Ngô Quốc Anh

April 25, 2013

The Cauchy formula for repeated integration

Filed under: Các Bài Tập Nhỏ, Giải Tích 2 — Ngô Quốc Anh @ 23:41

The Cauchy formula for repeated integration, named after Augustin Louis Cauchy, allows one to compress $n$ antidifferentiations of a function into a single integral.

Let $f$ be a continuous function on the real line. Then the $n$-th repeated integral of $f$ based at $a$,

$\displaystyle f^{(-n)}(x) = \int_a^x \int_a^{\sigma_1} \cdots \int_a^{\sigma_{n-1}} f(\sigma_{n}) \, \mathrm{d}\sigma_{n} \cdots \, \mathrm{d}\sigma_2 \, \mathrm{d}\sigma_1,$

is given by single integration

$\displaystyle f^{(-n)}(x) = \frac{1}{(n-1)!} \int_a^x\left(x-t\right)^{n-1} f(t)\,\mathrm{d}t.$

A proof is given by induction. Since $f$ is continuous, the base case follows from the Fundamental theorem of calculus

$\displaystyle\frac{\mathrm{d}}{\mathrm{d}x} f^{(-1)}(x) = \frac{\mathrm{d}}{\mathrm{d}x}\int_a^x f(t)\,\mathrm{d}t = f(x);$

where

$\displaystyle f^{(-1)}(a) = \int_a^a f(t)\,\mathrm{d}t = 0.$

March 10, 2012

An integral of 1/(1+|x|^2) over the whole Euclidean space

Filed under: Giải Tích 2, Giải Tích 3, Linh Tinh — Ngô Quốc Anh @ 2:46

Sometimes, we need a precise value for following

$\displaystyle\int_{{\mathbb{R}^n}} {\frac{{dx}}{{{{(1 + |x{|^2})}^\alpha }}}}, \quad \alpha>\frac{n}{2}.$

As such, I am going to calculate it and place the result here for future works.

In order to evaluate the above integral, we need to use the so-called co-area formula. We first write

$\displaystyle\int_{{\mathbb{R}^n}} {\frac{{dx}}{{{{(1 + |x{|^2})}^\alpha }}}} = \int_0^{ + \infty } {\left( {\int_{\partial {B_0}(r)} {\frac{{dS}}{{{{(1 + {r^2})}^\alpha }}}} } \right)dr}.$

Note that

$\displaystyle \begin{gathered} \int_{\partial {B_0}(r)} {\frac{{dS}}{{{{(1 + {r^2})}^\alpha }}}} = \frac{1}{{{{(1 + {r^2})}^\alpha }}}\int_{\partial {B_0}(r)} {dS} \hfill \\ \qquad\qquad\qquad= \frac{1}{{{{(1 + {r^2})}^\alpha }}}\text{Area}({B_0}(r)) = \frac{{2{\pi ^{\frac{n}{2}}}}}{{\Gamma \left( {\frac{n}{2}} \right)}}\frac{{{r^{n - 1}}}}{{{{(1 + {r^2})}^\alpha }}}\end{gathered}.$

Therefore,

$\displaystyle\int_{{\mathbb{R}^n}} {\frac{{dx}}{{{{(1 + |x{|^2})}^\alpha }}}} = \frac{{2{\pi ^{\frac{n}{2}}}}}{{\Gamma \left( {\frac{n}{2}} \right)}}\int_0^{ + \infty } {\frac{{{r^{n - 1}}}}{{{{(1 + {r^2})}^\alpha }}}} dr = {\pi ^{\frac{n}{2}}}\frac{{\Gamma \left( {\alpha - \frac{n}{2}} \right)}}{{\Gamma (\alpha )}}$

November 5, 2011

Filed under: Giải Tích 2, Giải Tích 5, Liên Kết — Tags: — Ngô Quốc Anh @ 0:26

This is not mathematics. I just found an equation so that we can draw a heart in 3D. Indeed, the following equation

$\displaystyle {\left( {{x^2} + \frac{9}{4}{y^2} + {z^2} - 1} \right)^3} - {x^2}{z^3} - \frac{9}{{80}}{y^2}{z^3} = 0$

will generate a heart. I have tried and the following pictures show that fact.

