Evaluating the improper integral $\int_0^{\infty} \frac{\sin x}{x+x^2} \ dx$ Evaluating the improper integral $$\int_0^{\infty} \frac{\sin x}{x+x^2} \ dx$$
I'm trying to determine if the integral exists.
I can't seem to deal with $$\lim_{a\to 0^+} \int_a^\infty \frac{\sin x}{x + x^2} \ dx,$$ could someone help with the limit above?
Edit: doing it by parts does seem to work, but wondering if there is a neater way to evaluate the limit at 0
 A: We only seek to prove convergence, not to find the actual value.
Let $f(x) = \frac{\sin(x)}{x(x+1)}$.  Note that:
$$\int_0^\infty f(x) \;dx = \int_0^1f(x)\;dx + \int_1^\infty f(x)\;dx$$
(if the rhs converges)
Note that:
$$\int_0^1 f(x)\;dx \le \int_0^1 \frac{x}{x(1+x)}\;dx$$
(via Taylor series for $\sin(x)$)
Thus:
$$\int_0^1 f(x)\;dx \le \int_0^1 \frac{dx}{1+x} = \ln(2)$$
So, the first integral converges.  Moving on to the second integral:
$$\int_1^\infty \frac{\sin(x)}{x(x+1)}\;dx \le \int_1^\infty \frac{1}{x(x+1)}\;dx \le \int_1^\infty \frac{1}{x^2}\;dx = 1$$
So, the second integral converges.  Thus:
$$\begin{align}\int_0^\infty f(x) \;dx &= \int_0^1f(x)\;dx + \int_1^\infty f(x)\;dx \\
&\le \ln(2) + 1 
\end{align}$$
Thus, the integral converges.
A: $$\lim_{x \rightarrow 0} \frac{\sin(x)}{x + x^2} = \lim_{x \rightarrow 0}\left(\frac{\sin(x)}{x}\right) \cdot \left(\frac{1}{x + 1}\right) = 1 \cdot 1 = 1$$
Since the integrand has a limit at zero, the integral $\int_0^1f(x)\, dx$ converges. At infinity you can use:
$$\left|\frac{\sin(x)}{x + x^2}\right| \leq \frac{1}{x + x^2} \leq \frac{1}{x^2}$$
A: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
$\ds{\int_{0}^{\infty}{\sin\pars{x} \over x + x^{2}}\,\dd x:\ {\large ?}}$

Let's $\ds{\ a > 0}$:
  \begin{align}
&\color{#c00000}{\int_{a}^{\infty}{\sin\pars{x} \over x + x^{2}}\,\dd x}
=-\Im\int_{a}^{\infty}\expo{-\ic x}\pars{{1 \over x} - {1 \over x + 1}}\,\dd x
\\[3mm]&=-\Im\bracks{%
\int_{a}^{\infty}{\expo{-\ic x} \over x}\,\dd x
-\int_{a + 1}^{\infty}{\expo{-\ic\pars{x - 1}} \over x}\,\dd x}
\\[3mm]&=-\Im\bracks{%
\int_{1}^{\infty}{\expo{-\ic ax} \over x}\,\dd x
-\expo{\ic}\int_{1}^{\infty}{\expo{-\ic\pars{a + 1}x} \over x}\,\dd x}
=-\Im\bracks{{\rm E}_{1}\pars{\ic a} - \expo{\ic}{\rm E}_{1}\pars{a + 1}}
\end{align}
  where $\ds{{\rm E}_{1}\pars{z}}$ is the
  Exponential Integral Function.

\begin{align}
&\color{#c00000}{\int_{a}^{\infty}{\sin\pars{x} \over x + x^{2}}\,\dd x}
=-\Im{\rm E}_{1}\pars{\ic a} + \cos\pars{1}\Im{\rm E}_{1}\pars{\ic\bracks{a + 1}}
+ \sin\pars{1}\Re{\rm E}_{1}\pars{\ic\bracks{a + 1}}
\end{align}

Since
  $\ds{\lim_{a \to 0^{+}}{\rm E}_{1}\pars{\ic a} = -\,{\pi \over 2}}$, we'll get:
  \begin{align}
&\color{#00f}{\int_{0}^{\infty}{\sin\pars{x} \over x + x^{2}}\,\dd x
=\half\pi + \cos\pars{1}\Im{\rm E}_{1}\pars{\ic}
+ \sin\pars{1}\Re{\rm E}_{1}\pars{\ic}}
\end{align}

A: To show that the integral exists, we use limit comparison test with $f(x) = \sqrt x$ in the interval $(0, a)$ where, $\sin(x) > 0 \, \forall x \in (0, a] $.
$$\int_0^\infty\frac{\sin(x)}{x}\,dx = \frac \pi 2 \tag{1}$$
It is done here, for the other one, set $y = x+1$
$$\int_0^\infty \frac{\sin(x)}{x+1}\, dx = \int_{1}^\infty \frac{\sin(y-1)}{y}= \frac{\sin(y) \cos (1) - \sin(1)\cos(y)}{y} dy \tag{2}$$
For first integral of $(2)$, 
$$ \int_{1}^\infty \frac{\sin(y) }{y} dy = \int_{0}^\infty \frac{\sin(y)}y \,dy -  \int_{0}^1 \frac{\sin(y)}y \,dy \tag{3}$$
The second term of $(3)$can be evaluated interms of Sine Integral. Similarly for the second term of the $(2)$.
