# Integral inequalities (4)

Problem. Let $\displaystyle f: [0,1] \longrightarrow \mathbb{R}$ be a differentiable function such that $|f'(x)| \le 1$ for all $x \in [0,1].$ For an integer $n \ge 0,$ let $\displaystyle I_n:=\int_0^1 x^nf(x) \ dx.$
Show that for any integer $n \ge 0,$

$\displaystyle |I_{n+1}| \le \frac{1}{(n+1)(n+2)}\sum_{k=0}^n\left((k+1)|I_k|+\frac{1}{k+2}\right)-\frac{1}{(n+2)(n+3)}.$

(We assume that $f'$ is integrable).

Solution. Integration by parts gives

$\displaystyle I_k=\frac{f(1)}{k+1}-\frac{1}{k+1}\int_0^1x^{k+1}f'(x) \ dx.$

So

\displaystyle \begin{aligned} \frac{1}{(n+1)(n+2)}\sum_{k=0}^n (k+1)I_k=\frac{f(1)}{n+2}-\frac{1}{(n+1)(n+2)}\int_0^1\sum_{k=0}^nx^{k+1}f'(x) \ dx \\ =I_{n+1}+\frac{1}{n+2}\int_0^1x^{n+2}f'(x) \ dx - \frac{1}{(n+1)(n+2)}\int_0^1\sum_{k=0}^nx^{k+1}f'(x) \ dx\end{aligned}

and therefore

$\displaystyle I_{n+1}=\frac{1}{(n+1)(n+2)}\left(\sum_{k=0}^n(k+1)I_k + \int_0^1\left(\sum_{k=0}^nx^{k+1}-(n+1)x^{n+2}\right)f'(x) \ dx \right).$

Thus, since

$\displaystyle \sum_{k=0}^n x^{k+1}-(n+1)x^{n+2}=\sum_{k=0}^n(x^{k+1}-x^{n+2}) \ge 0$

and $|f'(x)| \le 1,$ on $[0,1],$ we have

\displaystyle \begin{aligned} |I_{n+1}| \le \frac{1}{(n+1)(n+2)}\left(\sum_{k=0}^n(k+1)|I_k|+\int_0^1 \left(\sum_{k=0}^nx^{k+1}-(n+1)x^{n+2}\right) dx\right) \\ = \frac{1}{(n+1)(n+2)}\sum_{k=0}^n\left((k+1)|I_k|+\frac{1}{k+2}\right)-\frac{1}{(n+2)(n+3)}. \ \Box \end{aligned}

Example. Let’s check the above result for the function $f(x)=x, \ \ x \in [0,1].$
Then, for any integer $n \ge 0,$ we have $\displaystyle I_n:=\int_0^1 x^nf(x) \ dx = \frac{1}{n+2}$ and so

\displaystyle \begin{aligned} \frac{1}{(n+1)(n+2)}\sum_{k=0}^n\left((k+1)|I_k|+\frac{1}{k+2}\right)-\frac{1}{(n+2)(n+3)} \\ =\frac{1}{(n+1)(n+2)}\sum_{k=0}^n\left(\frac{k+1}{k+2}+\frac{1}{k+2}\right)-\frac{1}{(n+2)(n+3)} \\ =\frac{1}{n+2}-\frac{1}{(n+2)(n+3)}=\frac{1}{n+3}=I_{n+1}.\end{aligned}

Exercise. Show that if $f: [0,1] \longrightarrow \mathbb{R}$ satisfies the conditions given in the above problem and if $\displaystyle \left|\int_0^1 f(x) \ dx\right| \le \frac{1}{2},$ then $\displaystyle \left|\int_0^1x^nf(x) \ dx\right| \le \frac{1}{n+2}$ for all integers $n \ge 0.$
Hint. Induction!

# An application of linear algebra in Calculus

This post is for those who have some knowledge of linear algebra.

Definition. For an integer $n \ge 1,$ the $n\times n$ Hilbert matrix is defined by $H_n=[a_{ij}],$ where

$\displaystyle a_{ij}=\frac{1}{i+j-1}, \ \ 1 \le i,j \le n.$

It is known that $H_n$ is invertible and if $H_n^{-1}=[b_{ij}],$ then $\displaystyle \sum_{i,j}b_{ij}=n^2.$ We are going to use these two properties of Hilbert matrices to solve the following calculus problem.

