# How to calculate $\lim\limits_{n \rightarrow +\infty} \int_{0}^{+\infty} \frac{1}{(1+\frac{x}{n})^n x^{1/n} } dx$ using Beppo Levi theorem?

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!

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}

Hint: (DCT)

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

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