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Questions tagged [beta-function]

For questions about the Beta function (also known as Euler's integral of the first kind), which is important in calculus and analysis due to its close connection to the Gamma function. It is advisable to also use the [special-functions] tag in conjunction with this tag.

0 votes
0 answers
28 views

Since I have to teach a course involving some statistical hypothesis tests, while not being a statistician myself, I am looking for some simple, but possibly accurate, bounds for cumulative ...
Riccardo Pengo's user avatar
1 vote
0 answers
47 views

Solving a problem I found several integrals that look like $$I(z) = \int \frac{f(x,y)}{(x y)^{1 + z}} dx dy$$ where the integral is over the triangle $\left\{(x,y) | x>0, y>0, x+y<1\right\}$. ...
Gaussian97's user avatar
8 votes
1 answer
208 views

Using the integral, $$\frac{1}{\binom{n}{r}} = (n+1)\int_0^1 x^r(1-x)^{n-r} \mathrm dx,$$ I want to solve some binomial questions. For instance, it seems, I might be unclear as to what I require. I am ...
Cuckoo Beats's user avatar
4 votes
1 answer
117 views

Where can I find the derivation of this formula? https://functions.wolfram.com/GammaBetaErf/Beta/06/03/0001/ $$B(a, b) = \sum^\infty_{k=0}\frac{(1 - b)_k}{(a + k) k!} ;\quad \Re(a) > 0 \;\text{ and ...
kkapitonets's user avatar
2 votes
1 answer
97 views

Considering the incomplete beta function $$B_x(a, b) = \int_0^x y^{a-1}(1-y)^{b-1} \:dy,$$ I am interested in deriving the following asymptotic expansion that I determined using Mathematica $$B_{1-x}(...
VerwirrterStudent's user avatar
2 votes
1 answer
127 views

Working on a problem I came across this series, \begin{equation} \frac{1}{2}+(\frac{n-1}{n})\sum_{k=1}^{\infty}\frac{x^{-\frac{k(n-1)}{n}}}{k(k+1)B\left(k,-\frac{k}{n}\right)} \end{equation} Removing ...
Francisco Alvarado's user avatar
1 vote
1 answer
79 views

I am dealing with integrals in the form $\mathcal{I} = \int_0^{1} dx \, x^\alpha \, (1-x)^\beta \, log(x)$ And $\mathcal{J} = \int_0^{1} dx \, x^\alpha \, (1-x)^\beta \, Li_n(x)$ . I have written the ...
Rrapo Hasimaj's user avatar
6 votes
2 answers
385 views

Context Some numerical work suggests that the following integral is true: $$\begin{align*} I&=\int_{0}^{1}\arcsin{(x^2)}\left(\frac{\log{(\sqrt{1-x^4}+1)}}{x^2}-\frac{2x(x(1-x^4)^{1/4}-1)}{(1-x^4)^...
User-Refolio's user avatar
  • 1,411
2 votes
4 answers
352 views

I'm trying to tackle integrals of the form $$ \int_{1}^{\infty} \frac{1}{x^{2k}+1} \ dx $$ for large $k \in \mathbb{Z}^{+}$. An easily computable general solution is desired, rather than, say, a ...
Sup2.0's user avatar
  • 103
2 votes
2 answers
109 views

Prove the given identity: $$ \sum_{k=0}^{2n} (-1)^k \beta\left(\frac{x+k}{2n+1}\right) = (2n+1) \beta(x) $$ Where $$ \beta(x) = \sum_{j=0}^{\infty} \frac{(-1)^j}{x+j} $$ The proof is more ...
user avatar
4 votes
6 answers
220 views

I'm stuck on this problem and would appreciate any help. I'm trying to compute the integral $$ \int_0^1 \frac{\ln(x)}{\sqrt{x - x^2}}\, dx. $$ To approach it, I define a parametric function $$ F(a) = \...
MiguelCG's user avatar
  • 826
3 votes
0 answers
83 views

I want to compute the sum: $$\sum_{0}^{n}k^{i}(n-k)^{j}$$ If I simply expand the polynomial, I get: $=\sum_{k=0}^{R}\sum_{l=0}^{j}k^{i}\binom{j}{l}R^{l}(-1)^{j-l}k^{j-l}\\ =\sum_{l=0}^{j}\binom{j}{l}...
soup's user avatar
  • 31
0 votes
1 answer
66 views

I found this beta function representation integral very pleasing, so I tried to work it out—but something went horribly wrong. I'm wondering if I am wrong, or if the book is. I've been working on this ...
EM4's user avatar
  • 1,277
3 votes
5 answers
294 views

Hmm, that is what we have. A "close form" using Beta function: $$\begin{align}\int_0^\pi \left(\ln(\sin x)\right)^k\text dx=\frac{1}{2^k} \frac{\partial^k}{\partial a^k} \text B\left(a, \...
welikestudying's user avatar
0 votes
1 answer
59 views

For finding when the median of a beta distribution is $\frac{1}{2}$, this answer says: If a $\mathrm{Beta}(a,b)$ distribution has $a>b$ then $\mathbb P(X \le \frac12) \lt \frac12$ and its median ...
Starlight's user avatar
  • 2,674

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