# Ngô Quốc Anh

## September 8, 2007

### “Tích phân suy rộng” vs. “Hàm Gamma”

Filed under: Các Bài Tập Nhỏ — Ngô Quốc Anh @ 19:36

Problem. For each $m,n>0$ prove that

$\displaystyle\int_{0}^{\infty}\ln x\cdot\sin (mx)\cdot e^{-nx}dx=\boxed{\frac{1}{m^{2}+n^{2}}\cdot\left(n\cdot\tan^{-1}\left(\frac{m}{n}\right)-m\cdot\gamma-\frac{m}{2}\cdot\ln (m^{2}+n^{2})\right)}$

and

$\displaystyle\int_{0}^{\infty}\ln x\cdot\cos (mx)\cdot e^{-nx}dx=\boxed{-\frac{1}{m^{2}+n^{2}}\cdot\left(m\cdot\tan^{-1}\left(\frac{m}{n}\right)+n\cdot\gamma+\frac{n}{2}\cdot\ln (m^{2}+n^{2})\right)}$.

Solution. Differentiate

$\displaystyle\int_{0}^{\infty}x^{z-1} e^{-Rx} dx =\frac{\Gamma(z)}{R^{z}}$

with respect to $z = 1$

$\displaystyle\int_{0}^{\infty}x^{z-1}\ln x e^{-Rx} dx =\frac{R^{z}\Gamma'(z)-\Gamma(z) R^{z}\ln R}{R^{2z}}$

and set $z = 1$

$\displaystyle\int_{0}^{\infty}\ln x e^{-Rx} dx =\frac{-\gamma-\ln R}{R}$

After that set $R = n+im$

$\displaystyle\begin{gathered} \int_0^\infty {\ln } x{e^{ - imx}}{e^{ - nx}}dx = \frac{{ - \gamma - \frac{1}{2}\ln ({n^2} + {m^2}) - i\arctan \frac{m}{n}}}{{n + im}} \hfill \\ \quad = \frac{{ - 1}}{{{n^2} + {m^2}}}\left[ {(n - im)\left( {\gamma + \frac{1}{2}\ln ({n^2} + {m^2}) + i\arctan \frac{m}{n}} \right)} \right] \hfill \\ \quad = \frac{{ - 1}}{{{n^2} + {m^2}}}\left( {n\gamma + \frac{n}{2}\ln ({n^2} + {m^2}) + m\arctan \frac{m}{n}} \right) \hfill \\ \qquad + \frac{i}{{{n^2} + {m^2}}}\left( {m\gamma + \frac{m}{2}\ln ({n^2} + {m^2}) - n\arctan \frac{m}{n}} \right). \hfill \\ \end{gathered}$

### “Tích phân xác định” vs. “Chuỗi số”

Filed under: Các Bài Tập Nhỏ — Ngô Quốc Anh @ 19:20

Đề bài. $\forall\,n\in\Bbb{N}^{+}$ hãy tính

$\int_{-\pi}^{\pi}\;\;\frac{\cos\,(n\cdot x)}{5+4\,\cos\,(x)}\;\;\textbf dx$

Cách giải 1. (sử dụng giải tích phức)Instead of finding

$\int_{-\pi}^{\pi}\frac{\cos nx}{5+4\cos x}dx$

rather find

$\int_{-\pi}^{\pi}\frac{e^{inx}}{5+4\cos x}dx$

If $|z|=1$ is a unit circle from $-\pi$ to $\pi$ we have that,

$\int_{-\pi}^{\pi}\frac{e^{inx}}{5+4\cos x}dx =\oint_{|z|=1}\frac{z^{n}}{5+2(z+z^{-1})}\cdot\frac{1}{iz}dz$

Which works out to be,

$\frac{1}{i}\oint_{|z|=1}\frac{z^{n}}{(2z+1)(z+2)}dz$

This contour integral has only a single pole within $|z|=1$ which is simple $z=-1/2$.
So,

$\mbox{res}(-1/2) =\lim_{z\to-1/2}(z+1/2)\cdot\frac{z^{n}}{(2z+1)(z+2)}=\frac{\left(-1/2\right)^{n}}{3}=\frac{(-1)^{n}}{3\cdot 2^{n}}$

By residue theorem together with the $1/i$ factor gives us,

$\int_{-\pi}^{\pi}\frac{\cos nx}{5+4\cos x}dx = (-1)^{n}\frac{\pi}{3\cdot 2^{n-1}}$

Cách giải 2. (sử dụng sai phân) Let

$I_{n}=\int_{-\pi}^{\pi}\frac{\cos nx}{5+4\cos x}\, dx$.

Using some trigonometry, $\cos (n+1)x+\cos (n-1)x = 2\cos nx\cos x$ and

$I_{n+1}+I_{n-1}\; =\;\int_{-\pi}^{\pi}\frac{2\cos nx\cos x}{5+4\cos x}\, dx\; =\;-\frac{5}{2}I_{n}$

Two roots of $2x^{2}+5x+2 = 0$ is

$x =-2\text{ or }-\frac{1}{2}$.

