Tiếp theo Phần 1 , Phần 2, Phần 3 , Phần 4, Phần 5, Phần 6, Phần 7 và Phần 8.

Ta tiếp tục bàn đến các vấn đề về tích phân của hàm logarit của các hàm lượng giác đã nói trong phần 8.
Bài 6:

\(\displaystyle \int_0^{\theta}\log|\cos x + \sin x|\,dx\)

Giải:

\(\displaystyle \int_0^{\theta}\log|\cos x + \sin x|\,dx=\int_0^{\theta}\log\,\left|\sqrt{2}\left(\sin\frac{\pi}{4}\cos x + \cos\frac{\pi}{4}\sin x\right)\right|\,dx=\)

\(\displaystyle -\frac{\theta}{2}\log 2+\int_0^{\theta}\log\,\left|2\sin\left(x+\frac{\pi}{4}\right)\right|\,dx=\)

\(\displaystyle -\frac{\theta}{2}\log 2+\frac{1}{2}\int_{\pi/2}^{2\theta+\pi/2}\log\,\left|2\sin\frac{y}{2}\right|\,dy=\)

\(\displaystyle -\frac{\theta}{2}\log 2-\frac{1}{2}\text{Cl}_2\left(\frac{\pi}{2}+2\theta\right)+\frac{1}{2}\text{Cl}_2\left(\frac{\pi}{2}\right)=\)

\(\displaystyle -\frac{\theta}{2}\log 2-\frac{1}{2}\text{Cl}_2\left(\frac{\pi}{2}+2\theta\right)+\frac{G}{2}\)

Do đó

\(\displaystyle \int_0^{\theta}\log|\cos x + \sin x|\,dx=-\frac{\theta}{2}\log 2-\frac{1}{2}\text{Cl}_2\left(\frac{\pi}{2}+2\theta\right)+\frac{G}{2}.\,\Box\)

Bài 7:

\(\displaystyle \int_0^{\theta}x\,\log(\sin x)\,dx\)

Giải:

\(\displaystyle \int_0^{\theta}x\,\log(\sin x)\,dx=\int_0^{\theta}x\,\log(2\sin x)\,dx-\frac{\theta^2}{2}\log 2=\)

\(\displaystyle \frac{1}{4}\int_0^{2\theta}y\,\log\left(2\sin \frac{y}{2}\right)\,dy-\frac{\theta^2}{2}\log 2=\)

\(\displaystyle \frac{1}{4}\int_0^{2\theta}x\,\left[-\frac{d}{dx}\text{Cl}_2(x)\right]\,dx-\frac{\theta^2}{2}\log 2=\)

\(\displaystyle -\frac{1}{4}x\,\text{Cl}_2(x)\biggr|_0^{2\theta}+\frac{1}{4}\int_0^{2\theta}\text{Cl}_2(x)\,dx-\frac{\theta^2}{2}\log 2\)

\(\displaystyle -\frac{\theta}{2}\,\text{Cl}_2(2\theta)-\frac{\theta^2}{2}\log 2+\frac{1}{4}\int_0^{2\theta}\text{Cl}_2(x)\,dx=\)

\(\displaystyle -\frac{\theta}{2}\,\text{Cl}_2(2\theta)-\frac{\theta^2}{2}\log 2+\frac{1}{4}\sum_{k=1}^{\infty}\frac{1}{k^2}\int_0^{2\theta}\sin kx\,dx=\)

\(\displaystyle -\frac{\theta}{2}\,\text{Cl}_2(2\theta)-\frac{\theta^2}{2}\log 2-\frac{1}{4}\sum_{k=1}^{\infty}\frac{\cos kx}{k^3}\Biggr|_0^{2\theta}=\)

\(\displaystyle -\frac{\theta}{2}\,\text{Cl}_2(2\theta)-\frac{\theta^2}{2}\log 2-\frac{1}{4}\sum_{k=1}^{\infty}\frac{\cos 2k\theta}{k^3}+\frac{1}{4}\sum_{k=1}^{\infty}\frac{1}{k^3}=\)

