Consider the integral $Z(0)=\int^{\infty}_{-\infty} dx e^{-x^2}= \sqrt{\pi}$. (a) Show that the integral $Z(j)=\int^{\infty}_{-\infty} dx e^{-x^{2}+jx}$, where $j$ is not a function of $x$, is $Z(j)=e^{j^{2}/4a} Z(0)$. (b) Show that $$\dfrac 1 {Z(0)}=\int x^{2n} e^{-x^2}= \dfrac {(2n-1)!!}{2^n},$$where $(2n-1)!!$ is defined as $(2n-1)(2n-3)\times\cdots\times3\times 1$. (c) What is the number of ways to form $n$ pairs from $2n$ distinct objects? Interpret the previous part of the problem in term of this answer.
Problem
Source: Bangladesh National Mathematical Olympiad 2016
Tags: combinatorics, calculus, contest problem, definite integrals, Integers
ubermensch
23.02.2019 19:43
What is $a$ in part a of the problem
MathPassionForever
23.02.2019 19:44
And suddenly $\frac{1}{Z(0)}$ depends on $n$?
Aniv
23.02.2019 20:39
This is the expression of the gamma distribution and the integral is $Z(0)=\int^{\infty}_{-\infty} e^{-x^2}dx=2\int^{\infty}_{0} e^{-x^2}dx =\int^{\infty}_{0} e^{-z}z^{\frac{1}{2}-1}dz=\Gamma(\frac{1}{2})$ ; consider $x^2=z$ again $\int_{-\infty}^{\infty} x^{2n} e^{-x^2}dx=\int_{0}^{\infty}e^{-z}z^{n+\frac{1}{2}-1}dz=\Gamma(n+\frac{1}{2})$
Olympus_mountaineer
23.02.2019 21:24
ubermensch wrote: What is $a$ in part a of the problem I am also thinking about it.The question was same as my post.
Olympus_mountaineer
24.02.2019 16:57
This solution is taken from Principal and Techniques of Combinatorics
Pick an arbitary element say $x$ from $2n$ numbers.The number of ways to select $x's$ partner say $y$ is $2n-1$ and {$x,y$} forms a pair.
Pick an arbitary element say $z$ from $2n-2$ numbers.The number of ways to select $z's$ partner say $w$ is $2n-3$ and they also form another pair
Olympus_mountaineer
10.03.2019 19:06
Aniv wrote: This is the expression of the gamma distribution and the integral is $Z(0)=\int^{\infty}_{-\infty} e^{-x^2}dx=2\int^{\infty}_{0} e^{-x^2}dx =\int^{\infty}_{0} e^{-z}z^{\frac{1}{2}-1}dz=\Gamma(\frac{1}{2})$ ; consider $x^2=z$ again $\int_{-\infty}^{\infty} x^{2n} e^{-x^2}dx=\int_{0}^{\infty}e^{-z}z^{n+\frac{1}{2}-1}dz=\Gamma(n+\frac{1}{2})$ Not understanding the last part of the solution.
Aniv
11.03.2019 05:58
Consider $x^2=z$ variable change, then the integral becomes gamma function