# Find $\lim_{n\to\infty}\int_0^{\frac {\pi}{3}}\frac {\sin^nx}{\sin^nx+\cos^nx}dx$

Evaluate $\lim_{n\to\infty}\int_0^{\frac {\pi}{3}}\frac {\sin^nx}{\sin^nx+\cos^nx}dx$

I tried using the substitution $u=\frac {\pi}{2}-x$ and maybe thought this function might be symmetrical in some way and got: $$\lim_{n\to\infty}\int_{\frac {\pi}{6}}^{\frac {\pi}{2}}\frac {\cos^nx}{\sin^nx+\cos^nx}dx$$

but it's not really helping me... any other ideas?

• Hint: if $a, b > 0$, then $$\lim_{n\to\infty} \frac{a^n}{a^n+b^n} = \begin{cases} 1, & a>b \\ \frac{1}{2}, & a = b \\ 0, & a < b \end{cases}$$ – Sangchul Lee Jun 7 '18 at 6:13
• @SangchulLee Yeah but $\cos x$ and $\sin x$ to $\infty$... I don't think they work like that – C. Cristi Jun 7 '18 at 6:16
• @SangchulLee Obviously useful, but is it sufficient? The limit function is not continuous, so convergence is not uniform. Can we still justify integrating the pointwise limit? If dominated convergence is a known tool, then... – Jyrki Lahtonen Jun 7 '18 at 6:17
• @JyrkiLahtonen, That is a valid objection. My initial guess was that this is a calculus-level question so that certain technicality may be deliberately ignored. If a basic level of analysis is also available, then writing $I_n$ for the integral and using the monotonicity of the integrand of $I_n$, with $a_\pm=\frac{\pi}{4}\pm\epsilon$ we have $$\left(\frac{\pi}{3}-a_+\right)\frac{\sin^n(a_+)}{\sin^n(a_+)+\cos^n(a_+)}\leq I_n\leq a_-\frac{\sin^n(a_-)}{\sin^n(a_-)+\cos^n(a_-)}+\frac{\pi}{3}-a_-.$$ This gives a bound on limsup/inf and then letting $\epsilon\to0$ yields the answer. – Sangchul Lee Jun 7 '18 at 6:28
• @C.Cristi, My intention was that you may take a leap of faith to interchange the order of integral and limit$$\lim_{n\to\infty}\int_{0}^{\frac{\pi}{3}}\frac{\sin^n x}{\sin^n x+\cos^n x}\,dx=\int_{0}^{\frac{\pi}{3}}\lim_{n\to\infty}\frac{\sin^n x}{\sin^n x+\cos^n x}\,dx$$ and then evaluate the inner limit. Of course, this is indeed a huge leap and deserved some justification for mathematical rigor. – Sangchul Lee Jun 7 '18 at 6:37

A heuristic solution. Interchanging the order of integral and limit, we have

$$\lim_{n\to\infty} \int_{0}^{\frac{\pi}{3}} \frac{\sin^n x}{\sin^n x+\cos^n x} \, dx = \int_{0}^{\frac{\pi}{3}} \lim_{n\to\infty} \frac{\sin^n x}{\sin^n x+\cos^n x} \, dx.$$

Now depending on the relative size of $\sin x$ and $\cos x$, we have

$$\lim_{n\to\infty}\frac{\sin^n x}{\sin^n x+\cos^n x} = \begin{cases} 0, & \text{if } 0 \leq x < \frac{\pi}{4} \\ \frac{1}{2}, & \text{if } x = \frac{\pi}{4} \\ 1, & \text{if } \frac{\pi}{4} < x \leq \frac{\pi}{3}. \end{cases} \tag{1}$$

So it follows that the limit is $\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} dx = \frac{\pi}{12}$.

Remark. Indeed we took a huge leap by assuming that the integral and the limit can be switched. This is indeed possible in our case, although a direct justification required some advance results such as dominated convergence theorem.

A preliminary analysis level solution. Fix any sufficiently small $\epsilon > 0$ and consider $a=\frac{\pi}{4}-\epsilon$ and $b = \frac{\pi}{4}+\epsilon$. If we write $I_n = \int_{0}^{\frac{\pi}{3}} \frac{\sin^n x}{\sin^n x+\cos^n x} \, dx$, then the integrand is monotone increasing in $x$ and hence

\begin{align*} I_n &\geq \int_{b}^{\frac{\pi}{3}} \frac{\sin^n x}{\sin^n x+\cos^n x} \, dx \\ &\geq \int_{b}^{\frac{\pi}{3}} \frac{\sin^n b}{\sin^n b+\cos^n b} \, dx \\ &= \left(\frac{\pi}{3}-b\right)\frac{\sin^n b}{\sin^n b+\cos^n b} \end{align*}

This gives

$$\liminf_{n\to\infty} I_n \geq \lim_{n\to\infty} \left(\frac{\pi}{3}-b\right)\frac{\sin^n b}{\sin^n b+\cos^n b} = \frac{\pi}{12}-\epsilon.$$

But since the LHS of the above inequality is a constant independent of $\epsilon$, letting $\epsilon \downarrow 0$ proves that $\liminf_{n\to\infty} I_n \geq \frac{\pi}{12}$. Similarly,

\begin{align*} I_n &\leq \int_{0}^{a} \frac{\sin^n x}{\sin^n x+\cos^n x} \, dx + \int_{a}^{\frac{\pi}{3}} \frac{\sin^n x}{\sin^n x+\cos^n x} \, dx \\ &\geq \int_{0}^{a} \frac{\sin^n a}{\sin^n a+\cos^n a} \, dx + \int_{a}^{\frac{\pi}{3}} dx \\ &= a\frac{\sin^n a}{\sin^n b+\cos^n a} + \left(\frac{\pi}{3}-a\right) \end{align*}

and hence

$$\limsup_{n\to\infty} I_n \leq \lim_{n\to\infty} \left[ a\frac{\sin^n a}{\sin^n b+\cos^n a} + \left(\frac{\pi}{3}-a\right) \right] = \frac{\pi}{12}+\epsilon.$$

Letting $\epsilon \downarrow 0$, we obtain $\limsup_{n\to\infty} I_n \leq \frac{\pi}{12}$. These together tell that

$$\liminf_{n\to\infty} I_n = \limsup_{n\to\infty} I_n = \frac{\pi}{12}$$

and therefore the limit of $I_n$ exists and has the value $\frac{\pi}{12}$.

Addendum. The following demonstrates graphs of the integrand for different values of $n$.

$\hspace{5em}$

The graph is already quite close to $\text{(1)}$ when $n=100$, which provides a sanity check.

• What happened with the cases when $x$ is in between $0\leq x \leq \frac {\pi}{4}$? – C. Cristi Jun 7 '18 at 7:08
• @C.Cristi, Are you fine with the claim $$a,b>0\quad\Rightarrow\quad\lim_{n\to\infty} \frac{a^n}{a^n+b^n} = \begin{cases} 1, & a>b \\ \frac{1}{2}, & a = b \\ 0, & a < b \end{cases}?$$ If then, simply consider $a=\sin x$ and $b=\cos x$ for each given $x\in(0,\frac{\pi}{4}]$. (The case $x=0$ is trivial.) – Sangchul Lee Jun 7 '18 at 7:10