# 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

• Since $x+x^2=x(x+1)$ perhaps a partial fractions decomposition would help? Something along the lines of $\frac 1{x+1}-\frac 1x$ – abiessu Apr 22 '14 at 18:43
• @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 – dani Apr 22 '14 at 18:52
• Since you say "I'm trying to determine if the integral exists," I have a way to determine if it is convergent posted below... – apnorton Apr 22 '14 at 19:12

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.

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

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

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


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}

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 $(2)$.