Prove the integral from 0 to x of sin t /(t2 + 1) is positive for positive x

Prove that for all $x \geq 0$ we have

$\int_0^x \frac{\sin t}{t+1} \, dt \geq 0.$

Before proceeding with the proof, we recall the second mean-value theorem for integrals (Theorem 5.5 on p. 219 of Apostol). For a continuous function $g$ on the interval $[a,b]$ if $f$ has a continuous derivative which never changes sign on the interval $[a,b]$ then there exists a $c \in [a,b]$ such that

$\int_a^b f(x) g(x) \, dx = f(a) \int_a^c g(x) \, dx + f(b) \int_c^b g(x) \, dx.$

Proof. Now, we want to apply the mean-value theorem above with $g(t) = \sin t$ and $f(t) = \frac{1}{t+1}$ . Since $\sin t$ is continuous everywhere $\mathbb{R}$ , it is continuous on any interval $[0,x]$ . Then,

$f'(t) = \frac{-1}{(1+t)^2}$

is continuous for all $t \neq -1$ (so, in particular, for all $t \geq 0$ ). Furthermore, since $(1+t)^2 > 0$ for all $t \geq 0$ we have that $f'(t) < 0$ for all $t \geq 0$ . Thus, $f'(t)$ is continuous and never changes sign in any interval $[0,x]$ . Therefore, we can apply the mean-value theorem to conclude there exists a $c \in [0,x]$ for any $x \geq 0$ such that

$\begin{align*} \int_0^x \left(\frac{1}{1+t}\right) \sin t \, dt &= \frac{1}{1+0} \int_0^c \sin t \, dt + \frac{1}{1+1} \int_c^x \sin t \, dt \\ &= (- \cos t) \Big \rvert_0^c + \frac{1}{2} (-\cos t) \Big \rvert_c^x \\ &= 1 - \cos c + \frac{1}{2} \cos c - \frac{1}{2} \cos x \\ &= 1 - \frac{1}{2} ( \cos c + \cos x). \end{align*}$

But since $\cos x \leq 1$ for all $x$ we know $\cos c + \cos x \leq 2$ for any $c$ and $x$ . Hence,

$\int_0^x \frac{\sin t}{1+t} \, dt = 1 - \frac{1}{2}(\cos c + \cos x) \geq 0. \qquad \blacksquare$

5 comments

June 20, 2017 at 12:16 pm

Andres Tellez says:

At the end, following the same procedure, you get (x(1-cosc)+1-cosc)/(x+1) which is always greater or equal to 0 for any choice of c or x, such that 0<=c<=x.

- June 20, 2017 at 12:18 pm
  
  Andres Tellez says:
  
  I’m sorry it should be (x(1-cosc)+1-cosx)/(x+1)
  
June 8, 2017 at 9:36 pm

Ketan says:

I agree with Batman…

i got 1 – (x/(x+1)) * cos (c) – cos(x) / (x + 1) > or = 0, with equality only holding when cos(c) = cos(x) = 1

May 4, 2016 at 6:42 pm

Batman says:

The integral limits were from 0 to x. this one is ok: f(a)=f(0) = 1/(1+0), but wasn’t the other supposed to be f(b) = f(x) = 1/(1+x)? You’ve written 1/(1+1), why?

- July 23, 2019 at 12:20 pm
  
  Artem says:
  
  Same question

Stumbling Robot

Prove the integral from 0 to x of sin t /(t² + 1) is positive for positive x

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