Question

Hey there! It's been a while since I've done a calculus course, but a friend has challenged me to this limit.

Limit as n -> inf
Integral from 0 to pi/2 of sin(x)^n

The answer should be 0, but I'm completely stumped on where to go from here. Feel free to drop partial hints - I like the feeling of finally figuring things out on my own.

Answer 1

Consider integrating by parts. First denote the integral by
\[I_n=\int_0^{\pi/2}\sin^nx\,\mathrm{d}x\]and letting
\[\begin{matrix}u=\sin^{n-1}x&&\mathrm{d}v=\sin x\,\mathrm{d}x\\
\mathrm{d}u=(n-1)\sin^{n-2}x\cos x\,\mathrm{d}x&&v=-\cos x\end{matrix}\]so that you have
\[I_n=\bigg[-\cos x\sin^{n-1}x\bigg]_{x=0}^{x=\pi/2}+(n-1)\int_0^{\pi/2}\cos^2x\sin^{n-2}x\,\mathrm{d}x\]The first term disappears, and you can rewrite the remaining integral via an identity as
\[\begin{align*}I_n&=(n-1)\int_0^{\pi/2}(1-\sin^2x)\sin^{n-2}x\,\mathrm{d}x\\[1ex]
&=(n-1)\left(\int_0^{\pi/2}\sin^{n-2}x\,\mathrm{d}x-\int_0^{\pi/2}\sin^nx\,\mathrm{d}x\right)\\[1ex]
&=(n-1)I_{n-2}-(n-1)I_n\\[1ex]
I_n&=\frac{n-1}{n}I_{n-2}
\end{align*}\]

Answer 2

You can continue in this fashion and observe a neat pattern:
\[\begin{align*}I_n&=\frac{n-1}{n}I_{n-2}\\[1ex]&=\frac{(n-1)(n-3)}{n(n-2)}I_{n-4}\\[1ex]&=\cdots\\[1ex]&=\frac{(n-1)(n-3)(n-5)\cdots3\times1}{n(n-2)(n-4)\cdots4\times2}I_0\end{align*}\]where
\[I_0=\int_0^{\pi/2}\,\mathrm{d}x=\frac{\pi}{2}\]

Answer 3

Thanks for that!
The fraction has the same degree on top and bottom. Wouldn't the fraction then limit to 1?

Answer 4

It might seem so, but it's actually not the case!

You can make two observations here:

First, for all \(n\), you have that \(I_n\ge0\), since \(0\le\sin x\le1\) for \(0\le x\le\dfrac{\pi}{2}\).

Second, consider the quantity \(I_nI_{n+1}\). The first observation tells us this product is also non-negative. Indeed,
\[\begin{align*}I_{n-1}I_n&=\frac{(n-2)(n-4)(n-6)\cdots4\times2}{(n-1)(n-3)(n-5)\cdots3\times1}\times\frac{(n-1)(n-3)(n-5)\cdots3\times1}{n(n-2)(n-4)\cdots4\times2}\\[1ex]
&=\frac{1}{n}\end{align*}\]For large \(n\), the difference between \(I_{n-1}\) and \(I_n\) is small, so we can make a somewhat non-rigorous claim that \(I_n\) behaves somewhat like \(\dfrac{1}{\sqrt n}\). What happens as \(n\to\infty\)?

Answer 5

That makes sense. Can you explain why you decided to consider \[I_n I_{n+1}\]

Answer 6

Dominated convergence theorem

Answer 7

Sorry, I meant to write \(I_{n-1}I_n\) throughout, not \(I_nI_{n+1}\) (though the result would be the same with \(\dfrac{1}{n+1}\) in the end in place of \(\dfrac{1}{n}\)). This product lets you cancel all but the remaining factor in the denominator.