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Integral Calculus lecture notes: Advanced topic
Eugene Kritchevski, Vanier College, last updated on June 21, 2016
The sequence $I_n$
We consider the sequence $\set{I_0,I_1,I_2,\cdots}$ of positive numbers $I_n$ given by the definite integral
$$I_n=\int_{-1}^{1}\of{\sqrt{1-x^2}}^n\,dx$$
The applet below shows the graph of $y=\of{\sqrt{1-x^2}}^n$ and computes an approximate numerical value of $I_n$. Try to increase $n$ by moving the slider and by setting higher values for the maximal exponent. Can you guess the value of $\lim_{n\to\infty} I_n$?
Note that $I_n$ is a strictly decreasing sequence, that is
$$I_0 > I_1 > I_2 > I_3 >\cdots$$
Here is a possible proof: For all $x$ in $[-1,1]$, except $x=0$ or $x=\pm 1$, we have $0 < \sqrt{1-x^2} < 1$ and so $\of{\sqrt{1-x^2}}^n > \of{\sqrt{1-x^2}}^{n+1}$, which implies that $I_n > I_{n+1}$.
The first few values of $I_n$ are
$$I_0=\int_{-1}^{1}\of{\sqrt{1-x^2}}^0\,dx=\int_{-1}^{1}1\,dx=2$$
$$I_1=\int_{-1}^{1}\of{\sqrt{1-x^2}}^1\,dx=\textrm{ unit semicircle area }=\frac{\pi}{2}\simeq 1.5708$$
$$I_2=\int_{-1}^{1}\of{\sqrt{1-x^2}}^2\,dx=\int_{-1}^{1}\of{1-x^2}\,dx=\of{x-\frac{x^3}{3}}\Big|_{-1}^{1}=\frac{4}{3}\simeq 1.3333$$
For higher $n$, we can compute $I_n$ using a reduction formula. Integration by parts with
$$\byparts{u=\of{\sqrt{1-x^2}}^n=\of{1-x^2}^{\frac{n}{2}}\qquad du=\frac{n}{2}\of{1-x^2}^{\frac{n}{2}-1}(-2x)\,dx}{dv=1 \,dx\qquad v=x}$$
gives
$$\begin{equation*}
\begin{split}
I_n
&=\int_{x=-1}^{x=1}u\,dv = uv\Big|_{x=-1}^{x=1}-\int_{x=-1}^{x=1}v\,du \\
&= \of{1-x^2}^{\frac{n}{2}}\cdot x\Big|_{-1}^{1} - \int_{-1}^{1} x\cdot\frac{n}{2}\of{1-x^2}^{\frac{n}{2}-1}(-2x)\,dx \\
&=(0-0) -n \int_{-1}^{1} (-x^2)\of{1-x^2}^{\frac{n}{2}-1}\,dx \\
&=-n \int_{-1}^{1} (1-x^2-1)\of{1-x^2}^{\frac{n}{2}-1}\,dx \\
&=-n \of{\int_{-1}^{1} (1-x^2)\of{1-x^2}^{\frac{n}{2}-1}\,dx - \int_{-1}^{1}\of{1-x^2}^{\frac{n}{2}-1}\,dx} \\
&=-n \of{\int_{-1}^{1} \of{1-x^2}^{\frac{n}{2}}\,dx - \int_{-1}^{1}\of{1-x^2}^{\frac{n-2}{2}}\,dx} \\
&=-n \of{I_n -I_{n-2}} \\
&=-n I_n +n I_{n-2} \\
\end{split}
\end{equation*}
$$
Rearranging the equation $I_n = -n I_n +n I_{n-2}$ gives $(n+1) I_n= n I_{n-2}$ or, equivalently,
Reduction formula for $I_n$
$$ I_n=\frac{n}{n+1}I_{n-2}$$
Starting from $I_0$ and $I_1$, all the integrals $I_n$ can now be computed recursively by steps of $2$ thanks to the rule $I_{n-2} \xrightarrow[]{\cdot \frac{n}{n+1}} I_n$.
