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I devised this proof that

$$\tag{1} \int_0^x \frac{\sin t}{t+1}dt > 0 \text{ ; } \forall x >0$$

The idea is to prove that the area from $(0,\pi)$ is greater than the absolute value of the negative area in $(\pi, 2\pi)$, and so on, so that the final area is always positive.

$f(x) = \dfrac{\sin x}{x+1}$ is positive if $\sin x >0$ and negative if $\sin x <0$. This is to say

$$f(x) >0 \Leftrightarrow x \in \bigcup_{k=0}^{\infty}(2k\pi,(2k+1)\pi)$$

$$f(x) <0 \Leftrightarrow x \in \bigcup_{k=1}^{\infty}((2k-1)\pi,2k\pi)$$

If we prove that $$\tag{2} |f(x)| > |f(x+\pi)|$$ for every $x$ then we prove $(1)$.

But,

$|f(x)| =\left| \dfrac{\sin x}{x+1} \right|$

$|f(x+\pi)| =\left| \dfrac{\sin (x+\pi)}{x+\pi+1} \right|=\left| \dfrac{\sin x}{x+\pi+1} \right|$

Thus $(2)$ is proven, and we then have that in general,

$$ |f(x+n \pi)| > |f(x+(n+1) \pi)|$$, thus

$$\int\limits_{\left( {2k} \right)\pi }^{\left( {2k + 1} \right)\pi } {\frac{{\sin t}}{{t + 1}}dt} + \int\limits_{\left( {2k + 1} \right)\pi }^{\left( {2k + 2} \right)\pi } {\frac{{\sin t}}{{t + 1}}dt} > 0$$

and then

$$ \int_0^x \frac{\sin t}{t+1}dt > 0 \text{ ; } \forall x >0$$

Is it right? And if it is right - is it understandable?

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That should be $\geq$. ;-) Take $x = 0$... –  WimC Feb 26 '12 at 19:09
    
The idea seems quite solid, but you might want to mention induction somewhere in there, since you are inducting on $k$. –  Alex Becker Feb 26 '12 at 19:12
    
@WimC Problem states "$x>0$". –  Alex Becker Feb 26 '12 at 19:12
    
@AlexBecker: sorry, I meant in $|f(x)| > |f(x+\pi)|$. –  WimC Feb 26 '12 at 19:15
    
@WimC Ah yes, that's true. But it's only on a discrete set, so ignorable for integrals. Suppose that requires proof though. –  Alex Becker Feb 26 '12 at 19:17

2 Answers 2

up vote 4 down vote accepted

You're on the right track.. you also have to show that ${\displaystyle \int_{2k\pi}^x {\sin(t) \over t + 1}\,dt > 0}$ for all $2k\pi < x < 2(k+1)\pi$, since you have $$ \int_{0}^x {\sin(t) \over t + 1}\,dt = \sum_{i = 0}^{k-1} \bigg(\int_{2i\pi}^{(2i + 1)\pi} {\sin(t) \over t + 1}\,dt + \int_{(2i+1)\pi}^{(2i + 2)\pi} {\sin(t) \over t + 1}\,dt\bigg) + \int_{2k\pi}^x {\sin(t) \over t + 1}\,dt $$ You've shown the first sum is positive, but you still have to show the last term is positive too. For that part, I suggest showing that ${\displaystyle \int_{2k\pi}^x {\sin(t) \over t + 1}\,dt}$ increases as $x$ goes from $2k\pi$ to $(2k + 1)\pi$, and then decreases as $x$ goes from $(2k + 1)\pi$ to $(2k + 2)\pi$. From what you've done already, you know it will not decrease all the way to zero.


And now for the "slick trick" solution: Note that the derivative of $1 - \cos(x)$ is $\sin(x)$, and the derivative of ${\displaystyle {1 \over t + 1}}$ is ${\displaystyle -{1 \over (t + 1)^2}}$. So integrating by parts you have $$\int_{0}^x {\sin(t) \over t + 1}\,dt = {1 - \cos(t) \over t + 1}\bigg|_{t = 0}^{t = x} + \int_0^x {1 - \cos(t) \over (t + 1)^2}\,dt$$ $$= {1 - \cos(x) \over x + 1} + \int_0^x {1 - \cos(t) \over (t + 1)^2}\,dt$$ Since $1 - \cos(t) \geq 0$ for all $t$, the first term is nonnegative. Similarly, the integrand of the second term is nonnegative and thus the resulting integral is positive for $x > 0$.

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Yeah! That's great. Thanks. For your first remark, notice that the last term containing $x$ won't be "important" since if it is positive, we're ok, and if it is negative, it won't change the sign of the preceding positive sum, since we have shown it is not greater than it's preceding term. –  Pedro Tamaroff Feb 26 '12 at 19:25

The idea and the execution are fine. I would prefer a small variant. We want to prove that the integral from $2k\pi$ to $(2k+2)\pi$ is positive. Look at the integral from $(2k+1)\pi$ to $(2k+2)\pi$. The first half of the interval is fine. For the second half, make the change of variable $t=s+\pi$. Note that $\sin t=-\sin s$. It follows that $$\int_{t=(2k+1)\pi}^{(2k+2)\pi} \frac{\sin t}{t+1}dt=-\int_{s=2\pi}^{(2k+1)\pi}\frac{\sin s}{s+\pi+1}ds.$$

Now that the dummy variable $s$ has done its duty, replace it by $t$. Then our integral from $2k\pi$ to $(2k+2)\pi$ is equal to $$\int_{t=2k\pi}^{(2k+1)\pi} \left(\frac{\sin t}{t+1}-\frac{\sin t}{t+\pi +1}\right)dt.$$ If we wish, we can rewrite this as $$\int_{t=2k\pi}^{(2k+1)\pi} \frac{\pi \sin t}{(t+1)(t+\pi +1)}dt.$$ The integrand is non-negative, and the result follows.

One reason I prefer this version of your argument is that now it is easy to get a reasonable estimate of the integral from $2k\pi$ to $(2k+2)\pi$.

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