April 22, 2011

On Costa-Hardy-Rellich inequalities

This note is to concern a recent result by David G. Costa [here]. Here the statement

Theorem 1.1. For all $a,b\in \mathbb R$ and $u \in C^\infty_0(\mathbb R^N\backslash\{0\})$ one has

$\displaystyle\left| {\frac{{N - 2 - \gamma }}{2}\int_{\mathbb R^N} {\frac{{|\nabla u{|^2}}}{{|x{|^\gamma }}}dx} + \gamma \int_{\mathbb R^N} {\frac{{{{(x \cdot \nabla u)}^2}}}{{|x{|^{\gamma + 2}}}}dx} } \right| \leqslant {\left( {\int_{\mathbb R^N} {\frac{{|\Delta u{|^2}}}{{|x{|^{2b}}}}dx} } \right)^{\frac{1}{2}}}{\left( {\int_{\mathbb R^N} {\frac{{|\nabla u{|^2}}}{{|x{|^{2a}}}}dx} } \right)^{\frac{1}{2}}}$

where $\gamma=a+b+1$. In addition, if $\gamma \leqslant N-2$, then

$\displaystyle\widehat C\int_{\mathbb R^N} {\frac{{{{(x \cdot \nabla u)}^2}}}{{|x{|^{\gamma + 2}}}}dx} \leqslant {\left( {\int_{\mathbb R^N} {\frac{{|\Delta u{|^2}}}{{|x{|^{2b}}}}dx} } \right)^{\frac{1}{2}}}{\left( {\int_{\mathbb R^N} {\frac{{|\nabla u{|^2}}}{{|x{|^{2a}}}}dx} } \right)^{\frac{1}{2}}}$

where the constant $\widehat C=|\frac{N+a+b-1}{2}|$ is sharp.

Here’s the proof.

April 1, 2011

Several interesting limits from a paper by Chang-Qing-Yang

Recently, I have learnt from my friend, ZJ, the following result

Assume that $F:\mathbb R \to \mathbb R$ is absolutely integrable. Then

$\displaystyle\begin{gathered} \mathop {\lim }\limits_{t \to \pm \infty } {e^{2t}}\int_t^{ + \infty } {F(x){e^{ - 2x}}dx} = 0, \hfill \\ \mathop {\lim }\limits_{t \to \pm \infty } {e^{ - 2t}}\int_{ - \infty }^t {F(x){e^{ - 2x}}dx} = 0. \hfill \\ \end{gathered}$

The result seems reasonable by the following observation, for example, we consider the first identity when $t \to +\infty$. Then the factor

$\displaystyle\int_t^{ + \infty } {F(x){e^{ - 2x}}dx}$

decays faster then the exponent function $\exp (2t)$. This may be true, of course we need to prove mathematically, because the integrand contains the term $\exp (-2x)$ which turns out to be a good term since $x \geqslant t$. So here is the trick in order to solve such a problem.

February 8, 2010

The two-dimensional isodiametric inequality

Filed under: Các Bài Tập Nhỏ, Giải Tích 2, Nghiên Cứu Khoa Học — Tags: — Ngô Quốc Anh @ 0:00

The isodiametric inequality states that, of all bodies of a given diameter, the sphere has the greatest volume. A proof can be found, e.g., in Lawrence C. Evans & Ronald F Gariepy: Measure theory and fine properties of functions. This note is about a particularly simple and beautiful proof of the isodiametric inequality in two dimensions.

I originally posted this as a plea for help to discover the origin of the proof. I have since learned that it is found in Littlewood’s miscellany, on p. 32.

The diameter of a body is defined as the supremum of the distances between two points in the body.

Clearly, we need only showthe isodiametric inequality for convex bodies, since taking the convex closure does not increase the diameter, nor does it decrease the area. Next, we may move the (convex) body so that it lies in the upper half plane, with the origin at its boundary. Thus we may describe the body in polar coordinates by

$r\leqslant f(\theta), \quad 0\leqslant\theta\leqslant\pi$.

We now apply a bit of first year calculus to write the area of the body as

$\displaystyle A = \frac{1}{2}\int_0^\pi {f{{(\theta )}^2}d\theta }$.

We now split the integral in two halves and change the variable in the second half so that both halves can be written as the single integral

$\displaystyle A = \frac{1}{2}\int_0^{\frac{\pi }{2}} {\left( {f{{(\theta )}^2} + f{{\left( {\theta + \frac{\pi }{2}} \right)}^2}} \right)d\theta }$.

But here we recognize the integrand as the squared hypothenuse of the right triangle in the figure. By definition, the hypothenuse cannot be greater than the diameter d of the region

$\displaystyle f{(\theta )^2} + f{\left( {\theta + \frac{\pi }{2}} \right)^2} \leqslant {d^2}$.

Thus

$\displaystyle A \leqslant \pi {\left( {\frac{d}{2}} \right)^2}$

which is the isodiametric inequality, and so the proof is complete.

September 26, 2009

How to calculate limit by using definition of definite integral?

