What is lim_(n→∞) ∫_(-∞)^∞〖〖 x〗^n e^(-n|x|) dm 〗?

  • Context: MHB 
  • Thread starter Thread starter Jack3
  • Start date Start date
Click For Summary

Discussion Overview

The discussion revolves around the limit of the integral \(\lim_{n\rightarrow\infty}\int_{-\infty}^{\infty} x^{n} e^{-n|x|}\,dx\). Participants explore the mathematical evaluation of this limit, including the use of substitutions and properties of the gamma function.

Discussion Character

  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant suggests that the integral can be rewritten as \(\int_{-\infty}^{\infty} |x|^{n} e^{-n|x|}\,dx = 2\int_{0}^{\infty} x^{n} e^{-nx}\,dx\).
  • Another participant provides a substitution \(y=nx\) to simplify the integral, leading to the expression \(\frac{2}{n^{n+1}}\Gamma(n+1)\).
  • A claim is made that \(\lim_{n\rightarrow \infty}\frac{2n!}{n^{n+1}}=0\), suggesting that the limit of the original integral approaches zero.
  • There is a suggestion for the original poster to learn LaTeX for clearer communication of mathematical expressions.

Areas of Agreement / Disagreement

Participants do not explicitly agree on the final outcome of the limit, although one participant claims it approaches zero. The discussion includes multiple viewpoints on the evaluation process without a consensus on the interpretation of the limit.

Contextual Notes

Assumptions include that \(n\) is a positive integer. The discussion does not resolve the implications of the limit or the conditions under which the integral is evaluated.

Jack3
Messages
9
Reaction score
0
What is lim_(n→∞) ∫_(-∞)^∞〖〖 x〗^n e^(-n|x|) dm 〗?

What is lim_(n→∞) ∫_(-∞)^∞〖〖 x〗^n e^(-n|x|) dm 〗?
Find the limit and prove your answer.
 
Last edited:
Physics news on Phys.org
Re: What is lim_(n→∞) ∫_(-∞)^∞〖〖 x〗^n e^(-n|x|) dm 〗?

Jack said:
What is lim_(n→∞) ∫_(-∞)^∞〖〖 x〗^n e^(-n|x|) dm 〗?
Find the limit and prove your answer.

Hi Jack, :)

I suggest you should learn some LaTeX before posting questions since the characters that you use makes it difficult to understand what your question is. We have a nice http://www.mathhelpboards.com/f26/ that you can use to learn LaTeX.

Is this your integral?

\[\lim_{n\rightarrow\infty}\int_{-\infty}^{\infty}\left|x\right|^{n} e^{-n|x|}\,dx\]

Kind Regards,
Sudharaka.
 
Re: What is lim_(n→∞) ∫_(-∞)^∞〖〖 x〗^n e^(-n|x|) dm 〗?

Sudharaka said:
Hi Jack, :)

I suggest you should learn some LaTeX before posting questions since the characters that you use makes it difficult to understand what your question is. We have a nice http://www.mathhelpboards.com/f26/ that you can use to learn LaTeX.

Is this your integral?

\[\lim_{n\rightarrow\infty}\int_{-\infty}^{\infty}\left|x\right|^{n} e^{-n|x|}\,dx\]

Kind Regards,
Sudharaka.

YEs。
 
Re: What is lim_(n→∞) ∫_(-∞)^∞〖〖 x〗^n e^(-n|x|) dm 〗?

\[\int_{-\infty}^{\infty}\left|x\right|^{n} e^{-n|x|}\,dx=2\int_{0}^{\infty}x^{n} e^{-nx}\,dx\]

Substitute \(y=nx\) and we get,

\begin{eqnarray}

\int_{-\infty}^{\infty}\left|x\right|^{n} e^{-n|x|}\,dx&=&\frac{2}{n^{n+1}}\int_{0}^{\infty}y^{n} e^{-y}\,dy\\

&=&\frac{2}{n^{n+1}}\Gamma(n+1)

\end{eqnarray}

I am assuming that \(n\) is a positive integer. Then,

\[\int_{-\infty}^{\infty}\left|x\right|^{n} e^{-n|x|}\,dx=\frac{2\Gamma(n+1)}{n^{n+1}}=\frac{2n!}{n^{n+1}}\]

It could be shown that, \(\displaystyle\lim_{n\rightarrow \infty}\frac{2n!}{n^{n+1}}=0\).

\[\therefore \lim_{n\rightarrow \infty}\int_{-\infty}^{\infty}\left|x\right|^{n} e^{-n|x|}\,dx=\lim_{n\rightarrow \infty}\frac{2n!}{n^{n+1}}=0\]
 

Similar threads

  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
6
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K