\(\displaystyle -\frac{\theta}{2}\,\text{Cl}_2(2\theta)-\frac{\theta^2}{2}\log 2-\frac{1}{4}\text{Cl}_3(2\theta)+\frac{1}{4}\zeta(3)\)

vì

\(\displaystyle \text{Cl}_{2m+1}(\varphi)=\sum_{k=1}^{\infty}\frac{\cos k\varphi}{k^{2m+1}}\)

Do đó

\(\displaystyle \int_0^{\theta}x\,\log(\sin x)\,dx=\frac{1}{4}\left[\zeta(3)-2\theta^2\log2-2\theta\text{Cl}_2(2\theta)-\text{Cl}_3(2\theta)\right].\,\Box\)

Bài 8:

\(\displaystyle \int_0^{\theta}x\log(\cos x)\,dx\)

Giải:

Đặt  \(\displaystyle x=\pi/2-y\,\) ta có

\(\displaystyle \int_0^{\theta}x\log(\cos x)\,dx=-\int_{\pi/2}^{\pi/2-\theta}\left(y-\frac{\pi}{2}\right)\log(\sin y)\,dy=\)

\(\displaystyle -\int_{\pi/2}^{\pi/2-\theta}x\,\log(\sin x)\,dx+\frac{\pi}{2}\int_{\pi/2}^{\pi/2-\theta}\log(\sin x)\,dx=\)


\(\displaystyle -\frac{1}{4}\Big[\zeta(3)-2\theta^2\log2-2\theta\text{Cl}_2(2\theta)-\text{Cl}_3(2\theta)\Big]_{\pi/2}^{\pi/2-\theta}
+\frac{\pi}{2}\Big[-\frac{1}{2}\text{Cl}_2(2\theta)-\theta\log 2\Big]_{\pi/2}^{\pi/2-\theta}\)


\(\displaystyle \frac{\theta^2}{2}\log 2-\frac{\theta}{2}\text{Cl}_2(\pi-2\theta)+\frac{1}{4}\text{Cl}_3(\pi-2\theta)-\frac{1}{4}\text{Cl}_3(\pi)\)

và

\(\displaystyle \text{Cl}_3(\pi)=\sum_{k=1}^{\infty}\frac{\cos \pi k}{k^3}=\sum_{k=1}^{\infty}\frac{(-1)^k}{k^3}=-\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^3}=-\eta(3)=\)

\(\displaystyle -\left(1+\frac{1}{2^3}-\frac{1}{3^3}+\frac{1}{4^3}-\cdots\right)=-\zeta(3)\left(1-\frac{2}{2^3}\right)=-\frac{3}{4}\zeta(3)\)

Kết quả

\(\displaystyle \int_0^{\theta}x\log(\cos x)\,dx=\)


\(\displaystyle \frac{3}{16}\zeta(3)+\frac{\theta^2}{2}\log 2-\frac{\theta}{2}\text{Cl}_2(\pi-2\theta)+\frac{1}{4}\text{Cl}_3(\pi-2\theta).\,\Box\)

Bài 9:

\(\displaystyle \mathcal{I}(a\,;\, b)=\int_0^{\theta}\log(a\cos x +b\sin x)\,dx\)

Giải:

Để ý

(1) \(\displaystyle \,\,\,\,\,\,\,\,\,\,a\cos x+b\sin x=\sqrt{a^2+b^2}\cos(x\pm\varphi);\,\,\,\,\, \varphi=\tan^{-1}(\mp b/a)\)

(2) \(\displaystyle \,\,\,\,\,\,\,\,\,\,a\cos x+b\sin x=\sqrt{a^2+b^2}\sin(x\pm\varphi);\,\,\,\,\, \varphi=\tan^{-1}(\pm a/b)\)

Tích phân đã cho có thể viết lại

\(\displaystyle \mathcal{I}(a\,;\, b)=\int_0^{\theta}\log(a\cos x +b\sin x)\,dx\)

có thể viết dưới dạng

\(\displaystyle \frac{\theta}{2}\log(a^2+b^2)+\int_0^{\theta}\log\left[\sin(x-\varphi)\right]\,dx\)

trong đó

\(\displaystyle \varphi=\tan^{-1}(- a/b)=-\tan^{-1}(a/b)\)

