41 votes
Accepted

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

Using Beta function, one can express: $$\sum_{n=0}^{\infty}\frac{2^n(5n^5+5n^4+5n^3+5n^2-9n+9)}{(2n+1)(2n+2)(2n+3){2n\choose n}}$$ $$=\frac{5n^5+5n^4+5n^3+5n^2-9n+9}{(n+1)^2}\cdot\frac{1}{2}\...
Mc Cheng's user avatar
  • 2,298
26 votes
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Prove: $\int_0^2 \frac{dx}{\sqrt{1+x^3}}=\frac{\Gamma\left(\frac{1}{6}\right)\Gamma\left(\frac{1}{3}\right)}{6\Gamma\left(\frac{1}{2}\right)}$

An elementary solution: Consider the substitution $$t = \frac{{64 + 48{x^3} - 96{x^6} + {x^9}}}{{9{x^2}{{(4 + {x^3})}^2}}}$$ $t$ is monotonic decreasing on $0<x<2$, and $$\tag{1}\frac{{dx}}{{\...
pisco's user avatar
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18 votes
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Solve the integral using beta functions : $\int_0^1 \frac{(1-x^4)^{3/4}}{(1+x^4)^2}\,dx$

Apply the substitution $x^4 = \frac{1-u}{1+u}$. Then $$ I := \int_{0}^{1} \frac{(1-x^4)^{3/4}}{(1+x^4)^2} \, dx = \frac{1}{2^{9/4}} \int_{0}^{1} u^{3/4}(1-u)^{-3/4} \, du = \frac{1}{2^{9/4}} B(\tfrac{...
Sangchul Lee's user avatar
11 votes

Integrals related to $\int_0^{\pi} \left(\frac{\sin(\alpha u)^\alpha \sin((1-\alpha) u)^{1-\alpha}}{\sin u} \right)^{\rho/\alpha}du$

The claim seems highly nontrivial, this integral definitely deserves more attention. It is equivalent to prove $$\int_0^\pi {{{\left[ {\frac{{\sin^a {{(ax)}}\sin^{1-a} {{((1 - a)x)}}}}{{\sin x}}} \...
pisco's user avatar
  • 18.9k
10 votes

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

Here is an answer based upon the arcsine function. We start with the following formula valid for $u\in(0,2)$ \begin{align*} \sum_{n=0}^\infty&\frac{2^{n-1}}{(2n+1)(2n+3)\binom{2n}{n}}u^n\\ &...
Markus Scheuer's user avatar
10 votes
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Writing the Beta Function in terms of the Gamma Function

$$f(\alpha, \beta, t) = \int_0^t x^{\alpha -1} (t-x)^{\beta -1}\;dx = \big(t^{\alpha - 1} * t^{\beta -1}\big)(t)$$ so using the fact that convolution is multiplicative in the Laplace domain, \begin{...
adfriedman's user avatar
  • 3,641
9 votes
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$ \int_0^\frac{\pi}{2}\ln^n\left(\tan(x)\right)\:dx$

$$\begin{split} I_{2k} &= \int_0^\infty\frac{\ln^{2k}u}{u^2 + 1}du \\ &= \int_0^1\frac{\ln^{2k}u}{u^2 + 1}du +\int_1^{+\infty}\frac{\ln^{2k}\left(u\right)}{u^2 + 1}du \\ &=\int_0^1\frac{\...
Stefan Lafon's user avatar
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8 votes

About the relation between $\int_{0}^{\infty}\frac{\cosh((\alpha-\beta)x)}{(\cosh(x))^{\alpha+\beta }} \,dx$ and $B(\alpha,\beta)$

I will show you how to go from the second integral to the first one: so, in order to accomplish your task, you will have to take this answer and read it backwards ($(\alpha,\beta)B=(\beta,\alpha)B$ ...
Jack D'Aurizio's user avatar
8 votes

Prove: $\int_0^2 \frac{dx}{\sqrt{1+x^3}}=\frac{\Gamma\left(\frac{1}{6}\right)\Gamma\left(\frac{1}{3}\right)}{6\Gamma\left(\frac{1}{2}\right)}$

A hypergeometric solution: Modulo Beta function $I_0=\int_0^{\infty } \frac{1}{\sqrt{x^3+1}} \, dx=\frac{2 \Gamma \left(\frac{1}{3}\right) \Gamma \left(\frac{7}{6}\right)}{\sqrt{\pi }}$ one may ...
Infiniticism's user avatar
  • 8,604
8 votes

