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How can I prove this: if $x \ge 0, $ then $$\int_{0}^{x} \frac{\sin(t)}{t+1} \, dt \ge 0 $$

I attempted this problem using monotony of integral, but I didn't get anything really useful.

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  • $\begingroup$ Just some intuition... When $0 \leq x \leq \pi$, $\frac{\sin t}{t+1} dt \get$ for $t \in [0,x]$ and the result is immediate. For larger values of $x$, note that the division by $t+1$ makes the negatives values always smaller than the positive values that precede them. $\endgroup$ – PierreCarre Apr 17 at 20:55
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Here's an idea:

The monotony of the integral suffices to show that $$\int_0^x\frac{\sin(t)}{t+1}\mathrm{d}t\ge0\qquad\forall0\le x\le\pi$$ Next, use the substitution $t\rightarrow t+\pi$ to show that $$-\int_{\pi}^{2\pi}\frac{\sin(t)}{t+1}\mathrm{d}t=\int_0^{\pi}\frac{\sin(t)}{t+\pi+1}\mathrm{d}t\le\int_0^{\pi}\frac{\sin(t)}{t+1}\mathrm{d}t.$$ Monotony again shows that $$\int_{\pi}^x\frac{\sin(t)}{t+1}\mathrm{d}t\ge\int_{\pi}^{2\pi}\frac{\sin(t)}{t+1}\mathrm{d}t,\text{ hence }\int_0^x\frac{\sin(t)}{t+1}\mathrm{d}t\ge0\qquad\forall0\le x\le2\pi$$ Can you use the same idea to show that $\int_{2k\pi}^{x}\sin(t)/(t+1)\mathrm{d}t\ge0$ for all $2k\pi\le x\le 2(k+1)\pi$, where $k\in\mathbb{N}_0$ is arbitrary and then conclude the general inequality by summing?

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We can write the integral as $$\int_0^x\frac{\sin{(t)}}{t+1}dt=\int_0^\pi\frac{\sin{(t)}}{t+1}dt+\int_\pi^{2\pi}\frac{\sin{(t)}}{t+1}dt+\dots+\int_{2a\pi}^x\frac{\sin{(t)}}{t+1}dt$$ where $2a\pi\le x$ for some $a\in\mathbb{N}$. Then, the following inequality follows from the fact that $\frac1{t+1}$ is strictly decreasing $$\left|\int_{n\pi}^{(n+1)\pi}\frac{\sin{(t)}}{t+1}dt\right|\gt\left|\int_{(n+1)\pi}^{(n+2)\pi}\frac{\sin{(t)}}{t+1}dt\right|$$ so we must have that $$\left|\int_{n\pi}^{(n+1)\pi}\frac{\sin{(t)}}{t+1}dt\right|-\left|\int_{(n+1)\pi}^{(n+2)\pi}\frac{\sin{(t)}}{t+1}dt\right|\gt0$$ Now taking $n=2k$ for some $k\in\mathbb{N}$ we have that $$\int_{2k\pi}^{(2k+1)\pi}\frac{\sin{(t)}}{t+1}dt\gt0$$ $$\int_{(2k+1)\pi}^{(2k+2)\pi}\frac{\sin{(t)}}{t+1}dt\lt0$$ so the inequality becomes $$\int_{2k\pi}^{(2k+1)\pi}\frac{\sin{(t)}}{t+1}dt+\int_{(2k+1)\pi}^{(2k+2)\pi}\frac{\sin{(t)}}{t+1}dt\gt0$$ for all $k\in\mathbb{N}$. Thus the original integral simplifies down to $$\int_0^x\frac{\sin{(t)}}{t+1}dt=b+\int_{2a\pi}^x\frac{\sin{(t)}}{t+1}dt$$ where $b\gt0$ for some $b\in\mathbb{R}$. We can finally conclude that for $x\in[(2a+1)\pi,(2a+2)\pi)$ $$\int_{2a\pi}^x\frac{\sin{(t)}}{t+1}dt\ge\int_{2a\pi}^{(2a+1)\pi}\frac{\sin{(t)}}{t+1}dt+\int_{(2a+1)\pi}^{(2a+2)\pi}\frac{\sin{(t)}}{t+1}dt\gt0$$ and for $x\in[2a\pi,(2a+1)\pi)$ we have that $$\frac{\sin{(t)}}{t+1}\ge0\implies\int_{2a\pi}^x\frac{\sin{(t)}}{t+1}dt\ge0$$ hence the original integral is positive.