Ta có
\(\displaystyle \int_0^{\theta}\log\left[\sin(x-\varphi)\right]\,dx=\int_0^{\theta}\log\left[2\sin(x-\varphi)\right]\,dx-\theta\log 2\)

Thay \(\displaystyle x-\varphi=y/2\,\Rightarrow\)

\(\displaystyle \frac{1}{2}\int_{-2\varphi}^{2\theta-2\varphi}\log\left(2\sin\frac{y}{2}\right)\,dy-\theta\log 2=\frac{1}{2}\left[\text{Cl}_2(-2\varphi)-\text{Cl}_2(2\theta-2\varphi)\right]-\theta\log 2\)

Mặt khác, hàm Clausen bậc 2 là hàm lẻ nên

\(\displaystyle \text{Cl}_2(-\phi)=\sum_{k=1}^{\infty}\frac{\sin (-k\phi)}{k^2}=-\sum_{k=1}^{\infty}\frac{\sin (k\phi)}{k^2}=-\text{Cl}_2(\phi)\)

Do đó

\(\displaystyle \int_0^{\theta}\log\left[\sin(x-\varphi)\right]\,dx=-\frac{1}{2}\left[\text{Cl}_2(2\varphi)+\text{Cl}_2(2\theta-2\varphi)\right]-\theta\log 2\)

Vậy

\(\displaystyle \int_0^{\theta}\log(a\cos x +b\sin x)\,dx=\)


\(\displaystyle \frac{\theta}{2}\log(a^2+b^2)-\frac{1}{2}\left[\text{Cl}_2(2\varphi)+\text{Cl}_2(2\theta-2\varphi)\right]-\theta\log 2.\,\Box\)

Bài 10:

\(\displaystyle I_0=\int_0^1\log\Gamma(x)\,dx\)

Giải:

Theo công thức Euler cho hàm Gamma ta có

\(\displaystyle \Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin\pi x}\)


Lấy logarit hai vế, rồi lấy tích phan từ 0 tới 1 ta được

\(\displaystyle \int_0^1\log\Gamma(x)\,dx+\int_0^1\log\Gamma(1-x)\,dx=\log\pi-\int_0^1\log(\sin \pi x)\,dx\)

Thay \(\displaystyle x\to 1-y\,\) tích phân thứ hai


\(\displaystyle 2\int_0^1\log\Gamma(x)\,dx=2\,I_0=\log\pi-\int_0^1\log(\sin \pi x)\,dx\)

Mặt khác

\(\displaystyle \int_0^1\log(\sin \pi x)\,dx=\int_0^1\log(2\sin \pi x)\,dx-\log 2\int_0^1\,dx=\)


\(\displaystyle \int_0^1\log(2\sin \pi x)\,dx-\log 2\)
Tiếp theo, đặt \(\displaystyle \pi x=y/2\,\) tích phân trên trở thành


\(\displaystyle \int_0^1\log(2\sin \pi x)\,dx=\frac{1}{2\pi}\int_0^{2\pi}\log\left(2\sin\frac{y}{2}\right)\,dx=\)


\(\displaystyle -\frac{1}{2\pi}\text{Cl}_2(2\pi)=-\frac{1}{2\pi}\sum_{k=1}^{\infty}\frac{\sin 2\pi k}{k^2}=0\)

Do đó

\(\displaystyle 2\int_0^1\log\Gamma(x)\,dx=\log\pi +\log 2\)
Nói cách khác
\(\displaystyle \int_0^1\log\Gamma(x)\,dx=\log\sqrt{2\pi}.\,\Box\)
Bài tập luyện tập:
1. Tính $$\displaystyle \mathbb{T}(m\,;\, n)=\int_{0}^{\pi/4}\log^n(\tan x)\log^m[-\log(\tan x)]\,dx$$

2. Tính $$\displaystyle \mathbb{T}(m\,;\, n)=\int_0^1\frac{(\log x)^n\log^m(-\log x)}{1+x^2}\,dx$$

3. Tính $$\displaystyle S_m(\theta)=\int_0^{\theta}x^m\log(\sin x)\,dx$$

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