Proving that $I=\int_0^\infty \frac{n x^{n-1}}{(1+x)^{n+1}}dx<+\infty$

The integrand has a direct antiderivative, which we can see by rewriting the terms like so $$I = \int_0^\infty n\left(\frac{x}{x+1}\right)^{n-1}\frac{dx}{(x+1)^2} = \int_0^\infty n\left(1-\frac{1}{x+1}...
Ninad Munshi's user avatar
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8 votes
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Is Beta function essential to evaluating the integral $\int_0^{\infty} \frac{d x}{\left(1+x^4\right)^n}$?

Any standard method (including scaling $x$ in the first display equation in the question statement) yields $$\int_0^\infty \frac{dx}{a + x^4} = \frac{\pi}{2 \sqrt 2} a^{-3 / 4} .$$ If $n$ is a ...
Travis Willse's user avatar
7 votes
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Compute $\int_0^1x^m(1-x^n)^pdx$

Hint. As @Did has noticed, one may recall the Euler Beta integral result: $$ B(a,b)=\int\limits_0^1 t^{a-1}(1-t)^{b-1}\mathrm{d}t=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}, \quad a>0,\,b>0. $$ ...
Olivier Oloa's user avatar
7 votes
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Integrating $\int_{0}^{1} x^a (c-x)^b dx $

Enforcing $x =ct$ gives $$\int_{0}^{1} x^a (c-x)^b dx= c^{a+b+1}\int_{0}^{c} t^a (1-t)^b dt =\color{red}{c^{a+b+1} \mathrm {B} (c;\,a,b) }$$ where $\mathrm {B} (c;\,a,b)$is the incomplete beta ...
Guy Fsone's user avatar
  • 23.8k
7 votes

Prove that $\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{1}{(k+1)(k+2)}=\frac{1}{n+2}$

$$\frac{1}{(k+1)(k+2)}=\frac{1}{k+1}-\frac{1}{k+2} = \int_{0}^{1} x^k (1-x)\,dx $$ hence $$\begin{eqnarray*} \sum_{k=0}^{n}(-1)^k\binom{n}{k}\frac{1}{(k+1)(k+2)} &=& \int_{0}^{1}(1-x)\sum_{k=0}...
Jack D'Aurizio's user avatar
7 votes

Quick evaluation of the Gamma function?

$$ \Gamma(z+1) = z\Gamma(z) $$ That means $$ \frac{\Gamma(4 + 2)}{\Gamma(4)} = \frac{5\Gamma(4 + 1)}{\Gamma(4)} = \frac{4\times 5 \Gamma(4)}{\Gamma(4)} = 20 $$
caverac's user avatar
  • 19.2k
7 votes

Solve the integral using beta functions : $\int_0^1 \frac{(1-x^4)^{3/4}}{(1+x^4)^2}\,dx$

We can use the Binomial Theorem in the form $$ (1+x)^{-\alpha}=\sum_{k=0}^\infty\frac{(-1)^k\Gamma(k+\alpha)}{\Gamma(\alpha)\Gamma(k+1)}x^k\tag1 $$ to get $$ \begin{align} \int_0^1\frac{(1-x^4)^{3/4}}{...
robjohn's user avatar
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7 votes
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Integrals related to the reciprocal beta function

Let's recall (and modify) one of the known approaches to get the first of these integrals. Assume (temporarily) that $0<\Re a<\Re b$. Let $\gamma_+$ be the boundary of $$\{z:|z|<1,0<\arg z&...
metamorphy's user avatar
  • 38.8k
7 votes

How to evaluate the integral $I = \int_o^{\infty} \frac{x}{\sqrt{e^{2\pi\sqrt{x}}-1}}dx$?

Too long for a comment. If we consider the problem of $$I_n=\int \frac{t^n}{\sqrt{e^t-1}}dt$$ for $n>1$ they all write as $$I_n=-2 t^n \sin ^{-1}\left(e^{-t/2}\right)-4ne^{-t/2} J_n$$ where $J_n$ ...
Claude Leibovici's user avatar
7 votes
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How to evaluate $\sum_{k=1}^{\infty}\frac{B\left(k, \frac{1}{2}\right)}{(2k+1)^2}$, where $B(x, y)$ is the Beta function?

