## My article – Euler’s formula about Ap´ery’s constant

Note : This is written in Korean.

## Twenty-fifth quiz – Projection matrix

Quiz. Let $E_1, \cdots, E_k$ $n\times n$ complex matrices. Suppose that

$E_1 ^2=E_1, E_2 ^2=E_2, \cdots, E_k ^2= E_k$

$E_1+E_2+\cdots+E_k=I$

Prove or disprove the following proposition;

[ $\forall i, j(i\neq j)\; ; \; E_i E_j=0$ ]

## Twenty-fourth quiz – Matrix

Problem. $A$ is a $n\times n$ matrix which satisfies

$\bullet\;A^2=I,\;\;\;\;\;\bullet\;{\rm rank} (A+I)=1$

Find all possible values of ${\rm trace} A$.

## Twenty-third quiz – Integral equation

Quiz.

A continuous function $f:[0, 1]\rightarrow \mathbb{R}$ satisfies the following equality;

$\displaystyle \int_{0}^{1}f(x)dx=\frac{1}{3}+\int_{0}^{1}(f(x^2))^2 dx$

Find $f$.

$\phantom{aa}$

Hint. AM-GM inequality

## An integral similar to Gauss..

Problem. (제 29회 전국 대학생수학경시대회 제 2분야 8번)

Calculate $\displaystyle \int_{0}^{\infty} e^{-x^2} \cos (ax) dx$ where $a\in \mathbb{R}$.

$\phantom{a}$

My solution.

Let $\displaystyle f(a):=\int_{0}^{\infty}e^{-x^2}\cos (ax) dx$

We first expand $\cos(ax)$ to Taylor series, and exchange the summation and integral;

$\displaystyle f(a)=\int_{0}^{\infty}e^{-x^2}\sum_{n\ge 0}\frac{(-1)^{n}(ax)^{2n}}{(2n)!}dx=\sum_{n\ge 0}\left(\frac{(-1)^n a^{2n}}{(2n)!}\int_{0}^{\infty}e^{-x^2} x^{2n} dx\right)$

We can calculate the integral $\displaystyle \int_{0}^{\infty}e^{-x^2} x^{2n} dx$ using Gamma function; (I love this function 🙂 )

$\displaystyle \int_{0}^{\infty}e^{-x^2} x^{2n} dx\overset{s=x^2}{=}\int_{0}^{\infty}s^n e^{-s} \frac{ds}{2\sqrt{s}}=\frac{1}{2}\int_{0}^{\infty}s^{n-1/2} e^{-s} ds=\frac{1}{2}\Gamma\left(n+\frac{1}{2}\right)$

$\displaystyle \Gamma\left(n+\frac{1}{2}\right)=\Gamma\left(\frac{1}{2}\right)\prod_{0\leq k< n}\frac{2k+1}{2}=\frac{(2n)!}{2^{2n} n!}\sqrt{\pi}$

$\displaystyle \therefore \int_{0}^{\infty}e^{-x^2} x^{2n} dx=\frac{1}{2}\frac{(2n)!}{2^{2n} n!}\sqrt{\pi}$

So

\displaystyle \begin{aligned}f(a)&=\sum_{n\ge 0}\left(\frac{(-1)^n a^{2n}}{(2n)!}\times \frac{1}{2}\frac{(2n)!}{2^{2n} n!}\sqrt{\pi}\right)\\&=\frac{\sqrt{\pi}}{2}\sum_{n\ge 0}\left(\frac{(-1)^n}{n!}\left(\frac{a^2}{2^2}\right)^n\right)=\frac{\sqrt{\pi}}{2}e^{-a^2 /4}\end{aligned}

$\displaystyle \therefore f(a)=\int_{0}^{\infty} e^{-x^2} \cos (ax) dx=\frac{\sqrt{\pi}}{2}e^{-a^2 /4}$

$\square$

Posted in Calc&Evaluation, Mathematics | 2 Comments

## bought a score…

A few days ago, I bought the score of Sinfonietta(composed by N.Kapustin), which is one of my favorite music 🙂

## 새우깡 + Happy things

새우깡 CF + Happy Things(by J Rabbit)

played by me

## Simple integral

Problem.

Prove that $\displaystyle \int_{0}^{1} \sqrt{1-t^2} \log t dt=-\frac{\pi}{8}(1+\log 4)$.

$\phantom{a}$

I saw this problem at http://sos440.tistory.com/34.

$\phantom{a}$

My solution.

Let $\displaystyle A:=\int_{0}^{1} \sqrt{1-t^2}\log t dt$

Substitute $t=\sin u$, then $A$ becomes

$\displaystyle A=\int_{0}^{\pi/2} \sqrt{1-\sin^2 u} \log(\sin u)\cdot (\cos u du)=\int_{0}^{\pi/2} \cos^2 {u} \log(\sin u) du$

Now we use the following formula(which is one of my favorite formulas);

$\displaystyle \log(\sin \theta)=-\log 2-\sum_{n\ge 1}\frac{\cos (2n\theta)}{n} \phantom{aa}\left(0<\theta<\frac{\pi}{2}\right)$

By this formula,

\displaystyle \begin{aligned}A&=\int_{0}^{\pi/2}\cos^2 u \cdot \left(-\log 2-\sum_{n\ge 1}\frac{\cos (2nu)}{n}\right)du\\&=-\log 2 \int_{0}^{\pi/2} \cos^2 u du-\int_{0}^{\pi/2}\cos^2 u \cdot \left(\sum_{n\ge 1}\frac{\cos (2nu)}{n}\right)du\end{aligned}

We can easily prove that $\int_{0}^{\pi/2}\cos^2 u du=\frac{\pi}{4}\log 2$; For the second integral, we interchange summation and integral… then

$\displaystyle A=-\frac{\pi}{4}\log 2-\sum_{n\ge 1}\left(\frac{1}{n}\int_{0}^{\pi/2} \cos^2 u \cos(2nu) du\right)$

Using basic trigonometric identities, we can prove that

This gives

$\displaystyle A=-\frac{\pi}{4}\log 2-\frac{\pi}{8}$

$\square$