Problem. Let $n \ge 1$ be an integer and let $f : [0,1] \longrightarrow \mathbb{R}$ be a continuous function. Suppose that $\displaystyle \int_0^1 x^kf(x) \ dx = 1$ for all  $0 \le k \le n-1.$ Show that $\displaystyle \int_0^1 (f(x))^2 dx \ge n^2.$

Solution. Since $H_n,$ the $n\times n$ Hilbert matrix, is invertible, there exist real numbers $p_0, p_1, \cdots , p_{n-1}$ such that

$\displaystyle \sum_{i=1}^n\frac{p_{i-1}}{i+j-1}=1, \ \ \ 1 \le j \le n.$

So the polynomial $\displaystyle p(x)=\sum_{k=0}^{n-1}p_kx^k$ satisfies the conditions

$\displaystyle \int_0^1x^k p(x) \ dx =1, \ \ \ 0 \le k \le n-1.$

Clearly $\displaystyle \sum_{k=0}^{n-1}p_k$ is the sum of all the entries of $H_n^{-1}$ and so $\displaystyle \sum_{k=0}^{n-1}p_k=n^2.$ Now let $f$ be a real-valued continuous function on $[0,1]$ such that

$\displaystyle \int_0^1x^kf(x) \ dx = 1, \ \ \ 0 \le k \le n-1.$

Let $p(x)$ be the above polynomial.Then since

$\displaystyle (f(x))^2-2f(x)p(x)+(p(x))^2 =(f(x)-p(x))^2 \ge 0,$

integrating gives

\displaystyle \begin{aligned} \int_0^1 (f(x))^2dx \ge 2\int_0^1f(x)p(x) \ dx -\int_0^1(p(x))^2dx=2\sum_{k=0}^{n-1}p_k \int_0^1 x^kf(x) \ dx - \\ \sum_{k=0}^{n-1}p_k\int_0^1x^kp(x) \ dx = 2\sum_{k=0}^{n-1}p_k-\sum_{k=0}^{n-1}p_k=\sum_{k=0}^{n-1}p_k =n^2. \ \Box \end{aligned}

# Weird looking integrals (1)

For a general form of the following problem, see Exercise 3.

Problem. Show that

$\displaystyle I:=\int_0^{2\pi} \exp\left(\frac{7+5 \cos x}{10+6\cos x}\right) \cos \left( \frac{\sin x}{10+6 \cos x} \right) dx=2\pi e^{2/3}.$

Solution. We make the substitution $\displaystyle \tan\left(\frac{x}{2}\right)=2\tan\left(\frac{t}{2}\right).$ Then

$\displaystyle I=2\int_0^{\pi} \exp\left(\frac{7+5 \cos x}{10+6\cos x}\right) \cos \left( \frac{\sin x}{10+6 \cos x} \right) dx=8e^{5/8}\int_0^{\pi}\exp\left(\frac{\cos t}{8}\right)\cos\left(\frac{\sin t}{8}\right)\frac{dt}{5-3\cos t}. \ \ \ \ \ (1)$

Now, since $\displaystyle \exp\left(\frac{\cos t}{8}\right)\cos\left(\frac{\sin t}{8}\right)$ is the real part of

$\displaystyle \exp\left(\frac{\cos t}{8}\right)\exp\left(\frac{i\sin t}{8}\right)=\exp\left(\frac{1}{8}e^{it}\right)$

and

$\displaystyle \exp\left(\frac{1}{8}e^{it}\right)=\sum_{n=0}^{\infty}\frac{1}{8^nn!}e^{nit}=\sum_{n=0}^{\infty}\frac{\cos(nt)}{8^nn!}+i\sum_{n=0}^{\infty} \frac{\sin(nt)}{8^nn!},$

we have

$\displaystyle \exp\left(\frac{\cos t}{8}\right)\cos\left(\frac{\sin t}{8}\right)=\sum_{n=0}^{\infty} \frac{\cos(nt)}{8^nn!}.$