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My idea is to fold $[0, 2\pi]$ into $[0,\pi]$ and then fold that into $[0,\pi/2]$ and show that the result is positive everywhere.

It seems to work.

Here are the details.

$\begin{array}\\ S(2\pi n) &=\int_{2\pi n}^{2\pi (n+1)} \frac{\sin(t)}{t+1}dt\\ &=\int_{0}^{2\pi} \frac{\sin(t+2\pi n)}{t+2\pi n+1}dt\\ &=\int_{0}^{\pi} \frac{\sin(t+2\pi n)}{t+2\pi n+1}dt+\int_{\pi}^{2\pi} \frac{\sin(t+2\pi n)}{t+2\pi n+1}dt\\ &=\int_{0}^{\pi} \frac{\sin(t+2\pi n)}{t+2\pi n+1}dt-\int_{0}^{\pi} \frac{\sin(t+2\pi n)}{t+2\pi n+1+\pi}dt\\ &=\int_{0}^{\pi} \sin(t+2\pi n)\left(\frac1{t+2\pi n+1}-\frac1{t+2\pi n+1+\pi}\right)dt\\ &=\int_{0}^{\pi} \frac{\sin(t+2\pi n)}{(t+2\pi n+1)(t+2\pi n+1+\pi)}dt\\ &=\int_{0}^{\pi/2} \frac{\sin(t+2\pi n)}{(t+2\pi n+1)(t+2\pi n+1+\pi)}dt+\int_{\pi/2}^{\pi} \frac{\sin(t+2\pi n)}{(t+2\pi n+1)(t+2\pi n+1+\pi)}dt\\ &=\int_{0}^{\pi/2} \frac{\sin(t+2\pi n)}{(t+2\pi n+1)(t+2\pi n+1+\pi)}dt +\int_{0}^{\pi/2} \frac{\sin(t+\pi/2+2\pi n)}{(t+2\pi n+1+\pi/2)(t+2\pi n+1+3\pi/2)}dt\\ &=\int_{0}^{\pi/2} \frac{\sin(t+2\pi n)}{(t+2\pi n+1)(t+2\pi n+1+\pi)}dt +\int_{0}^{\pi/2} \frac{\sin(\pi/2-t+\pi/2+2\pi n)}{(\pi/2-t+2\pi n+1+\pi/2)(\pi/2-t+2\pi n+1+3\pi/2)}dt\\ &=\int_{0}^{\pi/2} \frac{\sin(t+2\pi n)}{(t+2\pi n+1)(t+2\pi n+1+\pi)}dt -\int_{0}^{\pi/2} \frac{\sin(t+\pi/2+2\pi n)}{(\pi/2-t+2\pi n+1+\pi/2)(\pi/2-t+2\pi n+1+3\pi/2)}dt\\ &=\int_{0}^{\pi/2} \sin(t+2\pi n)\left(\frac1{(t+2\pi n+1)(t+2\pi n+1+\pi)} - \frac1{(\pi/2-t+2\pi n+1+\pi/2)(\pi/2-t+2\pi n+1+3\pi/2)}\right)dt\\ &=\int_{0}^{\pi/2} \sin(t+2\pi n)\frac{2 (2 π n + π + 1) (π - 2 t)}{((2 π n - t + π + 1) (2 π n - t + 2 π + 1) (2 π n + t + 1) (2 π n + t + π + 1))}dt\\ &\qquad\text{(according to Wolfy)}\\ &\gt 0 \qquad\text{since } \pi > 2t\\ \end{array} $

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There is an n such that $x - n\pi / 2 \leq \pi / 2$. Subdivide the interval $[0, x]$ into subintervals $[i \pi / 2, (i + 1) \pi / 2]$ and show that the integral over every second interval $[(2j + 1) \pi / 2, (2j + 2) \pi / 2]$ is smaller than the integral over the previous one. Then, try an induction over n.

