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I want show that this improper integral convergence: $$\int_{0}^{+\infty}\frac{\sin x}{x+\sin x}dx$$ please help me.

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Different ways how to do it. Please observe that as a limit for x→0 the function is defined. So as a definite integral from, say (0,1] the integral is convergent. From (1,→), perform integration by parts. You then get behind the integral sign a (x+sinx)² in the denominator. Using Comparison theorem you can find that these integrals are convergent because x+sinx is greater than 0.5x –  imranfat May 1 '13 at 18:23

5 Answers 5

up vote 7 down vote accepted

For $0\lt x\le\pi$, $0\le\dfrac{\sin(x)}{x+\sin(x)}=\dfrac{\sin(x)/x}{1+\sin(x)/x}\le\dfrac12$, so $$ 0\le\int_0^{\pi}\frac{\sin(x)}{x+\sin(x)}\,\mathrm{d}x\le\frac\pi2 $$

Integrate by parts to get $$ \begin{align} \int_\pi^\infty\frac{\sin(x)}{x+\sin(x)}\,\mathrm{d}x &=\left.-\frac{\cos(x)}{x+\sin(x)}\right]_\pi^\infty -\int_\pi^\infty\frac{\cos(x)(1+\cos(x))}{(x+\sin(x))^2}\,\mathrm{d}x\\[6pt] &=-\frac1\pi-\int_\pi^\infty\frac{\cos(x)(1+\cos(x))}{(x+\sin(x))^2}\,\mathrm{d}x \end{align} $$ and the last integral converges since the absolute value of the numerator of the integrand is bounded by $2$ and the denominator is greater than $(x-1)^2$.

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The integral converges near $x=0$ because the limit of the integrand is $1$ there. And at infinity, the integrand approaches $\sin{x}/x$, which is integrable there. To see this, Taylor expand the denominator:

$$\frac{\sin{x}}{x+\sin{x}} \approx \frac{\sin{x}}{x} - \frac{\sin^2{x}}{x^2}$$

both of which are integrable at infinity. Further terms are still smaller.

Is there an $x \ne 0$ such that the denominator vanishes? Let's see: $x > \sin{x}$ so that there will be no cancellation outside of $x=0$. Thus, we have convergence.

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You are being careless at infinity, methinks. –  Harald Hanche-Olsen May 1 '13 at 18:24
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Better now, yes. I am not surprised that you can do it, but consider your audience. Also, there is the example $\sum_n (-1)^n/(\sqrt{n}+(-1)^n)$, which diverges. This shows that you can't be too careful. –  Harald Hanche-Olsen May 1 '13 at 18:29
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@GitGud: thanks. My wife likes the new look, but she is really tiring of the bats. –  Ron Gordon May 1 '13 at 18:35
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$$\frac{1}{x+\sin{x}} = \frac{1}{x} \frac{1}{1+(\sin{x}/x)} \approx \frac{1}{x} \left( 1 - \frac{\sin{x}}{x}\right )$$ –  Ron Gordon May 1 '13 at 19:23
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@RonGordon: ah, I think what may have been causing some confusion is that the Taylor series you are using is for $$\frac{u}{1+u}=u-u^2+u^3-u^4+u^5-u^6+\dots$$which converges for $|u|\lt1$ and using $u=\sin(x)/x$ in $$\frac{\sin(x)}{x+\sin(x)}=\frac{\sin(x)/x}{1+\sin(x)/x}$$ I was not certain of that at first since you mentioned Taylor expanding the denominator. –  robjohn May 1 '13 at 20:20

We have $$\lim_{x\to 0}\frac{\sin x}{x+\sin x}=\lim_{x\to 0}\frac{ x}{x+ x}=\frac{1}{2}$$ hence the integral $$\int_0^1\frac{\sin x}{x+\sin x}dx$$ is convergent

Moreover, by integration by part we have $$\int_1^A\frac{\sin x}{x+\sin x}dx=- \left[\frac{\cos x}{x+\sin x}\right]_1^A-\int_1^A\frac{\cos x(\cos x+1)}{(x+\sin x)^2}dx$$ and with $$|\frac{\cos x}{x+\sin x}|\leq |\frac{1}{x-1}|\to 0,\quad x\to\infty$$ and $$|\frac{\cos x(\cos x+1)}{(x+\sin x)^2}|\leq \frac{2}{(x-1)^2}$$ and the integral $$\int_2^\infty \frac{1}{(x-1)^2}dx$$ is convergent so the integral $$\int_1^\infty\frac{\sin x}{x+\sin x}dx$$ is also convergent and the desired result follows

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The problem is at $+\infty$, as the function tends to $1/2$ at $0$ and it is continuous on $(0,+\infty)$. Now $$ \frac{\sin x}{x+\sin x}-\frac{\sin x}{x}=\frac{x\sin x-\sin x(x+\sin x)}{x(x+\sin x)}=-\frac{\sin^2 x}{x(x+\sin x)}\sim-\frac{\sin^2x}{x^2} $$ at $+\infty$, so the former is integrable. And for $\frac{\sin x}{x}$, it is easy to prove convergence of the improper integral with an integration by parts on $[1,x]$.

Note: or as pointed out by imranfat, you can directly do integration by parts of the original integrand on $[1,x]$. It amounts essentially to the same.

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For $(\sin x)/x$, the question you refer to is overkill if all you want to show is that the integral converges. It is easier to put $a_n=\int_{\pi n}^{\pi(n+1)}(\sin x)/x\,dx$ and show that $\sum a_n$ converges by the alternating series test. –  Harald Hanche-Olsen May 1 '13 at 18:28
    
@HaraldHanche-Olsen Or even better, an integration by parts. That's just the only reference I could find... I agree it does not make a great answer... I'll edit. –  1015 May 1 '13 at 18:33

The integral can be compared to the convergent integral $\int_0^\infty {\sin(x)\over x}dx={\pi\over2}$. The difference is given by $\int_0^\infty{\sin^2(x)\over {x(x+\sin(x))}}dx$, which is absolutely convergent, so the integral converges.

The convergence of $\int_0^\infty{\sin(x)\over x} dx$ is well-known, and can be established by splitting the integral up into integrals of the form $\int_{2k\pi}^{(2k+2)\pi} {\sin(x)\over x} dx$, with $k$ a positive integer. Each of these integrals has a positive and negative component, and it is clear that $$|{\int_{(2k)\pi}^{(2k+2)\pi} {\sin(x)\over x} dx}| < {2\over{(2k)\pi}}-{2\over{(2k+1)\pi}}={1\over{k(2k+1)\pi}}\,.$$ If the upper bound is not a multiple of $2\pi$, then there will be an additional contribution of $\int_{2n\pi}^{R+2n\pi}{\sin(x)\over x}dx$, with $n$ the last positive integer for which the upper limit is smaller than $2n$ and $R$ smaller than $2\pi$, but this integral is clearly less than $R\over{2n\pi}$. The integral $\int_0^{R+2n\pi}{\sin(x)\over x}dx$ is therefore smaller than $2\pi+\sum_{k=1}^{n}{1\over{k(2k+1)\pi}}+{R\over{2n\pi}}$, with the $2\pi$ coming from the region from $x=0$ to $x=2\pi$. Since the sum is convergent as $n$ tends to $\infty$ (limit comparison to the $p$-series with $p=2$ and the additional piece tends to zero in this limit, we can say for sure that this integral converges. Showing that it is equal to $\pi\over2$ is a different matter, but it can be done in multiple ways (contour integration and relation to the Gamma function are the two that come to mind, but I am sure there are others).

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