Then we can write

$I_{n}= A (-2)^{n}+B\left(-\frac{1}{2}\right)^{n}$.

But since $I_{n}$ is bounded, $A = 0$ and

$I_{n}\; =\; I_{0}\left(-\frac{1}{2}\right)^{n}\; =\;\frac{2\pi}{3}\left(-\frac{1}{2}\right)^{n}$.

Cách giải 3. (sử dụng chuỗi số) For

$|b| < 1\quad\sum_{k = 0}^{\infty}b^{k}\cos kx =\frac{1-b\cos x}{1-2b\cos x+b^{2}}\;\Rightarrow\; 2\sum_{k = 0}^{\infty}b^{k}\cos kx-1 =\frac{1-b^{2}}{1-2b\cos x+b^{2}}$

$\int_{-\pi}^{\pi}\frac{\cos nx}{1-2b\cos x+b^{2}}=\frac{1}{1-b^{2}}\int_{-\pi}^{\pi}\cos nx\left(2\sum_{k = 0}^{\infty}b^{k}\cos kx-1\right) dx$

$=\frac{1}{1-b^{2}}\left[2\sum_{k = 0}^{\infty}b^{k}\underbrace{\int_{-\pi}^{\pi}\cos nx\,\cos kx\, dx}_{\pi\,\delta_{nk}}-\underbrace{\int_{-\pi}^{\pi}\cos nx dx}_{ = 0}\right] =\frac{2\pi\, b^{n}}{1-b^{2}}$

In particular for

$b =-\frac{1}{2}\quad\int_{-\pi}^{\pi}\frac{\cos nx}{1+\cos x+\frac{1}{4}}\, dx =\frac{2\pi}{1-\frac{1}{4}}\left(-\frac{1}{2}\right)^{n}$

### Một áp dụng của biểu diễn Euler cho hàm Gamma

Filed under: Các Bài Tập Nhỏ — Ngô Quốc Anh @ 19:12

Biểu diễn Euler cho hàm Gamma được phát biểu như sau

$\Gamma\, :\quad\Gamma(1+z)=\prod_{n=1}^{\infty}\frac{\left(1+\frac{1}{n}\right)^{z}}{1+\frac{z}{n}}$

Đề bài. Chứng minh rằng

$\prod_{n=1}^{\infty}\;\;\frac{1+n^{6}}{n^{6}}\;\;=\;\;\boxed{\frac{\sinh\,\pi\cdot\left(\cosh\,\pi\;-\;\cos\,(\sqrt 3\,\pi)\right)}{2\,\pi^{3}}}$

Lời giải. Ký hiệu $\xi$ là $e^{\frac{2\pi i}{m}}$. Khi đó ta có

$\prod_{k=0}^{m-1}\Gamma(1-\xi^{k}z)=\prod_{n=1}^{\infty}\frac{\left(1+\frac{1}{n}\right)^{\overbrace{(-\xi^{0}-...-\xi^{m-1})}^{=0}z}}{\left(1-\frac{\xi^{0}z}{n}\right)\cdots\left(1-\frac{\xi^{m-1}z}{n}\right) }=\prod_{n=1}^{\infty}\left[1-\left(\frac{z}{n}\right)^{m}\,\right]^{-1}$

Nói riêng với $m=6$ và

$z=e^{\frac{i\pi}{6}}\quad\prod_{n=1}^{\infty}\left(1+\frac{1}{n^{6}}\right)^{-1}=\prod_{k=0}^{5}\Gamma\left(1-e^{(2k+1)\frac{i\pi}{6}}\right)$

$=\Gamma(1-i)\,\Gamma(1+i)\;\;\,\Gamma(1-e^{\frac{i\pi}{6}})\,\Gamma(1+e^{\frac{i\pi}{6}})\;\;\,\Gamma(1-e^{\frac{5i\pi}{6}})\,\Gamma(1+e^{\frac{5i\pi}{6}})$

$=\frac{i\pi}{\sin i\pi}\;\;\frac{e^{i\frac{\pi}{6}}\pi}{\sin\left(e^{i\frac{\pi}{6}}\pi\right)}\;\;\frac{e^{i\frac{5\pi}{6}}\pi}{\sin\left(e^{i\frac{5\pi}{6}}\pi\right)}=\frac{\pi^{3}}{\sinh\pi}\;\,\frac{-1}{\sin\left(e^{i\frac{\pi}{6}}\pi\right)\,\sin\left(e^{i\frac{5\pi}{6}}\pi\right) }$

$=\frac{\pi^{3}}{\sinh\pi}\;\;\frac{2}{\cos\left(\underbrace{\left(e^{i\frac{\pi}{6}}+e^{i\frac{5\pi}{6}}\right)}_{=i}\pi\right)-\cos\left(\underbrace{\left(e^{i\frac{\pi}{6}}-e^{i\frac{5\pi}{6}}\right)}_{=\sqrt{3}}\pi\right) }$

Từ đây có điều phải chứng minh.