Prove that $$\lim_{n\to\infty} \int_0^1 \frac{x^n}{\sqrt{1+x^n}}\, \mathrm dx=0.$$

For full disclosure, this is for a homework problem for a real analysis class. I am stuck even figuring out how to approach this problem. The only thing that comes to mind is that the derivative of $\arctan(x)$, however I highly doubt that is useful.

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    $\begingroup$ That's a nice problem for my students, good post! +1 $\endgroup$ – imranfat Nov 12 '16 at 21:00

Observe that $$ 0\leq \frac{x^n}{\sqrt{1+x^n}}\leq x^n$$ for all $x\in[0,1]$, hence $$ 0\leq \int_0^1\frac{x^n}{\sqrt{1+x^n}}\;dx\leq \int_0^1x^n\;dx=\frac{1}{n+1}$$ for each $n\geq 1$.

  • $\begingroup$ perfect short answer! $\endgroup$ – imranfat Nov 12 '16 at 21:01

For each positive integer $n$, let $$f_n(x) = \frac{x^n}{\sqrt{1+x^n}},\ x\in[0,1].$$ Then for all $n$ and $x$, $$|f_n(x)|=\frac{x^n}{\sqrt{1+x^n}}\leqslant x^n\leqslant 1. $$ It follows from dominated convergence that $$\lim_{n\to\infty}\int_0^1 \frac{x^n}{\sqrt{1+x^n}}\,\mathsf dx = \int_0^1 \lim_{n\to\infty} \frac{x^n}{\sqrt{1+x^n}}\,\mathsf dx.$$ For $x\in(0,1)$, we have $$\lim_{n\to\infty} f_n(x) = 0, $$ and hence $$\int_0^1 \lim_{n\to\infty} f_n(x)\,\mathsf dx =0.$$


carmichael561's solution is the most efficient way to solve this problem. Here is a much less efficient solution.

Using the integral defintion (analytically continued) of Gauss's hypergeometric function \begin{equation} {}_{2}\mathrm{F}_{1}(a,b;c;z) = \frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)} \int\limits_{0}^{1} t^{a-1} (1-t)^{c-b-1} (1-zt)^{-a} dt \end{equation} for $\mathrm{Re}\,c \gt \mathrm{Re}\,b \gt 0,\,\,|\mathrm{arg}(1-z)| \lt \pi$

We have, using the substitution $y=x^{n}$ \begin{align} \int\limits_{0}^{1} \frac{x^{n}}{\sqrt{1+x^{n}}} dx &= \frac{1}{n} \int\limits_{0}^{1} \frac{y^{1/n}}{\sqrt{1+y}} dy \\ &= \frac{\Gamma(1+1/n)\Gamma(1)}{n\Gamma(2+1/n)} \,{}_{2}\mathrm{F}_{1}\left(\frac{1}{2},1+\frac{1}{n};2+\frac{1}{n};-1 \right) \\ \tag{1} &= \frac{1}{n+1} \,{}_{2}\mathrm{F}_{1}\left(\frac{1}{2},1+\frac{1}{n};2+\frac{1}{n};-1 \right) \end{align}

Taking the limit of the hypergeometric function and applying the integral definition again yields \begin{align} \lim_{n \to \infty} {}_{2}\mathrm{F}_{1}\left(\frac{1}{2},1+\frac{1}{n};2+\frac{1}{n};-1 \right) &= {}_{2}\mathrm{F}_{1}\left(\frac{1}{2},1;2;-1 \right) \\ &= \frac{\Gamma(2)}{\Gamma(1)\Gamma(1)} \int\limits_{0}^{1} \frac{1}{\sqrt{1+t}} dt \\ &= 2(\sqrt{2} - 1) \end{align}

Substituting this result into equation 1, we have \begin{equation} \lim_{n \to \infty} \int\limits_{0}^{1} \frac{x^{n}}{\sqrt{1+x^{n}}} dx = 0*2(\sqrt{2} - 1) = 0 \end{equation}


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