习题2.3

第 1 题

计算下列积分.

(1) \(\int_{0}^{+\infty}\dfrac{e^{-ax^2}-e^{-bx^2}}{x}\text{d}{x}(a, b > 0)\);

(2) \(\int_{0}^{+\infty}xe^{-ax^2}\sin yx \text{d}{x}(a > 0)\);

(3) \(\int_{0}^{+\infty}\dfrac{\cos ax - \cos bx}{x^2}\text{d}{x}(a, b > 0)\)(提示: 将 \(\cos ax - \cos bx\) 写成积分的形式, 并且 \(\int_{0}^{+\infty}\dfrac{\sin x}{x} = \dfrac{\pi}{2}\)).

解

(1) 因为 \(\dfrac{e^{-ax^2}-e^{-bx^2}}{x} = x\int_a^be^{-x^2y}\text{d}{y}\), 故

\[ \begin{aligned} \int_{0}^{+\infty}\dfrac{e^{-ax^2}-e^{-bx^2}}{x}\text{d}{x} & = \int_{0}^{+\infty}x\text{d}{x}\int_a^be^{-x^2y}\text{d}{y} \\ & = \dfrac{1}{2}\int_{0}^{+\infty}\text{d}{x^2}\int_a^be^{-x^2y}\text{d}{y} \\ & = \dfrac{1}{2}\text{d}{\theta}\int_a^be^{-\theta y}\text{d}{y} \\ & = \dfrac{1}{2}\int_a^b\dfrac{1}{y}\text{d}{y} \\ & = \dfrac{1}{2}\ln\dfrac{b}{a}. \end{aligned} \]

(2)

\[ \begin{aligned} \int_{0}^{+\infty}xe^{-ax^2}\sin yx \text{d}{x} & = -\dfrac{1}{2a}\left(e^{-ax^2}\sin(yx)\bigg\vert^{+\infty}_{0} - y\int_{0}^{+\infty}e^{-ax^2}\cos(yx)\text{d}{x}\right) \\ & = \dfrac{y}{2a}\int_{0}^{+\infty}e^{-ax^2}\cos(yx)\text{d}{x} \\ & = \dfrac{y}{4a}\sqrt{\dfrac{\pi}{a}}e^{-\frac{y^2}{4a}} \end{aligned} \]

(3) 由于 \(\cos ax - \cos bx = x\int_a^b\sin yx\text{d}{y}\), 故

\[ \begin{aligned} \int_{0}^{+\infty}\dfrac{\cos ax - \cos bx}{x^2}\text{d}{x} & = \int_0^{+\infty}\dfrac{1}{x}\text{d}{x}\int_a^b\sin(yx)\text{d}{y} \\ & = \int_{0}^{+\infty}\dfrac{\sin yx}{yx}\text{d}{(yx)}\int_a^b\text{d}{y} \\ & = \dfrac{\pi}{2}(b - a). \end{aligned} \]

第 2 题

利用对参变量的求导, 求下列积分.

(1) \(\int_{0}^{+\infty}e^{-tx^2}x^{2n}\text{d}{x}(t > 0)\)(提示: 利用 \(\int_{0}^{+\infty}e^{-x^2}\text{d}{x} = \dfrac{\sqrt{\pi}}{2}\));

(2) \(\int_{0}^{+\infty}\dfrac{\text{d}{x}}{(y + x^2)^{n+1}} = \dfrac{\pi(2n - 1)!!}{2(2n)!!}y^{-\left(n + \frac{1}{2}\right)}(y > 0)\)(提示: 利用 \(\int_{0}^{+\infty}\dfrac{\text{d}{x}}{y + x^2}\) 的值).

解

(1) 考虑 \(I(t) = \int_{0}^{+\infty}e^{-tx^2}\text{d}{x}\), 则其 \(n\) 阶导数为 \(I^{(n)}(t) = \int_{0}^{+\infty}e^{-tx^2}x^{2n}\text{d}{x}\) 即为所求. 注意到

\[ \begin{aligned} I(t) & = \int_{0}^{+\infty}e^{-tx^2}\text{d}{x} \\ & = \dfrac{1}{\sqrt{t}}\int_{0}^{+\infty}e^{-(\sqrt{t}x)^2}\text{d}{(\sqrt{t}x)} \\ & = \dfrac{\sqrt{\pi}}{2}\cdot t^{-\frac{1}{2}} \end{aligned} \]

因此有 \(\int_{0}^{+\infty}e^{-tx^2}x^{2n}\text{d}{x} = I^{(n)}(t) = \dfrac{\sqrt{\pi}}{2}\cdot\dfrac{(2n-1)!!}{2^n}(-1)^nt^{-\left(n+\frac{1}{2}\right)}\).

(2) 考虑 \(I(y) = \int_{0}^{+\infty}\dfrac{\text{d}{x}}{y + x^2}\), 则 \(\dfrac{(-1)^n}{n!}I^{(n)}(y) = \int_{0}^{+\infty}\dfrac{\text{d}{x}}{(y + x^2)^{n+1}}\) 即为所求. 由于

\[ \begin{aligned} I(y) & = \int_{0}^{+\infty}\dfrac{\text{d}{x}}{y + x^2} \\ & = \int_{0}^{+\infty}\dfrac{\sqrt{y}\text{d}{\left(\dfrac{x}{\sqrt{y}}\right)}}{y\left(1 + \left(\dfrac{x}{\sqrt{y}}\right)^2\right)} \\ & = y^{-\frac{1}{2}}\arctan\left(\dfrac{x}{\sqrt{y}}\right)\bigg\vert_{x = 0}^{x = +\infty} \\ & = \dfrac{\pi}{2}y^{-\frac{1}{2}} \end{aligned} \]

因此 \(I^{(n)}(y) = \dfrac{\pi}{2}\cdot\dfrac{(2n-1)!!}{2^n}(-1)^ny^{-\left(n + \frac{1}{2}\right)}\). 故 \(\int_{0}^{+\infty}\dfrac{\text{d}{x}}{(y + x^2)^{n+1}} = \dfrac{(-1)^n}{n!}I^{(n)}(y) = \dfrac{\pi(2n - 1)!!}{2(2n)!!}y^{-\left(n + \frac{1}{2}\right)}\).