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What is lim_(n→∞) ∫_(-∞)^∞〖〖 x〗^n e^(-n|x|) dm 〗?
- Thread starter Jack
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- Feb 5, 2012
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Re: What is lim_(n→∞) ∫_(-∞)^∞〖〖 x〗^n e^(-n|x|) dm 〗?
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 collection of threads 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.
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 collection of threads 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.
- Thread starter
- #3
Re: What is lim_(n→∞) ∫_(-∞)^∞〖〖 x〗^n e^(-n|x|) dm 〗?
YEs。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 collection of threads 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.
- Feb 5, 2012
- 1,621
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, \(\displaystyle\lim_{n\rightarrow \infty}\frac{2n!}{n^{n+1}}=0\).
\[\therefore \lim_{n\rightarrow \infty}\int_{-\infty}^{\infty}\left|x\right|^{n} e^{-n|x|}\,dx=\lim_{n\rightarrow \infty}\frac{2n!}{n^{n+1}}=0\]
\[\int_{-\infty}^{\infty}\left|x\right|^{n} e^{-n|x|}\,dx=2\int_{0}^{\infty}x^{n} e^{-nx}\,dx\]
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, \(\displaystyle\lim_{n\rightarrow \infty}\frac{2n!}{n^{n+1}}=0\).
\[\therefore \lim_{n\rightarrow \infty}\int_{-\infty}^{\infty}\left|x\right|^{n} e^{-n|x|}\,dx=\lim_{n\rightarrow \infty}\frac{2n!}{n^{n+1}}=0\]