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Why is this integral always positive for $x \ge 0$?

$$\int_0^x \frac{\sin(t)}{t+1} dt$$

It's easy for me to see this when I imagine the graph of the integrand. However, this question appears in the chapter that intoduces integration by substitution and by parts, as well as some trig identities.

So I'm looking for a proof, preferrably using integration by parts.

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2 Answers 2

up vote 4 down vote accepted

Use integration by parts. Let $u=\dfrac{1}{1+t}$ and $dv=\sin t\,dt$. Then $du=-\dfrac{1}{(1+t)^2}\,dt$, and we can take $v=-\cos t$. Thus $$\int_0^x\frac{\sin t}{1+t}\,dt=\left.-\frac{\cos t}{1+t}\right|_0^x-\int_0^x\frac{\cos t}{(1+t)^2}\,dt.$$

To bound $ \int_0^x\frac{\cos t}{(1+t)^2}\,dt $, note that $|\cos t|\le 1$, and is not always $1$. And $\int_0^x \frac{1}{(1+t)^2}\,dt=1-\frac{1}{1+x}$ so $ \int_0^x\frac{\cos t}{(1+t)^2}\,dt $ has absolute value $\lt 1-\dfrac{1}{1+x}$. Thus $$\int_0^x\frac{\sin t}{1+t}\,dt \gt 1-\frac{|\cos x|}{1+x}-1+ \frac{1}{1+x}.$$

It follows that our integral is $\gt \dfrac{1-\cos x}{1+x}$, and in particular is $\gt 0$.

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It's easy for me to see this when I imagine the graph of the integrand.

This imagination can become a formal proof: We have $$\int_0^x \frac{\sin(t)}{t+1} dt=\pi \int_0^{\frac{x}{\pi}} \frac{\sin(\pi t)}{\pi t+1} dt$$ So consider: $$g(t)=\frac{\sin(\pi t)}{\pi t+1}$$

We have: $$\int_0^x g(t) dt=\int_{[x]}^x g(t) dt+\sum_{k=1}^{[x]}{(-1)^{k-1}\int_{k-1}^k |g(t)| dt}$$ So if $$a_k=\int_{k-1}^k |g(t)| dt$$ It's enough to show that:

$$0\leq \sum_{k=1}^{n}{(-1)^{k-1} {a_k}}=(a_1-a_2)+(a_3-a_4)+..$$ But it's clear because: $$a_{k+1}=\int_{k}^{k+1} |\frac{\sin(\pi t)}{\pi t+1}| dt\leq \int_{k-1}^{k} |\frac{\sin(\pi t)}{\pi t+1}| dt= a_k$$

However some details are missed.

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1  
Very nice! The $\pi$ substitution makes the proof a lot cleaner than the way I was about to do it. –  Mark Feb 14 '13 at 4:49

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