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.
636 questions
0 votes
0 answers
28 views
Good known bounds for cumulative distribution functions
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 ...
1 vote
0 answers
47 views
How to derive relations between integrals over a triangle without evaluating them explicitly?
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 $\left\{(x,y) | x>0, y>0, x+y<1\right\}$. ...
8 votes
1 answer
208 views
Using the integral $\frac{1}{\binom{n}{r}} = (n+1)\int_0^1 x^r(1-x)^{n-r}\mathrm dx$
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 ...
4 votes
1 answer
117 views
Representation of the beta Euler function by an infinite sum
Where can I find the derivation of this formula? https://functions.wolfram.com/GammaBetaErf/Beta/06/03/0001/ $$B(a, b) = \sum^\infty_{k=0}\frac{(1 - b)_k}{(a + k) k!} ;\quad \Re(a) > 0 \;\text{ and ...
2 votes
1 answer
97 views
Asymptotic expansion of incomplete Beta function
Considering the incomplete beta function $$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}(...
2 votes
1 answer
127 views
Closed form or representation as a hypergeometric series of $\sum _{k=1}^{\infty } \frac{x^{k-1}}{B\left(k,-\frac{k}{n}\right)}$
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 ...
1 vote
1 answer
79 views
Definite integrals involving logarithm and polylog [closed]
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 ...
6 votes
2 answers
385 views
How to prove $I=\Gamma(\frac{1}{4})^2\frac{\sqrt{2}(\pi+4)}{8\sqrt{\pi}}-\pi-4$.
Context Some numerical work suggests that the following integral is true: $$\begin{align*} I&=\int_{0}^{1}\arcsin{(x^2)}\left(\frac{\log{(\sqrt{1-x^4}+1)}}{x^2}-\frac{2x(x(1-x^4)^{1/4}-1)}{(1-x^4)^...
2 votes
4 answers
352 views
General solution to $\int_{1}^{\infty} \frac{1}{x^{2k}+1} \ dx$ via beta function
I'm trying to tackle integrals of the form $$ \int_{1}^{\infty} \frac{1}{x^{2k}+1} \ dx $$ for large $k \in \mathbb{Z}^{+}$. An easily computable general solution is desired, rather than, say, a ...
2 votes
2 answers
109 views
Prove $ \sum_{k=0}^{2n} (-1)^k \beta\left(\frac{x+k}{2n+1}\right) = (2n+1) \beta(x) $
Prove the given identity: $$ \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} $$ The proof is more ...
4 votes
6 answers
220 views
Calculating $\int_0^1 \frac{\ln(x)}{\sqrt{x - x^2}}\, dx$
I'm stuck on this problem and would appreciate any help. I'm trying to compute the integral $$ \int_0^1 \frac{\ln(x)}{\sqrt{x - x^2}}\, dx. $$ To approach it, I define a parametric function $$ F(a) = \...
3 votes
0 answers
83 views
Evaluating the sum $\sum_{0}^{n}k^{i}(n-k)^{j}$ and its connection to the beta function
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}...
0 votes
1 answer
66 views
Confusion on Beta Function Integral Representation
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 ...
3 votes
5 answers
294 views
How can I evaluate the close form of $\int_0^{\pi}\left(\ln\left(\sin x\right)\right)^k\text dx$
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, \...
0 votes
1 answer
59 views
Finding values that make the Beta function symmetric
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 ...