Inequality

Hey integration fans. I decided to collate some of my favourite and most evil integrals I've written into one big integration bee problem set. I've been entering integration bees since 2017 and I've been really getting hands on with the writing side of things over the last couple of years. I hope you'll enjoy!

50 replies

Calcul8er

Mar 2, 2025

Shikhar_

Yesterday at 7:10 PM

Putnam 1950 B1

centslordm 2

N Yesterday at 6:19 PM by KAME06

In each of $n$ houses on a straight street are one or more boys. At what point should all the boys meet so that the sum of the distances that they walk is as small as possible?

2 replies

centslordm

May 25, 2022

KAME06

Yesterday at 6:19 PM

Another integral

Martin.s 2

N Yesterday at 12:43 PM by MS_asdfgzxcvb

$I = \int_{0}^{\frac{1}{\sqrt{3}}} \frac{u \arctan(u)}{(1 - u^2) \sqrt{1 - 2 u^2}} \, du$

2 replies

Martin.s

Mar 9, 2025

MS_asdfgzxcvb

Yesterday at 12:43 PM

Some integrals and sums(series)

Martin.s 1

N Yesterday at 12:09 PM by Entrepreneur

Source: Inspired from silver08

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

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

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

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

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

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

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

1 reply

Martin.s

Dec 26, 2024

Entrepreneur

Yesterday at 12:09 PM

An integral

gaussiemann144 1

N Yesterday at 10:10 AM by vanstraelen

Given $\alpha, \beta$ -
$\alpha = \int_0^1 xe^{\frac{x^2-1}{2}} \cos(x) dx \quad \beta = \int_1^{3/2} e^{2(x^2-2x)} \sqrt{1-\cos(4x-4)} dx$
Find- $\frac{\alpha - \cos(1) + e^{-1/2}}{\beta}$

1 reply

gaussiemann144

Monday at 8:15 PM

vanstraelen

Yesterday at 10:10 AM

Ahlfors 3.3.1.2

centslordm 3

N Yesterday at 9:14 AM by Mathzeus1024

If $T_1 z = \frac{z + 2}{z + 3}, \qquad T_2 z = \frac z{z + 1},$ find $T_1 T_2z, \,T_2 T_1z$ and ${T_1}^{-1} T_2 z.$

3 replies

centslordm

Jan 8, 2025

Mathzeus1024

Yesterday at 9:14 AM

Inequality

anhduy98 3

N Dec 9, 2024 by mihaig

Given a real constant $k\in(0;2)$ and three real numbers $x,y,z$ satisfying $: x^2+y^2+z^2+k(xy+yz+zx)=1$ . Find the maximum value of the expression : $T=xyz.$

3 replies

anhduy98

Dec 8, 2024

mihaig

Dec 9, 2024

Inequality

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anhduy98

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anhduy98

#1 h Dec 8, 2024, 8:52 AM • 1 Y

Y by MihaiT

Given a real constant $k\in(0;2)$ and three real numbers $x,y,z$ satisfying $: x^2+y^2+z^2+k(xy+yz+zx)=1$ . Find the maximum value of the expression : $T=xyz.$

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mihaig

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mihaig

#4 h Dec 9, 2024, 10:47 AM

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More generally, here
https://artofproblemsolving.com/community/c6t243f6h3458088_since_november_4th_2023

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