Thus, by $(1),$

$\displaystyle I=8e^{5/8}\sum_{n=0}^{\infty}\frac{1}{8^nn!}\int_0^{\pi} \frac{\cos(nt)}{5-3\cos t} \ dt=\frac{4}{3}e^{5/8}\sum_{n=0}^{\infty}\frac{(-1)^n}{8^nn!}\int_0^{2\pi} \frac{\cos(nt)}{\cos t + \frac{5}{3}} \ dt. \ \ \ \ \ \ \ (2)$

But here (Problem 2) I showed that

$\displaystyle \int_0^{2\pi} \frac{\cos(nt)}{\cos t + c} \ dt=\frac{2\pi}{\sqrt{c^2-1}}(-c+\sqrt{c^2-1})^n,$

for integers $n \ge 0$ and real numbers $c > 1.$ Thus

$\displaystyle \int_0^{2\pi} \frac{\cos(nt)}{\cos t + \frac{5}{3}} \ dt=(-1)^n\frac{\pi}{2 \cdot 3^{n-1}}$

and so, by $(2),$

$\displaystyle I=2\pi e^{5/8} \sum_{n=0}^{\infty} \frac{1}{24^nn!}=2\pi e^{5/8}e^{1/24}=2\pi e^{2/3}. \ \Box$

Exercise 1 (for those who are familiar with complex analysis).
Show that $\displaystyle \exp\left(\frac{7+5 \cos x}{10+6\cos x}\right) \cos \left( \frac{\sin x}{10+6 \cos x} \right)$ is the real part of $\displaystyle \exp\left(\frac{e^{ix}+2}{e^{ix}+3}\right).$ Now apply Cauchy’s integration formula to give another solution for the above problem.

Exercise 2. Show that

$\displaystyle \int_0^{2\pi} \exp\left(\frac{3+3\cos x}{5+4\cos x}\right) \cos \left( \frac{\sin x}{5+4\cos x} \right) dx=2\pi \sqrt{e}.$

Exercise 3. Given real numbers $a,b$ with $b \ne \pm 1,$ let

$\displaystyle I(a,b):=\int_0^{2\pi}\exp\left(\frac{1+ab+(a+b)\cos x}{b^2+1+2b\cos x}\right) \cos\left(\frac{(b-a)\sin x}{b^2+1+2b\cos x}\right)dx.$

Show that

$\displaystyle I(a,b)=\begin{cases} 2\pi e & \text{if} \ b=0 \ \text{or} \ |b| < 1 \\ 2\pi e^{a/b} & \text{if} \ |b| > 1.\end{cases}$

Hint. Make the substitution $\displaystyle \tan\left(\frac{x}{2}\right)=\left|\frac{b+1}{b-1}\right|\tan \left(\frac{t}{2}\right)$ and just do as I did in the solution of the above problem.

Remark. We also have $I(a,\pm 1)=2\pi e^{\min\{1,\pm a\}}.$ That quickly follows if we make the substitution $\displaystyle \tan\left(\frac{\pi}{2}\right)=t$ and use the identity $\displaystyle \int_0^{\infty} \frac{\cos(\alpha x)}{x^2+1} \ dx=\frac{\pi}{2}e^{-|\alpha|},$ which holds for all real numbers $\alpha$ but I have not discussed it in this blog yet.

# Limit of integrals (16)

It’s nice to see examples where limit of integrals are used to evaluate integrals. In this post, I give one of those examples.

Problem 1. Show that $\displaystyle \lim_{n\to\infty} \int_0^{2\pi} \frac{\cos(nx)}{\cos x + c} \ dx = 0$ for any real number $c > 1.$

Solution. Use integration by parts with $\cos(nx) \ dx = dv$ and $\displaystyle \frac{1}{\cos x + c} = u$ to get

$\displaystyle \int_0^{2\pi} \frac{\cos(nx)}{\cos x + c} \ dx = -\frac{1}{n}\int_0^{2\pi} \frac{\sin (nx)\sin x}{(\cos x + c)^2} \ dx$

The result now follows because $\displaystyle \frac{\sin (nx)\sin x}{(\cos x + c)^2}$ is bounded. $\Box$

Problem 2. Given an integer $n \ge 0$ and real number $c > 1,$ show that

$\displaystyle J_n:=\int_0^{2\pi} \frac{\cos(nx)}{\cos x + c} \ dx=\frac{2\pi}{\sqrt{c^2-1}}(-c+\sqrt{c^2-1})^n.$

