# Evaluate a limit involving definite integral

Evaluate the following limit: $$\lim_{n \to \infty} \left[n - n^2 \int_{0}^{\pi/4}(\cos x - \sin x)^n dx\right]$$

I've tried to rewrite the expression as follows: $$\lim_{n \to \infty} \left[n - n^2 \sqrt{2}^n \int_{0}^{\pi/4}\sin^n \left( \frac{\pi}{4} - x \right) dx\right]$$

However, this doesn't seem to help too much. Thank you!

• Did you try recursive formula for $I_n=n\int_{0}^{\pi/4}(cos x-sinx)^ndx$ ? Jan 30 '17 at 15:51
• What about expanding the integrand with the Binomial Theorem? Jan 30 '17 at 15:54
• math.stackexchange.com/questions/1866304/… Jan 30 '17 at 15:55

Writing

$$(\cos x - \sin x)^n = \color{blue}{\frac{\cos x - \sin x}{\cos x + \sin x}} \cdot \color{red}{ (\cos x + \sin x)(\cos x - \sin x)^{n-1}}$$

and applying integrating by parts, we have

$$\int_{0}^{\frac{\pi}{4}} (\cos x - \sin x)^n \, dx = \frac{1}{n} - \frac{2}{n} \int_{0}^{\frac{\pi}{4}} \frac{(\cos x - \sin x)^n}{(\sin x + \cos x)^2} \, dx.$$

Plugging this back,

$$n - n^2 \int_{0}^{\frac{\pi}{4}} (\cos x - \sin x)^n \, dx = 2n \int_{0}^{\frac{\pi}{4}} \frac{(\cos x - \sin x)^n}{(\sin x + \cos x)^2} \, dx.$$

Now from the observation

$$n \int_{0}^{\frac{\pi}{4}} (\sin x + \cos x)(\cos x - \sin x)^n \, dx = \frac{n}{n+1},$$

we can apply the usual approximation-to-the-identity argument to obtain

\begin{align*} \lim_{n\to\infty} \left[ n - n^2 \int_{0}^{\frac{\pi}{4}} (\cos x - \sin x)^n \, dx \right] &= \lim_{n\to\infty} 2n \int_{0}^{\frac{\pi}{4}} \frac{(\cos x - \sin x)^n}{(\sin x + \cos x)^2} \, dx \\ &= \lim_{x \to 0^+} \frac{2}{(\sin x + \cos x)^3} \\ &= \color{red}{\boxed{2}}. \end{align*}

• +1 It means you performed the integration: Quite fine !!!. Jan 30 '17 at 21:44

We may go through Laplace's method. $\cos(x)-\sin(x)$ is positive and concentrated in a right neighbourhood of the origin; by expanding $\log(\cos(x)-\sin(x))$ as a Taylor series we get $\cos(x)-\sin(x)= e^{-x-x^2+O(x^3)}$. In particular the first two terms of the asymptotic expansion of $$I(n) = \int_{0}^{\pi/4}\left(\cos(x)-\sin(x)\right)^n\,dx$$ are the same as $$J(n) = \int_{0}^{+\infty}\exp\left(-nx-nx^2\right)\,dx = \frac{1}{n}-\frac{2}{n^2}+O\left(\frac{1}{n^3}\right)$$ and the wanted limit is $\color{red}{\large 2}$. In terms of hypergeometric functions we are stating that $$\lim_{n\to +\infty}\left[n-\frac{n^2}{n+1}\;\phantom{}_2 F_1\left(\frac{1}{2},1;\frac{n+3}{2};-1\right)\right]=2.$$

• Good work(+1)....hope this one doesn't attract so many downvotes then my attempt in the linked topic above :P Jan 30 '17 at 16:30


When $\ds{n \to \infty}$, the main contribution to the integral comes from values of $\ds{x \gtrsim 0}$ such that it's is a 'candidate' to be evaluated by means of the Laplace Method. Namely,

\begin{align} \int_{0}^{\pi/4}\cos^{n}\pars{x + {\pi \over 4}}\,\dd x & = \int_{0}^{\pi/4}\exp\pars{n\ln\pars{\cos\pars{x + {\pi \over 4}}}}\,\dd x \\[5mm] & \sim \int_{0}^{\infty}\exp\pars{n\bracks{-\,{\ln\pars{2} \over 2} - x}} \pars{1 - nx^{2}}\,\dd x \\[5mm] & = 2^{-n/2}\,\pars{{1 \over n} - {2 \over n^{2}}}\quad \mbox{as}\ n \to \infty\label{2}\tag{2} \end{align}

With \eqref{1} and \eqref{2}:

\begin{align} &\lim_{n \to \infty}\braces{n - n^{2}\int_{0}^{\pi/4}\bracks{\cos\pars{x} - \sin\pars{x}}^{\,n}\,\dd x} = \lim_{n \to \infty}\braces{n - 2^{n/2}\,n^{2} \bracks{2^{-n/2}\,\pars{{1 \over n} - {2 \over n^{2}}}}} \\[5mm] = &\ \bbx{\ds{2}} \end{align}