)\pi^2\sqrt{x}}\right)\right) \, dx \end{align*}$
$\begin{align*} 3. & \quad \Omega = \sum_{n=1}^{\infty} (-1)^{n-1} \left(\frac{a n + b}{n(n+1)}\right)^3 H_n^2 \\ & \text{where: } H_n = \sum_{k=1}^{n} \frac{1}{k} \quad \text{(Harmonic numbers)}, \quad a, b \in \mathbb{R}. \end{align*}$
$\begin{align*} 4. & \quad \int_{0}^{\infty} \frac{\ln\left(\sqrt{z^4 + z^2 + 1}\right) - \ln(z)}{z^{10} + 1} \cdot \frac{z^2 + 1}{z^4 + z^2 + 1} \, dz \end{align*}$
$\begin{align*} 5. & \quad \int_{0}^{\frac{\pi}{2}} \frac{\left(c(a_1 - a_2 \sin^2 x)(b_1 - b_2 \cos^2 x)\right)}{\left(\alpha_1 + \beta_1 \sin^2 x\right)\left(\alpha_2 + \beta_2 \cos^2 x\right)} \, dx \end{align*}$
$\begin{align*} 6. & \quad \Omega = \int_{0}^{\infty} e^{-(a+b+c)x} \prod_{n=1}^{\infty}\left(1 + \frac{(a-b)^2 x^2}{n^2}\right) \, dx, \quad a, b \in \mathbb{R}^{+}, \, 0 \leq c \in \mathbb{R}. \end{align*}$

$8. \int_{0}^{1} \frac{\tan^{-1}(x)}{1-x} \ln\left(\frac{1}{2} \left(\frac{1}{\sqrt{x}} + \sqrt{x}\right)\right) \, dx = \frac{\ln(2)}{4} - C - \frac{\pi^3}{192} + \frac{\pi}{32} (\ln(2))^2.$

$9. \int_{0}^{\frac{\pi}{4}} \int_{0}^{\frac{\pi}{4}} \frac{\ln^{2n}(\sin x) \sum_{k=0}^{\infty} \sum_{j=0}^{2n-1} \binom{j}{k-1} \left( \frac{\ln(\sec x)}{\ln(\sin x)} \right)^k}{\cot x \left( \cos^2 y + \tan x \cos y \sin y \right)} \, dy \, dx, \quad n \in \mathbb{Z}^+.$

$\text{Find: } 10. \sum_{n=1}^\infty \sum_{m=-\infty}^\infty \frac{1}{n^p m^2 (m^2 + 1)^3 (n+1)^q}, \quad 2 \leq p, q \in \mathbb{Z},$ $11. \sum_{n=1}^\infty \sum_{m=-\infty}^\infty \frac{(-1)^{n+m}}{n^p m^2 (m^2 + 1)^3 (n+1)^q}, \quad 2 \leq p, q \in \mathbb{Z}.$

$12. \int_{0}^{\frac{\pi}{4}} \frac{\sin x}{\cos(2x) + 2} \tan^{-1}\left(\frac{\cos x \cot(2x)}{\sqrt{2}}\right) dx = \frac{5\pi^2}{48\sqrt{2}} - \frac{\pi}{4\sqrt{2}} \cos^{-1}\left(\frac{1}{3}\right).$

This post has been edited 7 times. Last edited by Martin.s, Dec 26, 2024, 3:03 PM

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#2 h Tuesday at 12:09 PM

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$\int_{0}^{\infty} \frac{\ln\left(\sqrt{z^4 + z^2 + 1}\right) - \ln(z)}{z^{a} + 1} \cdot \frac{z^2 + 1}{z^4 + z^2 + 1}dz.$

This is equivalent to $\mathcal I=\frac 12\int_0^\infty\ln\left(\frac{z^4+z^2+1}{z^2}\right)\frac{z^2+1}{(z^a+1)(z^4+z^2+1)}dz.$ Substitute $z\to\frac 1z$ to obtain $\mathcal I=\frac 12\int_0^\infty\ln\left(\frac{z^4+z^2+1}{z^2}\right)\frac{z^a(z^2+1)}{(z^a+1)(z^4+z^2+1)}dz.$ Thus,
$\begin{align*} \mathcal I&=\frac 14\int_0^\infty\ln\left(\frac{z^4+z^2+1}{z^2}\right)\frac{z^2+1}{(z^4+z^2+1)}dz\\ &=\frac 14\int_0^\infty\ln\left(z^2+z^{-2}+1\right)\frac{1+z^{-2}}{z^2+z^{-2}+1}dz\\ &=\frac 14\int_{-\infty}^{\infty}\frac{\ln(t^2+3)}{t^2+3}dt,\;\left[t=z-z^{-1}\right]\\ &=\frac 12\int_{0}^{\infty}\frac{\ln(t^2+3)}{t^2+3}dt\\ &=\boxed{\frac{\pi\ln(12)}{4\sqrt 3}}. \end{align*}$ Where the last result follows from $\int_0^\infty\frac{\ln(t^2+n)}{t^2+n}dt=\frac{\pi\ln(4n)}{2\sqrt n}.$

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