Inspired by this answer, we write \begin{align}S&=\sum_{k\ge1}\frac{\operatorname B(k,1/2)}{(2k+1)^2}=\sum_{k\ge1}\frac{4^k}{k(2k+1)^2{2k\choose k}}=\sum_{k\ge1}\frac{\int_0^{\pi/2}\sin^{2k+1}x\,...
TheSimpliFire's user avatar
  • 27k
6 votes

Anti-treta function in terms of standard special functions

Why (3) is so different from (1) ? Because if we rewrite the double integrals as iterated ones, for (1) we get $$\int_0^1 \left(1-x_2\right)^{\alpha_3-1}\left({\color{blue}{\int_0^{x_2}x_1^{\alpha_1-...
Start wearing purple's user avatar
6 votes
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How to compute $\int_0^1x^a(1-x)^be^{cx}dx$?

This is essentially the the moment generating function of the beta distribution. The result is hypergeometric and cannot be further simplified.
Alex R.'s user avatar
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6 votes
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Closed form for this integral with Beta function $\int_0^\infty x \mathrm B (x,x)~dx$

There is a kind of series 'solution' which I will put up as an answer in hope that someone will offer a better one. Let's work with this form of the integral: $$2 \int_0^{ 1/ \ln 4} \frac{dv}{\sqrt{...
Yuriy S's user avatar
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6 votes
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Prove this integral $\int_0^\infty \frac{dx}{\sqrt{x^4+a^4}+\sqrt{x^4+b^4}}=\frac{\Gamma(1/4)^2 }{6 \sqrt{\pi}} \frac{a^3-b^3}{a^4-b^4}$

$$\int_0^\infty (\sqrt{x^4+a^4}-\sqrt{x^4+b^4}) dx \implies $$ $$\int_0^\infty (\sqrt{x^4+a^4}-x^2-(\sqrt{x^4+b^4}-x^2)) dx$$ Because the $\int_0^\infty (\sqrt{x^4+a^4}-x^2)dx $ is convergent so the ...
Zau's user avatar
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6 votes
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Beta function-like integral

The key is the linear-fractional transformation, $$x\mapsto\frac{1-t}{1+t}.$$ Then $1-x=\frac{2t}{1+t}$ and $1+x=\frac{2}{1+t}$, and we find $$\begin{align} \int_{0}^{1}\frac{x^{1-a}\left(1-x\right)...
David H's user avatar
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6 votes
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Expected value of the k-th order statistic from uniform random variables

I think you're messing up in this way: $\Gamma(m) = (m-1)$! for a positive integer $m$. I'm getting \begin{align*} E(X_{(k)}) &=\frac{n!}{(k-1)!(n-k)!}\int_0^1 x^{k}[1-x]^{n-k}dx \\[5pt] &= ...
user365239's user avatar
  • 1,978
6 votes

Prove the following beta integral identity

With the convention $$\widehat f(p)=\int f(x)\exp(-ix\cdot p)dx$$ I assume you know $$\widehat{|x|^{-\alpha}}(p)=c_{n-\alpha,n}|p|^{\alpha-n}\text{ where } c_{\alpha,n}=\pi^{n/2}2^\alpha\frac{\Gamma(\...
Dap's user avatar
  • 25.3k
6 votes
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Laplace transform to prove relationship between the Gamma and Beta functions

In the ensuing analysis, we will make use of the Laplace Transform Pair $$\begin{align} f(t) &\leftrightarrow F(s)\\\\ t^{a-1} &\leftrightarrow \frac{\Gamma(a)}{s^a}\tag1 \end{align}$$ ...
Mark Viola's user avatar
  • 178k
6 votes
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Quick evaluation of the Gamma function?

$$\Gamma(n) = (n-1)!$$ So $\Gamma(4+2)=\Gamma(6)=5!=120$ and $\Gamma(4) = 3! = 6$, thus $$\frac{\Gamma(4+2)}{\Gamma(4)} = \frac{120}{6} = 20$$
gd1035's user avatar
  • 590
6 votes
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Series of binomial coefficient denominators

No Closed Form As pointed out by @drhab, your sum is equivalent to $\frac{1}{n!}\sum_{k=i}^n\binom{n}{k}$. Given that $\sum_{k=0}^n\binom{n}{k}=2^n$, your sum can be rearranged to $$\frac{2^n}{n!}-\...
Jam's user avatar
  • 10.3k

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