Solution. First see that $\displaystyle J_0=\frac{2\pi}{\sqrt{c^2-1}}$ and $\displaystyle J_1=2\pi-cJ_0.$ For $n \ge 2$ we have

\displaystyle \begin{aligned} J_n=\int_0^{2\pi} \frac{2\cos x \cos((n-1)x)-\cos((n-2)x)}{\cos x + c} \ dx=-2cJ_{n-1}-J_{n-2}\end{aligned}

and therefore $\displaystyle J_n+2cJ_{n-1}+J_{n-2}=0,$ which has the characteristic polynomial $r^2+2cr+1=0$ with the roots $\displaystyle r=-c \pm \sqrt{c^2-1}.$ So

$\displaystyle J_n=\alpha (-c+\sqrt{c^2-1})^n+\beta (-c-\sqrt{c^2-1})^n$

for some constants $\alpha, \beta.$ Since, by Problem 1, $\displaystyle \lim_{n\to\infty}J_n=0,$ we must have $\beta=0$ (why ?) and so $\displaystyle J_n=\alpha (-c+\sqrt{c^2-1})^n.$ We also have $\displaystyle \alpha=J_0=\frac{2\pi}{\sqrt{c^2-1}}.$ and that completes the solution. $\Box$

Problem 3. Given integers $m,n$ with $0 \le m \le n$ and real numbers $a,b,c$ with $c > 1,$ show that

$\displaystyle I_{m,n}:=\int_0^{2\pi} \frac{(a\cos x + b)^m\cos(nx)}{\cos x + c} \ dx=\frac{2\pi (b-ac)^m}{\sqrt{c^2-1}}(-c+\sqrt{c^2-1})^n.$

Solution. We first show that, given integers $k,n$ with $n \ge 1$ and $0 \le k \le n-1,$

$\displaystyle \int_0^{2\pi}(\cos x + c)^k\cos (nx) \ dx =0. \ \ \ \ \ \ \ \ (1)$

The proof of $(1)$ is by induction over $k.$ It’s clear for $k=0$ and the claim follows from the following identity

\displaystyle \begin{aligned} (\cos x + c)^k\cos(nx)=\frac{1}{2}(\cos x + c)^{k-1}(\cos((n+1)x)+\cos((n-1)x)+2c\cos(nx)).\end{aligned}

Next, we show that

$\displaystyle I_{m,n}=(b-ac)^m \int_0^{2\pi} \frac{\cos(nx)}{\cos x + c} \ dx. \ \ \ \ \ \ \ \ \ (2)$

To prove $(2),$ in $I_{m,n},$ write $\displaystyle a\cos x +b = a(\cos x + c)+b-ac$ and then expand $(a\cos x +b)^m$ using the binomial theorem. Then divide by $\cos x + c$ and use $(1)$ to finish the proof. Now $(2)$ and Problem 2 together complete the solution. $\Box$

# Integral inequalities (3)

Problem. Suppose that $f: [a,b] \longrightarrow \mathbb{R}$ is a continuously differentiable function, i.e. $f$ is differentiable and $f',$ the derivative of $f,$ is continuous.
Show that if $f(a)=f(b)=0,$ then

$\displaystyle \int_a^b | f(x)f'(x) |dx\le \frac{b-a}{4}\int_a ^b (f'(x))^2dx.$

Solution. Let $\displaystyle c:=\frac{a+b}{2}.$ Then

\displaystyle \begin{aligned} \int_a^b |f(x)f'(x)| \ dx = \int_a^c|f(x)f'(x)| \ dx + \int_c^b |f(x)f'(x)| \ dx \\ =\int_a^c \left|\int_a^x f'(t) \ dt \right| |f'(x)| \ dx + \int_c^b \left|\int_x^b f'(t) \ dt \right| |f'(x)| \ dx \\ \le \int_a^c \left(\int_a^x|f'(t)| \ dt \right) |f'(x)| \ dx + \int_c^b \left(\int_x^b |f'(t)| \ dt\right)|f'(x)| \ dx \\ =\frac{1}{2}\left(\int_a^c |f'(t)| \ dt\right)^2 + \frac{1}{2}\left(\int_c^b |f'(t)| \ dt\right)^2 \\ \leq \frac{c-a}{2} \int_a^c(f'(t))^2 dt + \frac{b-c}{2} \int_c^b(f'(t)^2dt \\ =\frac{b-a}{4}\left(\int_a^c (f'(t)^2dt + \int_c^b (f'(t)^2dt\right) \\ =\frac{b-a}{4}\int_a^b (f'(t)^2dt.\end{aligned}

Notice that the inequality in the fifth line is true by Cauchy-Schwarz. $\Box$

# Solutions of tan(x) = x

The function $f(x):=\tan x - x$ is odd and so to study the roots of $f,$ we only need to study positive roots of $f,$ which is what we’re going to do in this post. We begin with a basic fact.