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This really begins to be challenging once $x$ exceeds $\pi$. So, let's first examine the case $x = 2 \pi$: $$ \int_{0}^{2\pi} {\sin(t) \over 1 + t} dt = \int_{0}^{\pi} {\sin(t) \over 1 + t} dt + \int_{\pi}^{2\pi} {\sin(t) \over 1 + t} dt. $$ We want to examine the absolute values of the integrands in the last two integrals; i.e., the quantity $$ \left| {\sin(t) \over 1 + t} \right| $$ on each of the intervals $[0, \pi]$ and $[\pi, 2\pi]$.

The function $$ \phi(t) = {1 \over 1 + t} $$ is strictly decreasing for $t > 0$, so: $$ \left|{1 \over 1 + t}\right| < \left|{1 \over 1 + t + \pi}\right| \quad \mbox{ for all $t \in [0, \pi]$}. $$ And, as $|\sin(t)| = |\sin(t + \pi)|$, we have: $$ \left|{\sin(t) \over 1 + t}\right| < \left|{\sin(t) \over 1 + t + \pi}\right| \quad \mbox{ for all $t \in [0, \pi]$}. $$ Therefore, $$ \int_{0}^{\pi} {\sin(t) \over 1 + t} dt + \int_{\pi}^{2\pi} {\sin(t) \over 1 + t} dt = \int_{0}^{\pi} \left|{\sin(t) \over 1 + t}\right| dt - \int_{0}^{\pi} \left|{\sin(t + \pi) \over 1 + t + \pi}\right| dt \geq 0. $$

For a general $x$, you just want the highest integer $k_{x}$ such that $$ 2 \pi k_{x} \leq x: $$ the above reasoning can be used to show that $$ \int_{0}^{2 \pi k_{x}} {\sin(t) \over 1 + t} dt \geq 0, $$ so you just want to examine the sign of $$ \int_{2 \pi k_{x}}^{x} {\sin(t) \over 1 + t} dt. $$

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First off, lets assume that x is in radians.

The issue here is that $sin(t)$ is a periodic function, and so will oscillate between 1 and -1 as t goes from 0 to x. This means that you will be integrating over positive and negative values, so its not clear immediately that $\int_0^x sin(t)/(t+1)dt \ge 0$

However you have three important facts you can use (I don't give a full proof here)

  1. $sin(t)$ is periodic which means you know when its behavior repeats
  2. The denominator increases monotonically
  3. you can break up integrals in to sums of intergals over their domain e.g. $\int_a^b f(x)dx = \int_a^c f(x)dx + \int_c^b f(x)dx $ where $c$ in $[a,b]$

Putting these together, you know that $sin(t)/(t+1)$ is positive from $(0,\pi)$ and negative from $(\pi,2\pi)$ ( more generally, positive on $(2n\pi,(2n+1)\pi)$ and negative on $((2n+1)\pi,2n\pi)$ for $ n=0,1,2,...$). However, the absolute value $sin(t)/(t+1)$ will decrease across these intervals.

This means that, $\int_{2n\pi}^{(2n+1)\pi} sin(t)/(t+1)dt$ will both be positive and larger in aboslute value than $\int_{(2n+1)\pi}^{2n\pi} sin(t)/(t+1)dt$ so that their sum will be positive. Thus over any interval of $2\pi$ (a full period) the integral will be positive. (you'll need to prove this part)

If you break up $[0,x]$ into intervals $2\pi$ you'll see that integral will never be negative.

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