ง่ายกว่าจริงๆด้วย - -a ข้อ 254. แค่คูณด้วย conjugate แล้วก็แยกอินทิเกรต ทีละตัวก็ได้ละ
มันจะได้ $$\frac{1}{2}(\int_{-1}^{1} \frac{dx}{x^2+1}+\int_{-1}^{1} \frac{dx}{x}) - \frac{1}{2}\int_{-1}^{1}\frac{\sqrt{x^4+3x^2+1}}{x(x^2+1)} \, dx$$
ให้ $$I = \int_{-1}^{1} \frac{dx}{x}$$
$$I = \lim_{a \to 0^{-}} \int_{-1}^{a} \frac{dx}{x} + \lim_{b \to 0^{+}} \int_{b}^{1} \frac{dx}{x}$$
$$= \lim_{a \to 0^{-}} \ln{\left| a \,\right| } - \ln{\left| -1 \,\right| } + \ln{\left| 1 \,\right| } - \lim_{b \to 0^{+}}\ln{\left| b \,\right| } = 0$$
ให้ $J = \int_{-1}^{1}\frac{\sqrt{x^4+3x^2+1}}{x(x^2+1)} \, dx$
จัดรูปใหม่ จะได้
$$\int_{-1}^{1} \frac{\sqrt{(x+\frac{1}{x})^2+1}}{x(x+\frac{1}{x})} \, dx$$
ให้ $u = \frac{1}{x}$ ---> $-xdu = \frac{dx}{x}$
ดังนั้น $$J = -\int_{-1}^{1}\frac{x\sqrt{(u+\frac{1}{u})^2+1}}{u+\frac{1}{u}} \, du$$
$$= -\int_{-1}^{1}\frac{\sqrt{(u+\frac{1}{u})^2+1}}{u(u+\frac{1}{u})} \, du = -J$$
$$J = -J$$
$$J = 0$$
ดังนั้น $$\frac{1}{2}(\int_{-1}^{1} \frac{dx}{x^2+1}+\int_{-1}^{1} \frac{dx}{x}) - \frac{1}{2}\int_{-1}^{1}\frac{\sqrt{x^4+3x^2+1}}{x(x^2+1)} \, dx = \frac{1}{2}\int_{-1}^{1}\frac{dx}{x^2+1} = \frac{\pi}{4}$$
ส่วนข้อ 253
Conjugate ตัวล่าง จะได้
$$\int_{-1}^{1} (\frac{\sqrt{x^2+1}-1}{x})^3 \, dx$$
$f(x) = -f(-x)$
ค่าอินทิกรัลจึงเท่ากับ $0$
ข้อ 251. ครับ
Let $u=x^n$ ---> $nx^ndx = \sqrt[n]{u}du$
Hence, $$\lim_{n \to \infty} \int_{0}^{1} \frac{nx^n}{a+x^n} \, dx = \lim_{n \to \infty} \int_{0}^{1} \frac{\sqrt[n]{u}}{a+u} \, du$$
$$= \int_{0}^{1} \frac{u^{\lim_{n \to \infty} \frac{1}{n}}}{a+u} \, du$$
$$= \int_{0}^{1} \frac{du}{a+u} = \ln{(\frac{a+1}{a})}$$
ข้อ 59. นะครับ
Consider $2\cos{nx} = \cos{((n-1)+1)x} + \cos{((n+1)-1)x}$
$= \cos{(n-1)x}\cos{x} - \sin{(n-1)x}\sin{x} + \cos{(n+1)x}\cos{x}+\sin{(n+1)x}\sin{x}$
$= \cos{(n-1)x}\cos{x}+\cos{(n+1)x}\cos{x} +\sin{x}(\sin{(n-1)x}-\sin{(n+1)x})$
$= \cos{(n-1)x}\cos{x}+\cos{(n+1)x}\cos{x} -\sin{x}(2\cos{nx}\sin{x})$
$= \cos{(n-1)x}\cos{x}+\cos{(n+1)x}\cos{x} -2\sin^2{x}\cos{nx}$
