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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

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  • $\begingroup$ Since $x+x^2=x(x+1)$ perhaps a partial fractions decomposition would help? Something along the lines of $\frac 1{x+1}-\frac 1x$ $\endgroup$ – abiessu Apr 22 '14 at 18:43
  • $\begingroup$ @abiessu I used that while integrating by parts which brings out the solution - but I feel as though there is a nicer way to evaluate the limit $\endgroup$ – dani Apr 22 '14 at 18:52
  • $\begingroup$ Since you say "I'm trying to determine if the integral exists," I have a way to determine if it is convergent posted below... $\endgroup$ – apnorton Apr 22 '14 at 19:12
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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.

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$$\lim_{x \rightarrow 0} \frac{\sin(x)}{x + x^2} = \lim_{x \rightarrow 0}\left(\frac{\sin(x)}{x}\right) \cdot \left(\frac{1}{x + 1}\right) = 1 \cdot 1 = 1$$

Since the integrand has a limit at zero, the integral $\int_0^1f(x)\, dx$ converges. At infinity you can use:

$$\left|\frac{\sin(x)}{x + x^2}\right| \leq \frac{1}{x + x^2} \leq \frac{1}{x^2}$$

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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\infty}{\sin\pars{x} \over x + x^{2}}\,\dd x:\ {\large ?}}$

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}

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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 $(2)$.

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