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Evaluate $\lim_{n \rightarrow \infty} \int_{0}^{1} \sqrt{\frac{1}{x}+n^2x^{2n}}dx$.

The integral is non asymptomatic. The convergence is non uniform at both $x=0$ and $x=1$. I'm not sure how to proceed.

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3 Answers 3

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We have \begin{align} I_n &= \int_0^1 \sqrt{\frac{1}{x} + n^2 x^{2n}} \mathrm{d} x \\ &\le \int_0^1 \left(\sqrt{\frac{1}{x}} + nx^n\right) \mathrm{d} x \\ &\le \int_0^1 \left(\sqrt{\frac{1}{x}} + (n+1)x^n\right) \mathrm{d} x \\ &= \left(2\sqrt{x} + x^{n+1}\right)\Big\vert_0^1 \\ &= 3. \end{align} Also, we have \begin{align} I_n &= \int_0^{1-\frac{1}{\sqrt{n}}} \sqrt{\frac{1}{x} + n^2 x^{2n}}\ \mathrm{d} x + \int_{1-\frac{1}{\sqrt{n}}}^1 \sqrt{\frac{1}{x} + n^2 x^{2n}}\ \mathrm{d} x\\ &\ge \int_0^{1-\frac{1}{\sqrt{n}}} \sqrt{\frac{1}{x}}\ \mathrm{d} x + \int_{1-\frac{1}{\sqrt{n}}}^1 nx^n \mathrm{d} x \\ &= 2\sqrt{1 - \frac{1}{\sqrt{n}}} + \frac{n}{n+1} \left[1 - \left(1-\frac{1}{\sqrt{n}}\right)^{n+1}\right]. \end{align} Combining the above results, we have $$2\sqrt{1 - \frac{1}{\sqrt{n}}} + \frac{n}{n+1} \left[1 - \left(1-\frac{1}{\sqrt{n}}\right)^{n+1}\right] \le I_n \le 3.$$ Note that $$\lim_{n\to \infty} 2\sqrt{1 - \frac{1}{\sqrt{n}}} + \frac{n}{n+1} \left[1 - \left(1-\frac{1}{\sqrt{n}}\right)^{n+1}\right] = 3.$$ By the squeeze theorem, we have $\lim_{n\to \infty} I_n = 3$. We are done.

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    $\begingroup$ Very simple approach. +1 $\endgroup$
    – Paramanand Singh
    May 8, 2020 at 14:08
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    $\begingroup$ @ParamanandSingh Thank you! $\endgroup$
    – River Li
    May 8, 2020 at 14:09
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Let \begin{align} I = \int^1_0 \sqrt{\frac{1}{x}+n^2x^{2n}}\ dx. \end{align} Observe \begin{align} I_2=I-\int^1_0 \frac{dx}{\sqrt{x}} = \int^1_0 \frac{\sqrt{1+n^2x^{2n}}-1}{\sqrt{x}}\ dx = \int^1_0 \frac{n^2x^{2n}}{\sqrt{x}(\sqrt{1+n^2x^{2n}}+1)}\ dx. \end{align} To evaluate $I_2$, we will divided the interval. Fix $\delta=\frac{1}{n^2}$. Then we see that \begin{align} I_2 = \int^1_{1-\delta}+\int^{1-\delta}_\delta+ \int^\delta_0 \cdots\ dx =J_1+J_2+J_3. \end{align}

For $J_2$, the integrand converges uniformly to zero pointwise every where on the interval. Hence $\lim_{n\rightarrow \infty} J_2 = 0$.

For $J_3$, we see that for $n$ sufficiently large we have the estimate \begin{align} J_3 \le&\ \int^\delta_0 \frac{\delta}{\sqrt{x}(\sqrt{1+\delta}+1)}\ dx \leq\ \frac{2\delta\sqrt{\delta}}{\sqrt{1+\delta}+1}. \end{align}

Finally, for $J_1$. Observe that \begin{align} |J_1-1| \leq&\ \int^1_{1-\delta} dx\ \left|\frac{n^2}{\sqrt{1-\delta}(\sqrt{1+n^2x^{2n}}+1)}-\frac{1}{\delta}\right|\\ \leq& \left|\frac{n^2\delta}{\sqrt{1-\delta}(\sqrt{1+n^2(1-\delta)^{2n}}+1)}-1\right|. \end{align}

Hence, if you take $n\rightarrow \infty$, we see that $J_1 \rightarrow 1$, $J_2\rightarrow 0$ and $J_3\rightarrow 0$.

Thus $\lim_{n\rightarrow \infty} I_2 =1$ which means $I = 3$ since $\int^1_0 dx/\sqrt{x} = 2$.

Remark: Heuristically, we see that for $f_n(x) \rightarrow \frac{1}{\sqrt{x}}$ pointwise for any fixed point on $(0, 1)$. In fact, on $[\delta, 1-\delta]$ the convergence is uniform. Hence the integral of $\lim_n \int f_n$ coincides with the integral of $1/\sqrt{x}$ on that interval.

However, outside of the interval things get tricky. For the interval $[0, \delta]$, this is not a big problem since $n$ doesn't affect the integrand as long as $\delta$ is small. In particular, I chose $\delta = n^{-2}$ which is small enough for this to work (in fact, $n^{-2}$ was chosen so that the next part would work).

The most tricky part is the interval $[1-\delta, 1]$ because the area of $f_n$ concentrates around $1$ making the integral over the small interval not negligible. In fact, the main point is that $f_n \rightarrow \frac{1}{\sqrt{x}}+\delta(x-1)$ where $\delta(x)$ is the Dirac delta function. In fact, we see that \begin{align} \lim_{n\rightarrow \infty} \int^1_0 f_n(x)\ dx = \int^1_0 \frac{1}{\sqrt{x}}+\delta(x-1)\ dx = 2+1 = 3 \end{align}

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    $\begingroup$ For J2 isn't it a problem that the integration region varies with n when you exploit uniform convergence? Could you clarify a bit this? $\endgroup$
    – Thomas
    Sep 19, 2021 at 20:12
  • $\begingroup$ Further, how did you guess that J1 was tending to 1? $\endgroup$
    – Thomas
    Sep 19, 2021 at 20:14
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Hint :

Let $f_{n} := \sqrt{\frac{1}{x}+n^{2}{x^{2n}}}$, when $x$ is fixed, since $x \in (0,1)$ we note that $\lim\limits_{n \to +\infty} f_{n} = \frac{1}{\sqrt x}$, which is integrable on $(0,1)$.

If we can find an integrable function $g$, such that $|f_{n}(x)|\leq g(x)$, we can conclude by Dominated convergence Theorem the $\lim\limits_{n\to +\infty} \int_{0}^{1} f_{n} = \int_{0}^{1} \lim\limits_{n\to +\infty} f_{n} = \int_{0}^{1} \frac{1}{\sqrt x} = 2$.

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    $\begingroup$ Here dominated convergence does not apply since $x \mapsto nx^n$ is not (a.s) dominated by any integrable function on $[0,1]$, for $\int^1_0nx^n\,dx=\frac{n}{n+1}\xrightarrow{n\rightarrow\infty}1$ whereas $nx^n\xrightarrow{n\rightarrow\infty}0$ a.s. It was proven by River Li that the desired limit is in fact 3 (not 2 ) $\endgroup$ May 8, 2020 at 15:12

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