Filed under: Các Bài Tập Nhỏ, Giải Tích 2 — Ngô Quốc Anh @ 14:23

Let $f : [a, b] \to \mathbb R$ be a continuous function, not necessarily nonnegative. Partition $[a, b]$ into $n$ consecutive sub-intervals $[x_{i-1}, x_i]$ ($i = 1, 2, ..., n$) each of length $\Delta x = \frac{b-a}{n}$, where we set $a=x_0$, $b=x_n$ and $x_1, x_2,...,x_{n-1}$ to be successive points between $a$ and $b$ with $x_k-x_{k-1}=\Delta x$. Let $c_k$ be any intermediate point in the sub-interval $[x_{k-1},x_k]$. Then the sum

$\displaystyle\sum\limits_{k = 1}^n {f\left( {{c_k}} \right)\Delta x}$

is called a Riemann sum for $f$ on $[a, b]$.

Suppose we let the number of partition in $P$ tends to infinity.

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {f\left( {{c_k}} \right)\Delta x} = I.$

We call $I$ the Riemann integral (or definite integral) of $f$ over $[a, b]$ and we write

$\displaystyle I = \int_a^b {f(x)dx} .$

In other words,

$\displaystyle\int_a^b {f(x)dx} = \frac{{b - a}}{n}\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {f\left( {{c_k}} \right)}$

if the limit on the right side exists.

If we put $c_k=x_{k-1}$ we the obtain

$\displaystyle\int_a^b {f(x)dx} = \frac{{b - a}}{n}\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {f\left( {a + (k - 1)\frac{{b - a}}{n}} \right)} .$

Example 1. Find

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{n + 1}} + \cdots + \frac{1}{{2n - 2}} + \frac{1}{{2n - 1}}} \right).$

Solution. Clearly

$\displaystyle\frac{1}{n} + \frac{1}{{n + 1}} + \cdots + \frac{1}{{2n - 2}} + \frac{1}{{2n - 1}} = \sum\limits_{k = 1}^n {\frac{1}{{n + (k - 1)}}} = \frac{1}{n}\sum\limits_{k = 1}^n {\frac{1}{{1 + \frac{{k - 1}}{n}}}} .$

Then if we choose $a=0$, $b=1$ we then get

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{n + 1}} + \cdots + \frac{1}{{2n - 2}} + \frac{1}{{2n - 1}}} \right) = \int_0^1 {\frac{{dx}}{{1 + x}}} .$

With this it is easy to see that

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{n + 1}} + \cdots + \frac{1}{{2n - 2}} + \frac{1}{{2n - 1}}} \right) = \ln 2$

since

$\displaystyle\int_0^1 {\frac{{dx}}{{1 + x}}} = \ln 2.$

If we put $c_k=x_k$ we the obtain

$\displaystyle\int_a^b {f(x)dx} = \frac{{b - a}}{n}\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {f\left( {a + k\frac{{b - a}}{n}} \right)} .$

Example 2. Find

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \left( {\frac{n}{{{n^2} + {1^2}}} + \frac{n}{{{n^2} + {2^2}}} + \cdots + \frac{n}{{{n^2} + {{(n - 1)}^2}}} + \frac{n}{{{n^2} + {n^2}}}} \right).$

Solution. Clearly,

$\displaystyle\frac{n}{{{n^2} + {1^2}}} + \frac{n}{{{n^2} + {2^2}}} + \cdots + \frac{n}{{{n^2} + {n^2}}} = \sum\limits_{k = 1}^n {\frac{n}{{{n^2} + {k^2}}}} = \frac{1}{n}\sum\limits_{k = 1}^n {\frac{1}{{1 + {{\left( {\frac{k}{n}} \right)}^2}}}}$

which yields

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \left( {\frac{n}{{{n^2} + {1^2}}} + \frac{n}{{{n^2} + {2^2}}} + \cdots + \frac{n}{{{n^2} + {n^2}}}} \right) = \int_0^1 {\frac{{dx}}{{1 + {x^2}}}} = \frac{\pi }{4}.$

Remark. It is worth mentioning that in general it is not true that

$\displaystyle\lim \left( {summation} \right) = summation\left( {\lim } \right).$

For example, we all know that for each fixed $k$

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \frac{1}{{n + k}} = 0$

but

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \left( {\sum\limits_{k = 1}^n {\frac{1}{{n + k}}} } \right) = \ln 2 \ne 0 = \sum\limits_{k = 1}^n {\left( {\mathop {\lim }\limits_{n \to \infty } \frac{1}{{n + k}}} \right)} .$

The point is

$\displaystyle\lim \left( {summation} \right) = summation\left( {\lim } \right)$

holds true only for finite summation.

November 6, 2008

Differentiability via weird argument

Filed under: Các Bài Tập Nhỏ, Giải Tích 1, Giải Tích 2 — Ngô Quốc Anh @ 16:37

Question. Suppose that $f:\mathbb R \to \mathbb R$  is such that $f$ is continuous and $|f|$ is differentiable. Should $f$ also be differentiable?

Proof. Let $g=|f|$. Since $y=f(x)$ satisfies the differentiable equation

$y^2=g(x)^2$,

$f$ is differentiable at all points with $f(x) \neq 0$ by the implicit function theorem and

$2yy'=2gg'$,

and

$\displaystyle f'(x)=\frac{g(x)g'(x)}{f(x)}$.