$$\boxed{I_0=2}\xrightarrow[]{\cdot \frac{2}{3}} \boxed{I_2=2\cdot \frac{2}{3}} \xrightarrow[]{\cdot \frac{4}{5}} \boxed{I_4=2\cdot \frac{2}{3} \cdot\frac{4}{5}}\xrightarrow[]{\cdot \frac{6}{7}} \boxed{I_6=2\cdot \frac{2}{3} \cdot\frac{4}{5}\cdot\frac{6}{7}} \xrightarrow[]{\cdot \frac{8}{9}} \boxed{I_8=2\cdot \frac{2}{3} \cdot\frac{4}{5}\cdot\frac{6}{7}\cdot\frac{8}{9}}\quad \textrm{ and so on} $$
$$\boxed{I_1=\frac{\pi}{2}}\xrightarrow[]{\cdot \frac{3}{4}} \boxed{I_3=\frac{\pi}{2}\cdot \frac{3}{4}} \xrightarrow[]{\cdot \frac{5}{6}} \boxed{I_5=\frac{\pi}{2}\cdot \frac{3}{4} \cdot\frac{5}{6}}\xrightarrow[]{\cdot \frac{7}{8}} \boxed{I_7=\frac{\pi}{2}\cdot \frac{3}{4} \cdot\frac{5}{6}\cdot\frac{7}{8}} \xrightarrow[]{\cdot \frac{9}{10}} \boxed{I_9=\frac{\pi}{2}\cdot \frac{3}{4} \cdot\frac{5}{6}\cdot\frac{7}{8}\cdot\frac{9}{10}}\quad \textrm{ and so on} $$
Behavior of $I_n$ as $n\to\infty$
Using the fact that $I_n$ is decreasing, the reduction formula, and the Squeeze Theorem, we get:
$$I_{n-2}> I_{n-1}>I_{n}$$
$$ \Downarrow$$
$$\frac{I_{n-2}}{I_{n}} > \frac{I_{n-1}}{I_{n}} > 1 $$
$$ \Downarrow$$
$$\frac{n+1}{n} > \frac{I_{n-1}}{I_{n}} > 1 $$
$$ \Downarrow $$
$$\boxed{(*)\qquad\lim_{n\to\infty} \frac{I_{n-1}}{I_{n}}=1}$$
Multiplying the equation $(n+1) I_n= n I_{n-2}$ by $I_{n-1}$ gives
$$(n+1)I_n I_{n-1}=nI_{n-1}I_{n-2}$$
Observe that the right hand side is simply the left hand side with $n$ replaced by $n-1$. In other words, if we let $a_n=(n+1)I_{n}I_{n-1}$ then we have $a_{n}=a_{n-1}$ for all $n\geq 1$. Then also $a_{n-1}=a_{n-2}$, $a_{n-2}=a_{n-3}$ and so on until $a_1=a_0$. Therefore
$a_n=a_0$, i.e. $(n+1)I_{n}I_{n-1}=(1+1)I_{1}I_{0}=2\pi$ and we conlude that
$$\boxed{(**)\qquad I_{n}I_{n-1}=\frac{2\pi}{n+1}}$$
Using $(*)$ and $(**)$, we can understand the behaviour of $I_n$ as $n\To\infty$. Dividing $(**)$ by $I_n^2$ the gives
$$\frac{I_{n-1}}{I_{n}}=\frac{2\pi}{n+1}\cdot\frac{1}{I_n^2}$$
Taking the limit and using $(*)$ gives
$$1=\lim_{n\to\infty}\of{\frac{2\pi}{n+1}\cdot\frac{1}{I_n^2}}.$$
We also have
$$\lim_{n\to\infty}\of{\frac{2\pi}{n}\cdot\frac{1}{I_n^2}}= \lim_{n\to\infty}\of{\frac{2\pi}{n+1}\cdot\frac{1}{I_n^2}\cdot \frac{n+1}{n}}=\lim_{n\to\infty}\of{\frac{2\pi}{n+1}\cdot\frac{1}{I_n^2}} \cdot \lim_{n\to\infty}\frac{n}{n+1}=1\cdot 1=1 $$
and taking the square roots gives
$$\lim_{n\to\infty}\of{\frac{\sqrt{\frac{2\pi}{n}}}{I_n}}=1$$
or, equivalently,
$$\boxed{\lim_{n\to\infty} \of{\frac{I_n}{\sqrt{\frac{2\pi}{n}}}}=1}.$$
The last limit means that as $n\to\infty$
$$I_n\sim \sqrt{\frac{2\pi}{n+1}}\sim \sqrt{\frac{2\pi}{n}}$$
Exercise: Using the limiting behavior of $I_n$ for even $n$, prove Wallis product formula
$$\frac{\pi}{2}=\lim_{m\to\infty}\prod_{k=1}^{m}\frac{2k}{2k-1}\cdot \frac{2k}{2k+1}=\lim_{m\to\infty}\frac{1}{2m+1}\cdot \frac{2^{4m} [m!]^4}{[(2m)!]^2}$$
The volume of a ball in the $n$-dimensional space
The ball $B_n(R)$, centered at the origin and having radius $R$, in $\Real^n$ is defined by
$$B_n(R)=\set{x_1^2+x_2^2+\cdots+x_n^2\leq R^2}$$
The unit ball is $B_n(1)$. The $n$-dimensional volume of $B_n(R)$ is denoted by $V_n(B_n(R))$. When $n=1$, we have $B_1(R)=[-R,R]$ and $V_1(B_1(R))=2R$. When $n=2$, $B_2(R)$ is a disk of radius $R$ in the plane and $V_2(B_2(R))=\pi R^2$. When $n=3$, $B_3(R)$ is a sphere (surface AND interior) of radius $R$ in the space and $V_3(B_3(R))=\frac{4}{3}\pi R^3$. Note that so far, for $1\leq n \leq 3$, we have
$$V_n(B_n(R))=c_n R^n=V_n(B_n(1))R^n$$
The same will be true true in all dimensions and the sequence of constants $c_n$ will have an interesting behavior.
