数术杂编(六)：费曼积分法两例

$$\begin{align*} I_1 & = \int_0^\infty \frac{1-\cos x}{x(1-e^x)} dx = \frac{1}{2}\log\left(\frac{\pi}{\sinh\pi}\right) \\ I_2 & = \int_0^\infty \frac{1 - \cos x}{x(1+e^x)} dx = \frac{1}{2}\log\left(\frac{\pi}{2}\coth\frac{\pi}{2}\right) \end{align*}$$

用费曼积分法，考虑如下含参积分：

$$\begin{align*} I(a) & = \int_0^\infty \frac{dx}{1-e^x} \int_0^a \sin (xy) dy = -\int_0^a dy \int_0^\infty \sin (xy) \sum_{n=0}^{\infty} e^{-(n+1)x} dx \\ & = -\sum_{n=0}^{\infty} \int_0^a dy \int_0^\infty e^{-(n+1)x}\sin (xy) dx = -\sum_{n=0}^{\infty} \int_0^a dy \int_0^\infty \mathcal{L}^{-1} [e^{-(n+1)x}] \mathcal{L} [\sin (xy)] dx \\ & = -\sum_{n=0}^{\infty} \int_0^a dy \int_0^\infty \delta (x-(n+1))\frac{y}{x^2+y^2} dx = -\sum_{n=0}^{\infty} \int_0^a dy \int_0^\infty \delta (x-(n+1)) \frac{y}{(n+1)^2 + y^2} dx \\ & = -\sum_{n=0}^{\infty} \int_0^a \frac{y}{(n+1)^2 + y^2} dy \int_0^\infty \delta (x-(n+1)) dx = -\sum_{n=0}^{\infty} \int_0^a \frac{y}{(n+1)^2 + y^2} dy \\ & = -\sum_{n=0}^{\infty} \left[\frac{1}{2}\log ((n+1)^2 +y^2)\right]_0^a = -\frac{1}{2}\sum_{n=0}^{\infty} \log\left( 1 + \frac{a^2}{(n+1)^2} \right) \end{align*}$$

则：

$$I_1 = I(1) = -\frac{1}{2} \sum_{n=0}^{\infty} \log\left( 1 + \frac{1}{(n+1)^2} \right) = \frac{1}{2}\log\left(\prod_{n=0}^{\infty}\frac{(n+1)^2}{1+(n+1)^2}\right) = \frac{1}{2}\log\left(\frac{\pi}{\sinh\pi}\right)$$

对$I_2$的处理也是类似的：

$$\begin{align*} I & = \int_0^\infty \frac{dx}{1+e^x} \int_0^1 \sin (xy) dy = \int_0^1 dy \int_0^\infty \frac{e^{-x}\sin(xy)}{1+e^x} dx = \int_0^1 dy \int_0^\infty \sin (xy) \sum_{n=0}^{\infty} (-1)^n e^{-(n+1)x} dx \\ & = \sum_{n=0}^{\infty} (-1)^n \int_0^1 dy \int_0^\infty e^{-(n+1)x} \sin (xy) dx = \sum_{n=0}^{\infty} (-1)^n \int_0^1 dy \int_0^\infty \mathcal{L}^{-1} [e^{-(n+1)x}] \mathcal{L} [\sin(xy)] dx \\ & = \sum_{n=0}^{\infty} (-1)^n \int_0^1 dy \int_0^\infty \delta (x-(n+1)) \frac{y}{x^2 + y^2} dx = \sum_{n=0}^{\infty} (-1)^n \int_0^1 dy \int_0^\infty \delta (x-(n+1)) \frac{y}{(n+1)^2 + y^2} dx \\ & = \sum_{n=0}^{\infty} (-1)^n \int_0^1 \frac{y}{(n+1)^2 + y^2} dy \int_0^\infty \delta (x-(n+1)) dx = \sum_{n=0}^{\infty} (-1)^n \int_0^1 \frac{y}{(n+1)^2 + y^2} dy \\ & = \sum_{n=0}^{\infty} (-1)^n \left[\frac{1}{2}\log ((n+1)^2 + y^2)\right]_0^1 = \frac{1}{2}\sum_{n=0}^{\infty} (-1)^n \log\left( 1 + \frac{1}{(n+1)^2}\right) = \frac{1}{2}\log\left(\prod_{n=0}^{\infty} (-1)^n \left( 1 + \frac{1}{(n+1)^2}\right)\right) \\ & = \frac{1}{2}\log\left(\frac{\pi}{2}\coth\frac{\pi}{2}\right) \end{align*}$$