Problem 1. Consider the function $f(x):=\tan x - x.$ For an integer $n \ge 1,$ let $U_n$ be the interval $\displaystyle \left(n\pi, n\pi+\frac{\pi}{2}\right).$ Show that

i) $f$ is increasing on any interval which is in its domain

ii) the set of positive roots of $f$ is a sequence $\displaystyle \{x_n : \ \ n \ge 1\}$ where $x_n \in U_n.$

Solution. i) $f'(x)=\tan^2x \ge 0.$

ii) Let $n \ge 1$ be an integer. We have $f(n\pi)=-n\pi < 0.$ Also, the limit of $f(x)$ as $x$ approaches $\displaystyle n\pi+\frac{\pi}{2},$ from the left, is $+\infty.$ Since $f$ is continuous in $U_n,$ the intermediate value theorem implies that $f(x_n)=0$ for some $\displaystyle x_n \in U_n.$ Since, by i), $f$ is increasing on $U_n, \ x_n$ is the unique root of $f$ in $U_n.$
Finally, $f$ has no other positive roots because $f$ is positive on $\displaystyle \left(0,\frac{\pi}{2}\right),$ and negative on the interval $\displaystyle \left((m-1)\pi+\frac{\pi}{2}, m\pi \right)$ for any integer $m \ge 1$ (why ?). $\ \Box$

Next, given integer $n \ge 1,$ we’re going to refine the basic result given in Problem 1, ii), by finding an interval which is much smaller than $U_n$ and still contains $x_n.$ But first we need to prove something.

Problem 2. Show that the function

$\displaystyle g(x):=x\cot x - \frac{\pi x}{2}+x^2, \ \ x \in (0,\pi),$

is decreasing.

Solution. We find the first and the second derivatives of $g;$ we have

$\displaystyle g'(x)=\cot x - x \cot^2x+x-\frac{\pi}{2}, \ \ \ g''(x)=\frac{(2x-\sin(2x))\cos x}{\sin^3x}.$

So $g''(x)$ and $\cos x$ have the same sign because both $\sin x$ and $2x-\sin(2x)$ are positive on the interval $(0,\pi).$ Thus the maximum of $g'$ occurs at $\displaystyle x=\frac{\pi}{2}$ and so $g'(x) \le g'(\pi/2)=0. \ \Box$

Problem 3. Let $\{x_n\}_{n\ge 1}$ be the sequence of positive roots of $f(x)=\tan x - x.$ Show that

$\displaystyle n\pi-\frac{1}{n\pi}+\frac{\pi}{2} < x_n < n\pi-\frac{2}{n\pi^2}+\frac{\pi}{2}.$

Solution. Let $\displaystyle y_n:=n\pi-\frac{1}{n\pi}+\frac{\pi}{2}$ and $\displaystyle z_n:=n\pi-\frac{2}{n\pi^2}+\frac{\pi}{2}.$ We are done if we show that $f(y_n) < 0$ and $f(z_n) > 0.$ Let

$\displaystyle g(x):=x\cot x - \frac{\pi x}{2}+x^2, \ \ x \in (0,\pi).$

By Problem 2, $g$ is decreasing and thus $\displaystyle g(x) < \lim_{x\to{0^+}}g(x)=1$ for all $x \in (0,\pi).$ In particular, $g(1/n\pi) < 1,$ i.e.

$\displaystyle \frac{1}{n\pi}\cot(1/n\pi)-\frac{1}{2n}+\frac{1}{n^2\pi^2} < 1,$

proving that $f(y_n) < 0.$
On the other hand, since $\displaystyle \frac{2}{n\pi^2} \le \frac{2}{\pi^2}$ for all $n \ge 1$ and $g$ is decreasing, we have $\displaystyle g(2/n\pi^2) \ge g(2/\pi^2),$ i.e.

