Show that the Sequence of $\mathrm{Si}((2m+1)\pi)_{m\in\mathbb{N}_0 }$ is monotonically decreasing

Question

Hello I am struggling in showing that the Sequence of f $\mathrm{Si}((2m+1)\pi)_{m\in\mathbb{N}_0 }$ is monotonically decreasing, where $\mathrm{Si}$ is the Sine integral defined by $\mathrm{Si}(x)=\int_{0}^{x}\frac{\sin(t)}{t}dt$. My Problem is, that i dont see how to show that \begin{align} \mathrm{Si}((2m+1)\pi)-\mathrm{Si}((2m+3)\pi)&=\int_0^{(2m+1)\pi}\frac{\sin(t)}{t}dt-\int_0^{(2m+3)\pi}\frac{\sin(t)}{t}dt>0 \end{align} which is equivalent to solving \begin{align} \int_{(2m+1)\pi}^{(2m+3)\pi}\frac{\sin(t)}{t}dt<0. \end{align} Obvioussly it holds for $m\in\mathbb{N}_0$, but i dont know how to prove it.

Answer 1

By substitution

$$\int \limits_{(2m+2)\pi}^{(2m+3)\pi} \frac{\sin(t)}{t} \, \text{d} t = \int \limits_{(2m+1)\pi}^{(2m+2)\pi} \frac{\sin(t+\pi)}{t+\pi} \, \text{d} t = \int \limits_{(2m+1)\pi}^{(2m+2)\pi} \frac{-\sin(t)}{t+\pi} \, \text{d} t$$

Since on this interval $-\sin(t) > 0$, we have that $$\frac{- \sin(t)}{t+\pi} < \frac{-\sin(t)}{t}$$ hence

$$\int \limits_{(2m+1)\pi}^{(2m+2)\pi} \frac{-\sin(t)}{t+\pi} \, \text{d} t < \int \limits_{(2m+1)\pi}^{(2m+2)\pi} \frac{-\sin(t)}{t} \, \text{d} t = -\int \limits_{(2m+1)\pi}^{(2m+2)\pi} \frac{\sin(t)}{t} \, \text{d} t$$

so finally

$$\int \limits_{(2m+1)\pi}^{(2m+3)\pi} \frac{\sin(t)}{t} \, \text{d} t = \int \limits_{(2m+2)\pi}^{(2m+3)\pi} \frac{\sin(t)}{t} \, \text{d} t + \int \limits_{(2m+1)\pi}^{(2m+2)\pi} \frac{\sin(t)}{t} \, \text{d} t < 0.$$