# Find $\lim_{n \rightarrow \infty} \left(n \int^{\frac{\pi}{4}}_0 (\cos(x)-\sin(x))^n \right)$

Find $$\lim_{n \rightarrow \infty} \left(n \int^{\frac{\pi}{4}}_0 (\cos(x)-\sin(x))^n \right)$$

I've managed to prove that the limit is in $(0,1]$ and I believe it is $1$ but I don't know how to prove it. Could you help me?

• Are you sure it's in (0,1]? $n-n^2$ is negative, but the integral is positive. How can the limit of a sequence of negative terms be positive? – Clement C. Feb 21 '17 at 1:41
• I think that $n-n^2$ should be just $n$. – Sungjin Kim Feb 21 '17 at 2:50
• You perhaps have a typo. The correct question is math.stackexchange.com/q/2121035/72031 – Paramanand Singh Feb 21 '17 at 4:18
• Oh, it was already answered with more precision in the link by Paramanand Singh. – Sungjin Kim Feb 21 '17 at 4:25
• @i707107: no worry you still get a +1 for your efforts from my side. – Paramanand Singh Feb 21 '17 at 4:30

Using $\cos x = \sin(\pi/2 - x)$ and $\sin A - \sin B = 2\cos (\frac{A+B}2)\sin (\frac{A-B}2)$, we have $$\cos x - \sin x = \sqrt 2 \sin(\frac{\pi}4 - x).$$ From this, we have $$\int_0^{\frac{\pi}4} (\cos x - \sin x)^n dx = 2^{n/2} \int_0^{\frac{\pi}4} \sin^n (\frac{\pi}4 - x) dx= 2^{n/2}\int_0^{\frac{\pi}4} \sin^n x dx := 2^{n/2} I_n .$$ Let $J_n = n 2^{n/2} I_n$. Then by the reduction formula, we have $$J_n = -1 + \frac{2(n-1)}{n-2}J_{n-2}.$$ Then $$J_n-1 = \frac 2{n-2} + 2(1+\frac 1{n-2} ) (J_{n-2} -1) .$$ It is easy to check that $J_n$ is bounded, so $J_n-1$ is also bounded. Taking $\limsup$ on the right, we have for $\alpha = \limsup (J_n-1)$, $$\alpha\geq 2\alpha.$$ Then take $\limsup$ on the left, we have $$\alpha \leq 2\alpha.$$ Thus, $\alpha = 2\alpha = 0$. Similarly for $\liminf$, we obtain $\liminf (J_n-1) = 0$. Therefore $\lim J_n = 1$.
This shows that $$\lim_{n\rightarrow\infty} n \int_0^{\frac{\pi}4} (\cos x - \sin x )^n dx = 1.$$