$\displaystyle \frac{2}{n\pi^2}\cot(2/n\pi^2)-\frac{1}{n\pi}+\frac{4}{n^2\pi^4} \ge \frac{2}{\pi^2}\cot(2/\pi^2)-\frac{1}{\pi}+\frac{4}{\pi^4}$

and that gives

$\displaystyle f(z_n) \ge n\left(\cot(2/\pi^2)-\frac{3\pi}{2}+\frac{2}{\pi^2}\right) > 0. \ \Box$

Example. For $n=20,$ Problem 3 gives $64.386 < x_{20} < 64.3926$ and the actual value of $x_{20}$ is approximately $64.387.$ So the lower and upper bounds that Problem 3 gives for $x_n$ are not that bad.

Exercise. Let $\{x_n\}_{n\ge 1}$ be the positive solutions of the equation $\tan(x)=x.$

i) Show that $\displaystyle \lim_{n\to\infty} (x_n-n\pi)=\frac{\pi}{2}$ and $\displaystyle \lim_{n\to\infty} \frac{x_n}{n}=\lim_{n\to\infty}(x_n-x_{n-1})=\pi.$

ii) Show that the series $\displaystyle \sum_{n=1}^{\infty}\frac{1}{x_n}$ is divergent.

iii) Let $\lfloor - \rfloor$ be the floor function. Is it true that $\displaystyle \lfloor x_n \rfloor = \left \lfloor n\pi-\frac{1}{n\pi}+\frac{\pi}{2} \right \rfloor$ for all $n \ge 1 ?$

# Integral inequalities (2)

Throughout this post, $\displaystyle f:[0,\infty) \longrightarrow [0,1]$ is an integrable function, i.e. $\displaystyle \int_0^xf(s) \ ds$ exists for any $x \ge 0.$ Every continuous function is integrable but the converse is not true, e.g. think of functions that are piecewise continuous.

Problem 1. Suppose that there exists $r \in [0,1)$ such that $\displaystyle f(x) \le \int_0^x (f(s))^rds$ for all $x \ge 0.$ Show that $\displaystyle f(x) \le x^{\frac{1}{1-r}}$ for all $x \ge 0.$

Solution. The idea is to prove that

$f(x) \le x^{1+r+ \cdots + r^n} \ \ \ \ \ \ \ \ \ (*)$

for all $x \ge 0$ and all integers $n \ge 0.$ Then taking limit of both sides of $(*)$ as $n\to\infty$ gives $f(x) \le x^{\frac{1}{1-r}},$ as required.
We prove $(*)$ by induction over $n.$ It is true for $n=0$ because

$\displaystyle f(x) \le \int_0^x(f(s))^rds \le \int_0^xds=x.$

Now suppose that $(*)$ is true for $n.$ Then

\displaystyle \begin{aligned} f(x) \le \int_0^x(f(s))^rds \le \int_0^x(s^{1+r+ \cdots + r^n})^rds=\int_0^x s^{e+r^2+ \cdots + r^{n+1}}ds \\ =\frac{1}{1+r + \cdots + r^{n+1}}x^{1+r+ \cdots + r^{n+1}} \le x^{1+r+ \cdots + r^{n+1}}. \ \Box \end{aligned}

Problem 2. Suppose that there exists $r \ge 0$ such that $\displaystyle f(x) \le r\int_0^xf(s) \ ds$ for all $x \ge 0.$ Show that $f(x)=0$ for all $x \ge 0.$

Solution. The claim is obviously true for $x=0.$ Now let $x > 0.$ We have

\displaystyle \begin{aligned} 0 \le f(x) \le r\int_0^{x} f(s_1) \ ds_1 \le r^2 \int_0^{x} \int_0^{s_1}f(s_2) \ ds_2 \ ds_1 \le \cdots \\ \le r^n \int_0^{x} \int_0^{s_1} \cdots \int_0^{s_{n-1}}f(s_n) \ ds_n \ ds_{n-1} \ \cdots \ ds_1 \\ \le r^n \int_0^{x} \int_0^{s_1} \cdots \int_0^{s_{n-1}} \ ds_n \ ds_{n-1} \ \cdots \ ds_1=\frac{(rx)^n}{n!}. \end{aligned}

