# Prove that a definite integral is positive

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.

• 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. – PierreCarre Apr 17 at 20:55

## 6 Answers

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?

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.

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.

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

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.

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 $$\int_{2 \pi k_{x}}^{x} {\sin(t) \over 1 + t} dt.$$

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.