then $2+2\cos{x}-\cos{(n-1)x}-2\cos{nx}-\cos{(n+1)x} = (-(\cos{(n+1)x}\cos{x}+\cos{(n+1)x}) + \cos{x} +1) + (-(\cos{(n-1)x}\cos{x}+\cos{(n-1)x})+\cos{x}+1) -2\sin^2{x}\cos{nx}$
$= (1+\cos{x})(2-\cos{(n+1)x}-\cos{(n-1)x}) - 2\sin^2{x}\cos{nx}$
$= 2(1+\cos{x})(1-\cos{nx}\cos{x}) - 2\sin^2{x}\cos{nx}$
$= 2(1+\cos{x}-\cos{nx}\cos{x}-\cos{nx}\cos^2{x}-\cos{nx}\sin^2{x})$
$= 2(1+\cos{x}-\cos{nx}\cos{x}-\cos{nx}) = 2(1-\cos{nx})(1+\cos{x})$
and $1-\cos{2x} = 2\sin^2{x} = 2(1-\cos{x})(1+\cos{x})$
hence
$$\int_{0}^{\pi} \frac{2+2\cos{x}-\cos{(n-1)x}-2\cos{nx}-\cos{(n+1)x}}{1-\cos{2x}}\, dx = \int_{0}^{\pi} \frac{1-\cos{nx}}{1-\cos{x}} \, dx$$
$$= \int_{0}^{\pi}\int_{0}^{x} \frac{n\sin{ny}}{1-\cos{x}} \, dydx$$
Apply Fubini's Theorem
$$= \int_{0}^{\pi}\int_{y}^{\pi} \frac{n\sin{ny}}{1-\cos{x}} \, dxdy$$
$$= n\int_{0}^{\pi} \sin{ny}\left[ -\cot{\frac{x}{2}} \,\right]_{y}^{\pi} \, dy $$
$$= n\int_{0}^{\pi} \frac{\sin{ny}\cos{\frac{y}{2}}}{\sin{\frac{y}{2}}} \, dy$$
$$= 2n\int_{0}^{\pi} \frac{\sin{ny}\cos^2{\frac{y}{2}}}{\sin{y}}\, dy$$
from my reply here :
http://www.mathcenter.net/forum/show...&postcount=302
$$= 2n\int_{0}^{\pi} \frac{\sin{ny}\cos^2{\frac{y}{2}}}{\sin{y}}\, dy = 2n
\int_{0}^{\pi} \frac{\sin{(n-2)y}\cos^2{\frac{y}{2}}}{\sin{y}} + 2\cos{(n+1)y}\cos^2{\frac{y}{2}} \, dy$$
For second integral, it's always be zero, considering only the first
$$= 2n\int_{0}^{\pi} \frac{\sin{(n-2)y}\cos^2{\frac{y}{2}}}{\sin{y}} \, dy$$
CASE 1 $n$ is odd
$$= 2n\int_{0}^{\pi}\frac{\sin{(n-2)y}\cos^2{\frac{y}{2}}}{\sin{y}} \, dy$$
$$= 2n\int_{0}^{\pi}\cos^2{\frac{y}{2}} \, dy = n\int_{0}^{\pi} \cos{y} + 1 \, dy$$
$$= n\left[ \sin{y}+y \,\right]_{0}^{\pi} = n\pi$$
CASE 2 $n$ is even
$$= 2n\int_{0}^{\pi}\frac{\cos^2{\frac{y}{2}}\sin{2y}}{\sin{y}} \, dy$$
$$= 2n\int_{0}^{\pi} 2\cos^2{\frac{y}{2}}\cos{y} \, dy$$
$$= 2n\int_{0}^{\pi} (\cos{y} - 1)\cos{y} \, dy$$
$$= 2n\int_{0}^{\pi} \cos^2{y} - \cos{y} \, dy$$
$$= 2n\left[ \frac{1}{2}\sin{y}\cos{y} + \frac{1}{2}y - \sin{y} \,\right]_{0}^{\pi} = n\pi $$