第5章 5.38

(1)

$\int_0^\infty \frac{dx}{e^x+e^{-x}}$

$e^x = t$ とおくと $dx = \dfrac{1}{t}dt$ $x:0\to\infty$ $t:1\to\infty$ これから

$= \int_1^\infty \frac{1}{t+\dfrac{1}{t}}\cdot\frac{1}{t}\,dt = \int_1^\infty \frac{1}{1+t^2}\,dt = \left[\tan^{-1}t\right]_1^\infty = \frac{\pi}{2}-\frac{\pi}{4} = \frac{\pi}{4}$

(2)

$\int_0^1 x\log x\,dx = \left[\frac{1}{2}x^2\log x\right]_0^1 - \frac{1}{2}\int_0^1 x\,dx = -\frac{1}{2}\left[\frac{1}{2}x^2\right]_0^1 = -\frac{1}{4}$

(3)

$\int_1^\infty \frac{\log x}{x^2}\,dx = \left[-\frac{1}{x}\log x\right]_1^\infty + \int_1^\infty \frac{1}{x^2}\,dx = \left[-\frac{1}{x}\right]_1^\infty = 1$

(4)

$\int_0^\infty \frac{\tan^{-1}x}{1+x^2}\,dx$

$\tan^{-1}x = t$ とおくと $\dfrac{1}{1+x^2}dx = dt$ $x:0\to\infty$ $t:0\to\dfrac{\pi}{2}$ これから

$= \int_0^{\frac{\pi}{2}} t\,dt = \left[\frac{1}{2}t^2\right]_0^{\frac{\pi}{2}} = \frac{\pi^2}{8}$

(5)

$\int_0^\infty x^2e^{-x}\,dx = \left[-x^2e^{-x}\right]_0^\infty + \int_0^\infty 2xe^{-x}\,dx = \left[-2xe^{-x}\right]_0^\infty + \int_0^\infty 2e^{-x}\,dx = 2\left[-e^{-x}\right]_0^\infty = 2$

(6)

$\int_0^1 \frac{\log(1+x^2)}{x^2}\,dx = \left[-\frac{1}{x}\log(1+x^2)\right]_0^1 + \int_0^1 \frac{2x}{x(x^2+1)}\,dx = -\log 2 + \int_0^1 \frac{2}{x^2+1}\,dx$

$= -\log 2 + 2\left[\tan^{-1}x\right]_0^1 = -\log 2 + \frac{\pi}{2}$

(7)

$\int_0^1\sqrt{\dfrac{x}{1-x}}dx$

$\sqrt{\dfrac{x}{1-x}}=t$ とおくと $dx=\dfrac{2t}{(1+t^2)^2}dt$ $x:0\to1$ $t:0\to\infty$ これから

$=\int_0^\infty t\cdot\dfrac{2t}{(1+t^2)^2}dt=\int_0^\infty\dfrac{2t^2}{(1+t^2)^2}dt=\int_0^\infty\left\{\dfrac{-2}{(1+t^2)^2}+\dfrac{2}{1+t^2}\right\}dt$

$=-2\left[-\dfrac{1}{2}\left(\dfrac{t}{t^2+1}+\tan^{-1}t\right)\right]_0^\infty+\left[2\tan^{-1}t\right]_0^\infty=-\dfrac{\pi}{2}+\pi=\dfrac{\pi}{2}$

(8)

$\int_0^2\dfrac{x}{\sqrt{2x-x^2}}dx=\int_0^2\dfrac{x}{\sqrt{1-(x-1)^2}}dx$

$x-1=t$ とおくと $dx=dt$ $x:0\to2$ $t:-1\to1$ これから

$=\int_{-1}^1\left\{\dfrac{t}{\sqrt{1-t^2}}+\dfrac{1}{\sqrt{1-t^2}}\right\}dt=\left[-\sqrt{1-t^2}\right]_{-1}^1+\left[\sin^{-1}t\right]_{-1}^1=\dfrac{\pi}{2}-\left(-\dfrac{\pi}{2}\right)=\pi$