My article – Irrational numbers

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

Note : This is written in Korean.

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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


Prove or disprove the following proposition;

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

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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.

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Twenty-third quiz – Integral equation


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.


Hint. AM-GM inequality

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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}.


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}


\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}


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bought a score…



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

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새우깡 + Happy things

새우깡 CF + Happy Things(by J Rabbit)

played by me

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Simple integral


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


I saw this problem at


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}


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My article about Leibniz rule and integral

A few days ago, I wrote an article about usefulness of Leibniz rule(it is sometimes called ‘differentiation under integral sign’) in calculating hard integrals. I submitted the article to the math club of my school; I’m still waiting for it to be accepted…

pdf file ; ML – Leibniz rule_2nd version (the article is written in Korean)

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