Thus $\displaystyle 0 \le f(x) \le \lim_{n\to\infty} \frac{(rx)^n}{n!}=0$ and so $f(x)=0. \ \Box$

Exercise 1. Solve Problem 2 using induction, i.e. show, by induction over $n,$ that $\displaystyle f(x) \le \frac{(rx)^n}{n!}$ for all $x \ge 0.$

Exercise 2.  Let $c$ be a real number. By the ratio test, the series $\displaystyle \sum_{n=0}^{\infty}\frac{c^n}{n!}$ converges (to $e^c$) and so, as a result, $\displaystyle \lim_{n\to\infty} \frac{c^n}{n!}=0.$
Using the definition of limit, prove that $\displaystyle \lim_{n\to\infty} \frac{c^n}{n!}=0.$

# Integral inequalities (1)

Problem. Show that if $\displaystyle f: [0,1] \longrightarrow (-\infty,1]$ is a continuous function such that $\displaystyle \int_0^1 f(x) \ dx = 0,$ then $\displaystyle \int_0^1 (f(x))^3 dx < \frac{1}{4}.$

Solution. Since $f(x) \le 1$ for all $x \in [0,1],$ we have $(f(x)-1)(2f(x)+1)^2 \le 0$ for all $x \in [0,1].$ Therefore

$\displaystyle 4(f(x))^3-3f(x)-1 \le 0 \ \ \ \ \ \ (*)$

and integrating $(*)$ gives $\displaystyle \int_0^1(f(x))^3 \ dx \le \frac{1}{4}.$
But can we have equality? The answer is no and here’s why. If $\displaystyle \int_0^1(f(x))^3 \ dx = \frac{1}{4},$ then $\displaystyle \int_0^1(4(f(x))^3-3f(x)-1) \ dx=0$ and thus, by $(*)$ and Problem 1 in this post, $\displaystyle 4(f(x))^3-3f(x)-1=0$ for all $x \in [0,1].$ Therefore $\displaystyle (f(x)-1)(2f(x)+1)^2=0$ for all $x \in [0,1]$ and hence, for every $x \in [0,1],$ either $f(x)=1$ or $\displaystyle f(x)=\frac{-1}{2}.$ Thus, since $f$ is continuous, either $\displaystyle f(x)=1$ for all $x \in [0,1]$ or $\displaystyle f(x)=\frac{-1}{2}$ for all $x \in [0,1].$ In either case, $\displaystyle \int_0^1 f(x) \ dx \ne 0$ and $\displaystyle \int_0^1 (f(x))^3 dx \ne \frac{1}{4}. \ \Box$

Exercise. Extend the result given in the above problem by proving that if $f:[a,b] \longrightarrow \mathbb{R}$ is continuous and if $\displaystyle 0 \ne M:=\max_{x \in [a,b]} f(x),$ then

$\displaystyle \int_a^b (f(x))^3dx < \frac{3}{4}M^2\int_a^bf(x) \ dx + \frac{b-a}{4}M^3.$

Hint. $(f(x)-M)(2f(x)+M)^2 \le 0$ for all $x \in [a,b].$

# The series \sum 1/(n^2 + a^2)

Problem. Show that $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2+a^2}=\frac{\pi a \coth(\pi a)-1}{2a^2}$ for all real numbers $a \ne 0.$

Solution. Since $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2+a^2}$ is an even function of $a,$ we may assume that $a > 0.$
First see that

$\displaystyle \int_0^{\infty}\sin(bx)e^{-ax} dx = \frac{b}{a^2+b^2}$

for all real numbers $a>0, \ b.$ So

\displaystyle \begin{aligned} \sum_{n=1}^{\infty} \frac{1}{n^2+a^2}=\sum_{n=1}^{\infty}\frac{1}{n}\int_0^{\infty}\sin(nx)e^{-ax} dx=\int_0^{\infty}\left(\sum_{n=1}^{\infty} \frac{\sin(nx)}{n}\right)e^{-ax}dx \\ =\sum_{k=0}^{\infty} \int_{2k\pi}^{(2k+1)\pi}\left(\sum_{n=1}^{\infty}\frac{\sin(nx)}{n}\right)e^{-ax}dx=\sum_{k=0}^{\infty}\int_0^{2\pi}\left(\sum_{n=1}^{\infty}\frac{\sin(nx)}{n}\right)e^{-a(2k\pi+x)}dx.\end{aligned}

