Articles of beta function

How to show $I_p(a,b) = \sum_{j=a}^{a+b-1}{a+b-1 \choose j} p^j(1-p)^{a+b-1-j}$

Show that $$I_p(a,b) = \frac{1}{B(a,b)}\int_0^p u^{a-1}(1-u)^{b-1}~du\\= \sum_{j=a}^{a+b-1}{a+b-1 \choose j} p^j(1-p)^{a+b-1-j}$$ when $a,b$ are positive integers. I have no idea how to proceed. Please help.

To evaluate integral using Beta function – Which substitution should i use?

$$\int_{0}^{1} \frac{x^{m-1}(1-x)^{n-1}}{(a+bx)^{m+n}}dx = \frac{B(m,n)}{(a+b)^ma^n}$$ I have to use some kind of substitution but i do not understand what i use and why ? Thanks

Sum: $\sum_{n=1}^\infty\prod_{k=1}^n\frac{k}{k+a}=\frac{1}{a-1}$

For the past week, I’ve been mulling over this Math.SE question. The question was just to prove convergence of $$\sum\limits_{n=1}^\infty\frac{n!}{\left(1+\sqrt{2}\right)\left(2+\sqrt{2}\right)\cdots\left(n+\sqrt{2}\right)}$$ but amazingly Mathematica told me it had a remarkably simple closed form: just $1+\sqrt{2}$. After some fiddling, I conjectured for $a>1$: $$\sum\limits_{n=1}^\infty\prod\limits_{k=1}^n\frac{k}{k+a}=\sum\limits_{n=1}^\infty\frac{n!}{(1+a)(2+a)\cdots(n+a)}=\frac{1}{a-1}$$ I had been quite stuck until today when I saw David H’s helpful […]

On the integral $\int_0^1\frac{dx}{\sqrtx\ \sqrt{1-x}\ \sqrt{1-x\,\beta^2}}=\frac{2\pi}{7\sqrt{2}\,\beta}$ and $\cos\frac{2\pi}{7}$

(This continues the post on integrals that use roots of reciprocal polynomials.) Given $N=7$. First, how do we show that the algebraic number $\beta$ that solves, $$\int_0^1\frac{dx}{\sqrt[4]x\ \sqrt[4]{1-x}\ \sqrt{1-x\,\beta^2}}=\frac{1}{7}\,\frac{2\pi}{\sqrt{2}\;|\beta|}\tag1$$ is a root of, $$\beta^2-\frac{2}{1+2\cos\tfrac{\pi}{14}}\beta+\frac{4}{2-\sqrt{2}\,\sec\tfrac{13\pi}{28}}=0$$ Second, the algebraic number $\gamma$ for the related integral, $$\int_0^1\frac{dx}{\sqrt[4]x\ \sqrt{1-x}\ \sqrt[4]{1-x\,\gamma^2}}=\frac{1}{7}\,\frac{2\pi}{\sqrt{2\gamma}}\tag2$$ is a root of, $$\gamma^2-2\cos\tfrac{2\pi}{7}\,\gamma+\frac{2-\sqrt{2}\,\csc\tfrac{9\pi}{28}}{4}=0$$ Equivalently, these roots $\beta$ […]

How to generalize Reshetnikov's $\arg\,B\left(\frac{-5+8\,\sqrt{-11}}{27};\,\frac12,\frac13\right)=\frac\pi3$?

We have, $$\arg z_1 = \frac{k\,\pi}3, \quad z_1 = \left(\tfrac{1+\sqrt{-3}}{2}\right)^k\tag1$$ $$\arg z_2=\frac{k\,\pi}3, \quad z_2 = \left( B\Big(\color{blue}{\tfrac{-5+8\,\sqrt{-11}}{27}};\,\tfrac12,\tfrac13\Big)\right)^k \tag2$$ for $k=1,2,3$ and incomplete beta function $B(z;a,b)$. The second with $k=1$ is by V. Reshetnikov. But that cannot be an isolated result. Reshetnikov states (without giving details) that $(2)$ is equivalent to, $$\begin{aligned}B\left(\frac19;\,\frac16,\frac13\right) &=\frac{\Gamma\big(\tfrac16\big)\,\Gamma\big(\tfrac13\big)}{2\sqrt{\pi}}\\ &=\frac12\,B\big(\tfrac16,\tfrac13\big) \end{aligned}\tag3$$ Thus, […]

