6
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Very simple approach. +1 $\endgroup$ – Paramanand Singh May 8 '20 at 14:08
  • 1
    $\begingroup$ @ParamanandSingh Thank you! $\endgroup$ – River Li May 8 '20 at 14:09
2
$\begingroup$

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}

$\endgroup$
0
$\begingroup$

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

$\endgroup$
  • 3
    $\begingroup$ 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 ) $\endgroup$ – Oliver Diaz May 8 '20 at 15:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.