Where $f(x)=0$, we note that $g$ has a minimum and $g'(x)=0$. This gives

$\displaystyle\lim_{t\to 0}\left|\frac{f(x+t)}{t}\right|=0$

and $f$ is differentiable by the definition.

Comment. The only points of interests are the zeros of $f$ since $f$ has the same sign in some neighborhood of points which are not roots. So there goes half the work. Geometrically it doesn’t make sense for $|f|$ to have anything but derivative 0 at roots of $f$ and this can be verified by taking left hand and right hand limits of the quotient.

September 17, 2008

L^2 differentiable?

Filed under: Các Bài Tập Nhỏ, Giải Tích 2, Giải Tích 3 — Ngô Quốc Anh @ 13:04

Let  and define .

a) Must  be differentiable at 0?
b) Must  have any differentiable points?
c) Let , show that  exists and determine what it is.

Solutions.

a) No. For example, let  for  so that  near zero. This  but  does not exist.

b) Yes. In fact,  must be differentiable almost everywhere, by the Lebesgue theorem on the differentiation of the integral. This theorem requires only that  which is true.

c) By the Schwarz inequality,

$f^2(x) = \left(\int_0^x}g(t)\,dt\right)^2\le \left(\int_0^x 1\,dt\right) \left(\int_0^xg^2(t)\,dt\right).$

At least that’s it for  Being careful about the other side, we determine that

$0\le\phi(x) = f^2(x)\le|x|\left|\int_0^xg^2(t)\,dt\right|.$

But since  is an integrable function we have (by an argument that uses the Dominated Convergence Theorem) that

$\lim_{x\to 0}\int_0^xg^2(t)\,dt = 0.$

Hence $\lim_{x\to0}\frac {\phi(x)}{x} = 0,$ so 

August 9, 2008

Tính tích phân suy rộng loại I

Filed under: Các Bài Tập Nhỏ, Giải Tích 2, Giải Tích 4 — Ngô Quốc Anh @ 13:16

Tính

$\displaystyle \int_{0}^{+\infty}{\frac{e^{-x^2}}{\left(x^2+\frac{1}{2}\right)^2}\; dx}$.

Lời giải. Đặt

$\displaystyle I(\alpha) = - \int_{0}^{\infty} \frac {e^{ - x^2}}{x^2 + \alpha} \, dx$.

Khi đó ta có

$\displaystyle\begin{gathered} \int_0^\infty {\frac{{{e^{ - {x^2}}}}}{{{x^2} + \alpha }}} \,dx = \int_0^\infty {{e^{ - {x^2}}}} \int_0^\infty {{e^{ - ({x^2} + \alpha )t}}} \,dtdx \hfill \\\qquad \qquad \qquad \; \,= \int_0^\infty {{e^{ - \alpha t}}} \int_0^\infty {{e^{ - (1 + t){x^2}}}} \,dxdt \hfill \\ \qquad \qquad \qquad \; \,= \int_0^\infty {\frac{{\sqrt \pi }}{2}} \frac{{{e^{ - \alpha t}}}}{{\sqrt {1 + t} }}\,dt, \hfill \\ \end{gathered}$

ta thấy

$\displaystyle\int_{0}^{\infty} \frac {e^{ - x^2}}{(x^2 + \frac {1}{2})^2} \, dx = I'(\tfrac{1}{2}) = \frac {\sqrt {\pi}}{2} \int_{0}^{\infty} \frac {t}{\sqrt {1 + t}} \, e^{ - \frac {t}{2}} \, dt$.

Nhưng

$\displaystyle \begin{gathered} \int_0^\infty {\frac{t}{{\sqrt {1 + t} }}} \,{e^{ - \frac{t}{2}}}\,dt = \int_0^\infty {\sqrt {1 + t} } \,{e^{ - \frac{t}{2}}}\,dt - \int_0^\infty {\frac{1}{{\sqrt {1 + t} }}} \,{e^{ - \frac{t}{2}}}\,dt \hfill \\ \qquad \qquad\qquad \qquad= \left[ { - 2\sqrt {1 + t} \,{e^{ - \frac{t}{2}}}} \right]_0^\infty + \int_0^\infty {\frac{1}{{\sqrt {1 + t} }}} \,{e^{ - \frac{t}{2}}}\,dt - \int_0^\infty {\frac{1}{{\sqrt {1 + t} }}} \,{e^{ - \frac{t}{2}}}\,dt \hfill \\ \qquad \qquad\qquad \qquad= 2. \hfill \\\end{gathered}$

Vậy

$\displaystyle\int_{0}^{\infty} \frac {e^{ - x^2}}{(x^2 + \frac {1}{2})^2} \, dx = \sqrt {\pi}$.

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