Proving a function defined as an integral is greater than 0

Question

fellow mathematicians. I was given the following exercise:

Given $x>0$, prove that $F(x)= \int_{0}^{x}{\frac{\sin(t)}{1+t}dt} > 0$

What I am trying to do is the following:

We can obtain the derivative by the fundamental theorem of calculus because $f(x)$ is continuous when $t\geq0$. Then:

$$ F'(x)=\frac{\sin(x)}{1+x} $$

Now I will find the maxima and minima of the function. We know that $F'(x)=0$ when $\sin(x) = 0$ and that happens in $0, \pi, 2\pi, 3\pi...$ i.e, in $k\pi$ when $k\in\mathbb{N}$. Moreover, $kx$ is a minimizer when $k$ is even. I claim that because:

$F''(x)=\frac{\cos(x)(1+x)-\sin(x)}{(1+x)^{2}}$ and

$F''(2\pi)=\frac{\cos(2\pi)(1+2\pi)-\sin(2\pi)}{(1+2\pi)^{2}} = 1 + 2\pi > 0 $

So $F(x)$ has a minimum in $x=2\pi$ and I will show that the minimum is greater than $0$

$$ F(2\pi)=\int_{0}^{2\pi}{\frac{\sin{t}}{1+t}} $$

And now I'm stuck. I tried to modif the expression using the fact that:

$$ \int_{ac}^{bc}{f(x)dx}=c\int_{a}^{b}{f(cx)dx} $$

But that did not work. Any suggestion? Thanks!

Answer 1

The sine function is periodic, with identical arches alternating in sign. Thanks to the increasing denominator, every positive contribution to the integral is followed by a negative one of smaller value, hence the integral does not return to zero.

You can prove it formally using that

$$\frac1{t+2k\pi+1}>\frac1{t+2k\pi+\pi+1}.$$

Answer 2

$\int_0^\pi \frac {\sin t}{1+t} \ dt > 0$ because $\frac {\sin t}{1+t} > 0$ everywhere in the interval.

$\int_{\pi}^{2\pi} \frac {\sin (t)}{1+t} \ dt = \int_{0}^{\pi} \frac {-\sin (u)}{1+\pi+t} \ du$

And

$|\int_{0}^{\pi} \frac {-\sin (t)}{1+\pi+t} \ du| < \int_0^\pi \frac {\sin t}{1+t} \ dt$ because $|\frac {-\sin (t)}{1+\pi+t}| < |\frac {\sin (t)}{1+t}|$ at every value of $t$ in the interval.

$F(2\pi) > 0$

And by induction:

$\int_{n\pi}^{(n+1)\pi} \frac {\sin t}{1+t} \ dt = \int_{(n-1)\pi}^{n\pi} \frac {\sin (u+\pi)}{1+\pi+u} \ du = -\int_{(n-1)\pi}^{n\pi} \frac {\sin (u)}{1+\pi+u} \ du$

We can extend this out for all $n.$

Answer 3

Good evening,

There are several ways to achieve this proof, but as you asked, I will mostly focus to complete your idea of proof. All you did seem correct, you just need to show eventually that : $$ F(2 \pi) = \int_0^{2\pi} \frac{ \sin (t) }{1+t} \mathrm d t > 0 $$

As you can observe on the graph, the function $ f : t \mapsto \frac{ \sin (t) }{1+t} $ has some kind of decreasing periodicity, and this will be used to conclude on your proof. In orange is the absolute value of the function $f$ over $[\pi, 2\pi]$

So, it can be easily proven that, $$\forall x \in ]0,\pi[, |f(x+\pi)| < f(x) $$

since the sinus part will be equal with the absolute value and the denominator will be higher. Therefore, we can integrate on both sides to get : $$ \int_0^{\pi} \left| \frac{ \sin (t+\pi) }{1+t+\pi} \right| \mathrm d t < \int_0^{\pi} \frac{ \sin (t) }{1+t} \mathrm d t $$ $$ \int_\pi^{2\pi} \left| \frac{ \sin (t) }{1+t} \right| \mathrm d t < \int_0^{\pi} \frac{ \sin (t) }{1+t} \mathrm d t $$ $$ \int_\pi^{2\pi} - \frac{ \sin (t) }{1+t} \mathrm d t < \int_0^{\pi} \frac{ \sin (t) }{1+t} \mathrm d t $$ $$ 0 < \int_0^{\pi} \frac{ \sin (t) }{1+t} \mathrm d t + \int_\pi^{2\pi} \frac{ \sin (t) }{1+t} \mathrm d t $$ $$ 0 < \int_0^{2\pi} \frac{ \sin (t) }{1+t} \mathrm d t $$

And you are done :)