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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,

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

4 comments

  1. 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.

  2. 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

  3. 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?

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