Mathematics
Robert Lee's user avatar
Robert Lee's user avatar
Robert Lee's user avatar
Robert Lee
  • Member for 3 years, 7 months
  • Last seen this week
Profile Activity
Stats
6,746
reputation
46k
reached
165
answers
75
questions
Communities
View all
About

It's not trivial.

Badges
View all badges
1
gold badge
11
silver badges
45
bronze badges
Top tags
View all tags
257
Score
98
Posts
41
Posts %
215
Score
89
Posts
37
Posts %
135
Score
48
Posts
20
Posts %
53
Score
17
Posts
7
Posts %
28
Score
23
Posts
10
Posts %
27
Score
13
Posts
5
Posts %
Top posts
View all questions and answers
All Questions Answers
Score Newest
question
22
How to prove $\int_0^1 \frac{\arctan^2(x)\ln\left(\frac{x}{(1-x)^2}\right)}x \, \mathrm{d}x=G^2$?
Mar 16, 2022
answer
15
Find the value of $\displaystyle \int \limits _{0}^{\infty} \dfrac{\mathrm dx}{\sqrt{x^n+a} + \sqrt{x^n+b}}$
Mar 9, 2022
answer
15
Integral of the shark function
Sep 11, 2021
question
15
How to prove $\int_{0}^{\infty} \frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \right)}{(1+x^2)^2}\, dx = \frac{\zeta(3)}{\pi}$?
Sep 17, 2021
question
12
Find matrix $A\in \mathcal{M}_n (\mathbb{N})$ such that $A^k =\left( \sum_{i=1}^{k}10^{i-1} \right)A$.
Sep 27, 2020
question
12
How to show $ \int_{-\infty}^{\infty} \frac{e^{-(x+1)^2}}{1+e^{-x}}\mathrm{d}x = \frac{\left(2\sqrt[4]{e} -1 \right)\sqrt{\pi}}{2e}$?
Sep 25, 2022
question
10
How can I prove that these definitions of curl are equivalent?
Jul 13, 2020
answer
9
What other tricks and techniques can I use in integration?
Apr 26, 2021
answer
8
How can I solve this defined integration in an efficient way?
May 11, 2022
answer
8
Find the minimum of $\sqrt{\cos x+3}+\sqrt{2\sin x+7}$ without derivative
Jan 6
Top network posts
View all network posts
8
How do I create this semi-enclosed arrow symbol?
6
Could Throg survive being locked in a jar?
5
Any TV news pieces/stickers?

Your privacy

By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.