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Gaussian integral
soruz   3
N Today at 8:25 AM by Mathzeus1024
Exist a method of calculation for $ \int e^{-x^2}\,dx $, with help of $ e^{i \phi}=cos \phi + i sin \phi $ and Moivre's formula.
3 replies
soruz
Oct 20, 2013
Mathzeus1024
Today at 8:25 AM
J
H
Limit conundrum
MetaphysicalWukong   4
N Today at 7:42 AM by MetaphysicalWukong
Source: UNSW
Why is the last statement not true? And how do we know the selected option is true?
4 replies
MetaphysicalWukong
Yesterday at 8:00 AM
MetaphysicalWukong
Today at 7:42 AM
J
H
Finding supremum of a weird function
pokoknyaakuimut   4
N Today at 6:56 AM by MihaiT
Find $\text{sup}\{2^{2x}+2^{\frac{1}{2x}}:x\in\mathbb{R}, x<0\}$. Easy to guess that the answer is $1$, but I haven't found the reason yet. :(
4 replies
pokoknyaakuimut
Feb 14, 2025
MihaiT
Today at 6:56 AM
J
H
real analysis
ay19bme   3
N Yesterday at 8:46 PM by ay19bme
...........................
3 replies
ay19bme
Yesterday at 4:19 PM
ay19bme
Yesterday at 8:46 PM
J
H
Integration Bee Kaizo
Calcul8er   50
N Yesterday at 7:10 PM by Shikhar_
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
J
H
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
J
H
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
J
H
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
J
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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
J
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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
J
a