Show that $\sum_{n=0}^{\infty}\frac{2^n(5n^5+5n^4+5n^3+5n^2-9n+9)}{(2n+1)(2n+2)(2n+3){2n\choose n}}=\frac{9\pi^2}{8}$

I don’t how prove this series and I have try look through maths world and Wikipedia on sum for help but no use at all, so please help me to prove this series. How to show that $$\sum_{n=0}^{\infty}\frac{2^n(5n^5+5n^4+5n^3+5n^2-9n+9)}{(2n+1)(2n+2)(2n+3)\dbinom{2n}{n}}=\frac{9\pi^2}{8}$$ I have tried the stuff in the tags.

A nice pattern for the regularized beta function $I(\alpha^2,\frac{1}{4},\frac{1}{2})=\frac{1}{2^n}\ $?

In this post, the problem was given integer/rational $N$, to solve for algebraic number $z$ in the equation, $$\begin{aligned}\frac{1}{N} &=I\left(z^2;\ a,b\right)\\[1.5mm] &= \frac{B\left(z^2;\ a,b\right)}{B\left(a,b\right)} \end{aligned} $$ using the beta function $B(a,b)$, incomplete beta $B(z;a,b)$ and regularized beta $I(z;a,b)$. It seems for some $a,b,$ there can be a pattern for the solutions. Given the equation for […]

Prove that $\int_0^1t^{p-1}(1-t)^{q-1}\,dt=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}$ for positive $p$ and $q$

I’m trying to prove that for $p,q>0$, we have $$\int_0^1t^{p-1}(1-t)^{q-1}\,dt=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}.$$ The hint given suggests that we express $\Gamma(p)\Gamma(q)$ as a double integral, then do a change of variables, but I’ve been unable thus far to express it as a double integral. Can anyone get me started or suggest an alternate approach? Note: This wasn’t actually […]

Why do these two integrals use roots of reciprocal polynomials?

There is the nice integral by V. Reshetnikov, $$\int_0^1\frac{dx}{\sqrt[3]x\ \sqrt[6]{1-x}\ \sqrt{1-x\,\alpha^2}}=\frac{1}{N}\,\frac{2\pi}{\sqrt{3}\;\alpha}\tag1$$ also discussed in this post. By direct analogy, we can consider its cousin, $$\int_0^1\frac{dx}{\sqrt[4]x\ \sqrt[4]{1-x}\ \sqrt{1-x\,\beta^2}}=\frac{1}{N}\,\frac{2\pi}{\sqrt{2}\;\beta}\tag{2a}$$ and the version investigated also by Reshetnikov, $$\int_0^1\frac{dx}{\sqrt[4]x\ \sqrt{1-x}\ \sqrt[4]{1-x\,\gamma^2}}=\frac{1}{N}\,\frac{2\pi}{\sqrt{2\gamma}}\tag{2b}$$ Given some integer/rational $N$, it entails finding algebraic numbers $\alpha, \beta, \gamma$ such that, $$\begin{aligned} \frac{1}{N}=I\left(\alpha^2;\ \tfrac12,\tfrac13\right)= \frac{B\left(\alpha^2;\ […]

Proving that $\int_0^\infty\Big(\sqrt{1+x^n}-x\Big)dx~=~\frac12\cdot{-1/n\choose+1/n}^{-1}$

How can we prove, without employing the aid of residues or various transforms, that, for $n>2$ $$\int_0^\infty\Big(\sqrt[n]{1+x^n}-x\Big)dx~=~\frac12\cdot{-1/n\choose+1/n}^{-1}$$ Motivation: In my previous question, thanks to Will Jagy’s simple but brilliant answer, I was able to express the area of the superellipse $x^n+y^n=r^n$, for odd values of $n=2k+1$, with $k\in\mathbb N^*$, as $A_n=r^2\displaystyle\cdot{2/n\choose1/n}^{-1}+r^2\cdot{-1/n\choose+1/n}^{-1}$, where the first term, […]