Skip to main content
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
Search options not deleted user 48831

Special functions, orthogonal polynomials, harmonic analysis, ordinary differential equations (ODE's), differential relations, calculus of variations, approximations, expansions, asymptotics.

14 votes
Accepted

Integral $\int_0^1 \int_0^1 \cdots \int_0^1\frac{x_{1}^2+x_{2}^2+\cdots+x_{n}^2}{x_{1}+x_{2}...

Here is another approach, which also gives the rational term. (I) To see how it works let $n\geq 2$ and consider first the simpler case \begin{align*} \mathbb{E}\bigg(\frac{1}{X_1+\ldots+X_n}\bigg)= …
esg's user avatar
  • 3,255
5 votes

(Sharp) inequality for Beta function

One can also use Jensen's inequality. Let (for $\sigma>0$) $G_\sigma$ denote a random variable with $\Gamma(1,\sigma)$-distribution, i.e. having Lebesgue density $$f_\sigma(t)=\frac{t^{\sigma-1}}{\G …
esg's user avatar
  • 3,255
24 votes

Binomial again, and again

Here is a proof which doesn't use the identity $\int_{-\infty}^\infty {n \choose x}\,dx= 2^n$: Using the representation ${ n \choose x}=\frac{1}{2\pi}\int_{-\pi}^\pi e^{-ixt}\left(1+e^{it}\right)^n\, …
esg's user avatar
  • 3,255
21 votes
Accepted

Asymptotic expansion of $\sum\limits_{n=1}^{\infty} \frac{x^{2n+1}}{n!{\sqrt{n}} }$

I sketch the arguments for $C(x)$, the arguments for $L(x)$ are essentially the same. The specific form of the sum suggests probabilistic arguments. Let $X_x$ be a $\mathrm{Poiss}(x^2)$-distributed …
esg's user avatar
  • 3,255