Improper Integral:$\int_{0}^{+\infty}\frac{\sin x}{x+\sin x}dx$ I want show that this improper integral convergence: $$\int_{0}^{+\infty}\frac{\sin x}{x+\sin x}dx$$ please help me.  
 A: 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 $(x-1)^2$.
A: 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.
A: 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.
A: 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
A: 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 $2\pi n$ 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).  
