Prove that $\int_0^x \dfrac {\sin t}{t+1}dt \geq 0 ~\forall~x\geq 0$

Attempt: Let $A(x) = \int_0^x \dfrac {\sin t}{t+1}dt$

Since, $\dfrac {\sin x}{x+1}$ is a continuous function,

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

I am not sure how to proceed ahead. Please guide me.

Thank you for your help.

EDIT: I am curious to know if this can be solved through the use of mean value theorem for integrals. Could somebody please show me a way using the same?



It suffices to show that, for any $k \ge 0$, $k \in \mathbb{Z}$, that

$$\int_{2 k \pi}^{(2 k+1) \pi} dt \frac{\sin{t}}{1+t} + \int_{(2 k+1) \pi}^{(2 k+2) \pi} dt \frac{\sin{t}}{1+t} \ge 0$$

The first integral is always positive, and the second integral is always negative. The sum is a worst-case scenario over a period of the sine. Any value of $x$ in the integration limit that represents less than a full period has the effect of making the above sum more positive.

We can rewrite the sum as

$$\int_0^{\pi} dt \frac{\sin{t}}{t+2 k \pi+1} - \int_0^{\pi} dt \frac{\sin{t}}{t+(2 k+1) \pi+1} $$

which simplifies to

$$\pi \int_0^{\pi} dt \frac{\sin{t}}{(t+2 k \pi +1)(t+(2 k+1) \pi+1)} $$

which is clearly $\ge 0$.

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  • $\begingroup$ Thank you for the answer :) $\endgroup$ – MathMan Jan 18 '15 at 15:21

Lemma 1: let $N=2k\pi,x-N<\pi$,then we have $$\int_{N}^{x}\dfrac{\sin{t}}{t+1}>0$$ proof: since $x-N<\pi$, then $\sin{x}$ does not change sign,use first mean value theorem for integration,we have $$\int_{N}^{x}\dfrac{\sin{t}}{t+1}=\dfrac{1}{\xi+1}\int_{N}^{x}\sin{t}=\dfrac{1}{1+\xi}(1-\cos{x})\ge 0$$ Lemma2: $$I=\int_{(2k-2)\pi}^{(2k-1)\pi}\dfrac{\sin{t}}{t+1}dt+\int_{(2k-1)\pi}^{2k\pi}\dfrac{\sin{t}}{t+1}dt>0$$ so $$I=\int_{0}^{\pi}\left(\dfrac{\sin{u}}{(2k-1)\pi-u+1}-\dfrac{\sin{u}}{2k\pi-u+1}\right)du>0$$ Now let $N\pi\le x<(N+1)\pi,N=2m$,and $N$ is odd number,since $$\int_{0}^{x}\dfrac{\sin{t}}{t+1}dt=\sum_{k=1}^{2m}\left(\int_{(2k-2)\pi}^{(2k-1)\pi}\dfrac{\sin{t}}{t+1}dt+\int_{(2k-1)\pi}^{2k\pi}\dfrac{\sin{t}}{t+1}dt\right)+\int_{2m\pi}^{x}\dfrac{\sin{t}}{t+1}>0$$ and the same simaler with $N=2m+1$

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  • 1
    $\begingroup$ You might mention the mean value theorem for integrals here, to help explain why this $\xi$ exists. $\endgroup$ – Ian Jan 18 '15 at 12:34
  • $\begingroup$ yes,use mean value theorem $\endgroup$ – math110 Jan 18 '15 at 12:34
  • $\begingroup$ Using mean value theorem, Can you explain a further more please? $\endgroup$ – MathMan Jan 18 '15 at 12:44

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