The volume in higher dimensions can be computed using cross sections. Recall from your integral calculus course that
$$V_3(B_3(R))=\int_{-R}^{R} V_2(C(t))dt$$
where $C(t)$ is the cross section of the sphere along $x_3=t$. In other words, we freeze $x_3=t$
$$x_1^2+x_2^3+x_3^2 \leq R^2 \Leftrightarrow x_1^2+x_2^3+t^2 \leq R^2 \Leftrightarrow x_1^2+x_2^3 \leq R^2 -t^2 $$
and the cross section $C(t)$ is the disk of radius $\sqrt{R^2-t^2}$ in the $(x_1,x_2)$-plane. Then
$$V_3(B_3(R))=\int_{-R}^{R} V_2(C(t))dt= \int_{-R}^{R} \pi \of{\sqrt{R^2-t^2}}^2 dt=\cdots=\frac{4}{3}\pi R^3$$
In higher dimensions, the cross-section $C(t)$ of $B_n(R)$ along $x_n=t$ is obtained by freezing $x_n=t$, where $-R\leq t \leq R$. We get
$$x_1^2+x_2^3+\cdots+x_{n-1}^2+x_n^2 \leq R^2 \Leftrightarrow x_1^2+x_2^3+\cdots+x_{n-1}^2+t^2 \leq R^2 \Leftrightarrow x_1^2+x_2^3+\cdots+x_{n-1}^2 \leq R^2 -t^2 $$
so that
$$C(t)=B_{n-1}\of{\sqrt{R^2-t^2}}.$$
The volume of $B_n(R)$ can be then computed by
$$V_n(B_n(R))=\int_{-R}^{R}V_{n-1} \of{B_{n-1}\of{\sqrt{R^2-t^2}}} dt$$
Exercise: Using a change of variables and induction, show that
$$V_n(B_n(R))=V_n(B_n(1))R^n$$
Convince yourself that this formula makes sense intuitively as $B_n(R)$ is obtained by scaling all the $n$ coordinates of $B_n(1)$ by the factor $R$ and so the volume scales by the factor $R^n$.
Let $c_n$ denote the volume of the unit ball in of radius $R$ in $\Real^n$
$$\boxed{c_n=V_n(B_n(1))}$$
Then for all $n$ we have
$$V_n(B_n(R))=c_n R^n$$ and so
$$c_n=V_n(B_n(1))=\int_{-1}^{1} c_{n-1} \of{\sqrt{1-t^2}}^{n-1} dt=c_{n-1}\int_{-1}^{1} \of{\sqrt{1-t^2}}^{n-1} dt$$
Therefore
$$\boxed{c_n=c_{n-1} I_{n-1}\qquad \textrm{ where } \quad I_n=\int_{-1}^{1}\of{\sqrt{1-x^2}}^n\,dx}$$
Since we already know a lot about the sequence of integrals $I_n$, we will be able to figure out what happens with the sequence $c_n$. Recall that $I_n\downarrow 0$. Also, one can check that $I_n\leq 1$ starting from $n=5$. Therefore the sequence $c_n$ is decreasing starting from $n=5$ and
$$\lim_{n\to\infty}c_n= 0$$
Let us obtain a precise formula for $c_n$. From $c_n=c_{n-1} I_{n-1}$ and $c_{n-1}=c_{n-2} I_{n-2}$, we get
$$c_n=I_{n-1}I_{n-2}c_{n-2}$$
Also, from $(**)$, we have $I_{n-1}I_{n-2}=\frac{2\pi}{n}$. Therefore
$$c_n=\frac{2\pi}{n} \cdot c_{n-2}$$
Starting from $c_1=2$ and $c_2=\pi$, all the terms of the sequence $c_n$ can now be computed recursively by steps of $2$ with the rule $c_{n-2} \xrightarrow[]{\cdot \frac{2\pi}{n}} c_{n-2}.$
