傅里叶正弦和余弦变换的导数恒等式

Problem

Show that

$$F_s[xf(x)] = -\frac{d}{ds}F_c(s).$$

Step 1: Start with the Fourier Cosine Transform

The Fourier Cosine transform of $f(x)$ is defined as

$$F_c(s) = \int_0^\infty f(x)\cos(sx)\,dx.$$

Step 2: Differentiate with Respect to $s$

Differentiating both sides with respect to $s$ gives

$$\frac{d}{ds}F_c(s) = \frac{d}{ds}\int_0^\infty f(x)\cos(sx)\,dx.$$

Differentiating inside the integral,

$$\frac{d}{ds}F_c(s) = \int_0^\infty f(x)\frac{d}{ds}\cos(sx)\,dx.$$

Since

$$\frac{d}{ds}\cos(sx) = -x\sin(sx),$$

we get

$$\frac{d}{ds}F_c(s) = \int_0^\infty f(x)(-x\sin(sx))\,dx.$$

Step 3: Identify the Fourier Sine Transform

Pulling out the negative sign,

$$\frac{d}{ds}F_c(s) = -\int_0^\infty xf(x)\sin(sx)\,dx.$$

The integral

$$\int_0^\infty xf(x)\sin(sx)\,dx$$

is the Fourier Sine transform of $xf(x)$, so

$$F_s[xf(x)] = \int_0^\infty xf(x)\sin(sx)\,dx.$$

Therefore,

$$\frac{d}{ds}F_c(s) = -F_s[xf(x)].$$

Hence,

$$F_s[xf(x)] = -\frac{d}{ds}F_c(s).$$