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I'm trying to find $\lim\limits_{n \rightarrow +\infty} \int\limits_{0}^{+\infty} \frac{1}{(1+\frac{x}{n})^n x^{1/n} } dx$. My numerical test with large numbers instead of the infinity shows that result is 1 (I suppose). Also I tried to apply some facts from real analysis. For example, Beppo Levi theorem.

So, can I just swap the integral and limit by this theorem?

Thank you for help!

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2 Answers 2

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You don't even need such an exchange. With $x=n\tan^{2}t$,$$\begin{align}\int_0^\infty\frac{x^{-1/n}dx}{\left(1+\frac{x}{n}\right)^{n}}&=\int_0^{\pi/2}2n^{1-1/n}\sin^{1-2/n}t\cos^{2n+2/n-3}tdt\\&=n^{1-1/n}\text{B}\left(1-\frac{1}{n},\,n+\frac{1}{n}-1\right)\\&=n^{1-1/n}\frac{\Gamma\left(1-\frac{1}{n}\right)\Gamma\left(n+\frac{1}{n}-1\right)}{\Gamma\left(n\right)}\\&\stackrel{n\to\infty}{\sim}1.\end{align}$$

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Hint: (DCT)

For all $n \geqslant 2$, we have $\displaystyle\left(1 + \frac{x}{n} \right)^{-n}x^{-1/n} \leqslant \begin{cases}\left(1 + \frac{x}{2} \right)^{-2}, &x \geqslant 1\\x^{-1/2}, &0 < x < 1 \end{cases}$

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  • $\begingroup$ What does abbreviation "DCT" mean? $\endgroup$
    – margo
    Jan 5, 2021 at 20:57
  • $\begingroup$ The dominated convergence theorem. If $|f_n(x)| \leqslant g(x)$ and $\int_0^\infty g$ exists, then $ \lim_{n \to \infty}\int_0^\infty f_n = \int_0^\infty \lim_{n \to \infty} f_n $ $\endgroup$
    – RRL
    Jan 5, 2021 at 21:00
  • $\begingroup$ @margo Not to be focused with another CT, which is more similar in spirit to the titular theorem. $\endgroup$
    – J.G.
    Jan 5, 2021 at 21:02

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