$$\boxed{c_1=2}\xrightarrow[]{\cdot \frac{2\pi}{3}} \boxed{c_3=2\cdot \frac{2\pi}{3}} \xrightarrow[]{\cdot \frac{2\pi}{5}} \boxed{c_5=2\cdot \frac{2\pi}{3} \cdot\frac{2\pi}{5}}\xrightarrow[]{\cdot \frac{2\pi}{7}} \boxed{c_7=2\cdot \frac{2\pi}{3} \cdot\frac{2\pi}{5}\cdot\frac{2\pi}{7}} \xrightarrow[]{\cdot \frac{2\pi}{9}} \boxed{c_9=2\cdot \frac{2\pi}{3} \cdot\frac{2\pi}{5}\cdot\frac{2\pi}{7}\cdot\frac{2\pi}{9}}\quad \textrm{ and so on} $$
$$\boxed{c_2=\pi}\xrightarrow[]{\cdot \frac{2\pi}{4}} \boxed{c_4=\pi\cdot \frac{2\pi}{4}} \xrightarrow[]{\cdot \frac{2\pi}{6}} \boxed{c_6=\pi\cdot \frac{2\pi}{4} \cdot\frac{2\pi}{6}}\xrightarrow[]{\cdot \frac{2\pi}{8}} \boxed{c_8=\pi\cdot \frac{2\pi}{4} \cdot\frac{2\pi}{6}\cdot\frac{2\pi}{8}} \xrightarrow[]{\cdot \frac{2\pi}{10}} \boxed{c_{10}=\pi\cdot \frac{2\pi}{4} \cdot\frac{2\pi}{6}\cdot\frac{2\pi}{8}\cdot\frac{2\pi}{10}}\quad \textrm{ and so on} $$
Conclusion:
For even dimensions:
$$c_{2m}=\frac{\pi^m}{m!} \qquad \textrm{ and so }\quad V_{2m}(B_{2m}(R))=\frac{(\pi R^2)^m}{m!}$$
For odd dimensions:
$$c_{2m+1}=2\cdot\frac{(2\pi)^m}{m!!} \qquad \textrm{ and so }\quad V_{2m+1}(B_{2m+1}(R))=2\cdot\frac{(2\pi)^m}{m!!}\cdot R^{2m+1}$$
Stirling's approximation to $n!$
The Stirling's formula gives the asymptotic behavior of $n!$
$$n!\sim \of{\frac{n}{e}}^n \sqrt{2\pi n}$$
For sequences with posisive terms, the notation $a_n\sim b_n$ means $\displaystyle\lim_{n\to\infty}\frac{a_n}{b_n}=1$.
Exercise:
(1) Using the trapezoidal rule (with error) for the integral
$$\int_1^n \ln(x)\,dx$$
show that there are constants $C_1< C_2$ such that for all $n\geq 1$
$$ n \ln(n) - n + \frac{1}{2}\ln(n) + C_1 \leq \ln(n!) \leq n \ln(n) - n + \frac{1}{2}\ln(n) + C_2$$
Equivalently, in exponential form
$$ e^{C_1}\of{\frac{n}{e}}^n \sqrt{n} \leq n!\leq e^{C_2}\of{\frac{n}{e}}^n \sqrt{n}$$
(2) Consider the sequence $$a_n= \frac{n!}{\of{\frac{n}{e}}^n \sqrt{n}}$$
From part (1), we have $e^{C_1}\leq a_n \leq e^{C_2}$, which means that $\of{\frac{n}{e}}^n \sqrt{n}$ gives the order of magnitude of $n!$. Prove that $a_n$ is decreasing sequence. Hint: simplify the ratio at $r_n=\frac{a_{n+1}}{a_n}$ and use Calculus to show that $r_n < 1$.
(3) Since $a_n$ is decreasing and bounded below by $e^{C_1}$, we know that $\lim_{n\to\infty} a_n= C$ for some constant $C$. Use Wallis formula to compute that $C=\sqrt{2\pi}$ and conclude the proof of Stirling's formula.
...add some exercises on Stirling's formula
Measure concentration
near the surface, near equators,...coming soon
Probability
...coming soon