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- Thread starter Jack
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- Feb 5, 2012

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

Find the limit and prove your answer.

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.

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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

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

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

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