Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with integration
Search options not deleted
user 512834
Questions related to various forms of integration including the Riemann integral, Lebesgue integral, Riemann–Stieltjes integral, double integrals, line integrals, contour integrals, surface integrals, integrals of differential forms, ...
6
votes
1
answer
395
views
Conjectured closed form of $\int_0^1 \frac{\text{Li}_2\left(\frac{x}{4}\right)}{4-x}\,\log\l...
.$$
Approach:
Utilize the substitution $t=x/4$, integration by parts, and the identity $\sum_{n=1}^{\infty}H_{n}^{(2)}x^{n}=\frac{Li_{2}(x)}{1-x}$. …
- 224
1
vote
0
answers
101
views
Conjectured closed form of $\int_0^1\frac{\ln^3(1+x)\,\ln^3x}x\mathrm{d}x$
I posted this question on Math Stack Exchange, but there were no helpful comments or answers
https://math.stackexchange.com/q/4874446/1298448
How to integrate $${\displaystyle \int_0^1\frac{\ln^3(1+x) …
- 224
1
vote
Conjectured closed form of $\int\limits_0^1 \frac{\ln y \operatorname{Li}_2 (-y)}{1-y^2} \, dy$
Note section:
\begin{align*}
\begin{gathered}
\therefore \sum_{n=0}^{\infty} \frac{(-1)^n H_n^{(2)}}{n^2}=-4 L i_4\left(\frac{1}{2}\right)+\frac{51}{16} \zeta(4)-\frac{7}{2} \ln (2) \zeta(3)+\ln ^2(2) …
- 224
5
votes
3
answers
818
views
Conjectured closed form of $\int\limits_0^1 \frac{\ln y \operatorname{Li}_2 (-y)}{1-y^2} \, dy$
I uploaded this question here and here from my old account.
Let $\psi^{(1)}$ be the trigamma function defined by
\begin{equation}
\tag{1}
\psi_1(z) = -\int\limits_0^1 \frac{x^{z-1} \ln x}{1-x} \, dx.
…
- 224