But we showed here (you should also see Exercise 2 in that post!) that for $x \in (0,2\pi),$ we have $\displaystyle \sum_{n=1}^{\infty}\frac{\sin(nx)}{n}=\frac{\pi-x}{2}.$ So

\displaystyle \begin{aligned} \sum_{n=1}^{\infty} \frac{1}{n^2+a^2}=\sum_{k=0}^{\infty}e^{-2k\pi a}\int_0^{2\pi}\left(\frac{\pi-x}{2}\right)e^{-ax}dx =\left(\frac{\pi(e^{-2\pi a}+1)}{2a}+\frac{e^{-2\pi a}-1}{2a^2}\right)\sum_{k=0}^{\infty}e^{-2k\pi a} \\ =\left(\frac{\pi(e^{-2\pi a}+1)}{2a}+\frac{e^{-2\pi a}-1}{2a^2}\right) \frac{1}{1-e^{-2\pi a}}=\frac{\pi(1+e^{-2\pi a})}{2a(1-e^{-2\pi a})}-\frac{1}{2a^2} \\ = \frac{\pi}{2a}\coth(\pi a)-\frac{1}{2a^2}=\frac{\pi a \coth(\pi a)-1}{2a^2}. \ \Box \end{aligned}

Remark. In the above solution, we, once again, allowed ourselves to integrate an infinite series term by term. As I have mentioned many times in this blog, this is not always allowed.

Exercise 1. Show that $\displaystyle \int_0^{\infty}\sin(bx)e^{-ax} dx = \frac{b}{a^2+b^2}$ for all real numbers $a>0, \ b.$
Hint. Use integration by parts twice.

Exercise 2. Evaluate $\displaystyle \lim_{a\to0} \frac{\pi a \coth(\pi a)-1}{2a^2}.$ Now, considering the result given in the above problem and the fact that $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}.$ that we proved here, does the value of the limit make sense?

# Limits involving binomial coefficients (2)

Problem. Show that $\displaystyle \lim_{n\to\infty}\frac{1}{n^2}\ln\left(\prod_{k=0}^n \binom{n}{k}\right)=\frac{1}{2}.$

Solution. We have

\displaystyle \begin{aligned} a_n:=\ln\left(\prod_{k=0}^n \binom{n}{k}\right)=\sum_{k=0}^n (\ln n! - \ln k! - \ln(n-k)!)=(n+1)\ln n! - 2\sum_{k=0}^n \ln k! \\ =(n+1)\sum_{i=1}^n \ln i - 2\sum_{k=1}^n\sum_{i=1}^k \ln i =(n+1)\sum_{i=1}^n \ln i - 2 \sum_{i=1}^n\sum_{k=i}^n\ln i \\ =(n+1)\sum_{i=1}^n \ln i - 2\sum_{i=1}^n(n-i+1)\ln i=\sum_{i=1}^n(2i-n-1)\ln i \\ =\sum_{i=1}^n (2i-n)\ln \left(\frac{i}{n}\right)+\sum_{i=1}^n(2i-n)\ln n - \sum_{i=1}^n \ln i \\ =\sum_{i=1}^n (2i-n)\ln \left(\frac{i}{n}\right)+n\ln n- \ln n!. \end{aligned}

But $\displaystyle \lim_{n\to\infty} \frac{n\ln n}{n^2}=\lim_{n\to\infty} \frac{\ln n!}{n^2}=0$ (because $\displaystyle n! \le n^n$) and so

\displaystyle \begin{aligned} \lim_{n\to\infty} \frac{a_n}{n^2}=\lim_{n\to\infty} \frac{1}{n^2}\sum_{i=1}^n (2i-n)\ln \left(\frac{i}{n}\right)=\lim_{n\to\infty} \frac{1}{n}\sum_{i=1}^n \left(\frac{2i}{n}-1\right)\ln \left(\frac{i}{n}\right) \\ =\int_0^1(2x-1)\ln x \ dx = \frac{1}{2}. \ \Box \end{aligned}