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I've tried to figure out if this improper integral convergse or diverges: $$\int_1^\infty\frac{\sin(x) +1}{x^2}.$$

I want to use the "direct comparison" (sorry but I'm italian and in my book it's called "Confronto Diretto"). My integrand function $f(x)$ is always equal/bigger than zero, because $\sin(x)\in [-1,1]$, so I can apply the method. I have to find a function always bigger/equal than mine like: $$g(x)=\frac{2}{x^2}$$ $G(x)$ converges and therefore also $F(x)$ converges. Is it right?

Thanks for the help

Robert Z
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D.T
  • 13

2 Answers2

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Yes, it is correct. For $x\geq 1$, $$0\leq\frac{\sin(x) +1}{x^2}\leq \frac{2}{x^2}.$$ which implies that $F(t)=\int_1^t\frac{\sin(x) +1}{x^2}dx$ is an increasing in $[2,+\infty)$ with a finite upper bound $$F(t)\leq 2\int_1^{+\infty}\frac{1}{x^2}dx=2.$$ Therefore the limit $\lim_{t\to+ \infty}F(t)=\int_1^{+\infty}\frac{\sin(x) +1}{x^2}dx$ exists and it is finite.

P.S. We may say that we are using the direct comparison test for integrals.

Robert Z
  • 145,942
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As a minor addendum, I will outline an approach for a numerical evaluation.
We have $\int_{1}^{+\infty}\frac{dx}{x^2}=1$ and by a useful property of the Laplace transform

$$\begin{eqnarray*} \int_{1}^{+\infty}\frac{\sin(x)}{x^2}\,dx &\stackrel{=}{=}& \int_{0}^{+\infty}\frac{\sin(1)\cos(x)+\cos(1)\sin(x)}{x^2}\,dx\\ &\stackrel{\mathcal{L}}{=}&\int_{0}^{+\infty}\frac{\sin(1)s^2+\cos(1)s}{s^2+1}e^{-s}\,ds\\&=&\sin(1)-\int_{0}^{+\infty}\frac{\sin(1)-\cos(1)s}{s^2+1}e^{-s}\,ds \end{eqnarray*}$$ where the last integral can be manipulated by integration by parts / estimated through the Cauchy-Schwarz inequality. For instance $$\left|\int_{0}^{+\infty}\frac{\sin(1)-\cos(1)s}{s^2+1}e^{-s}\,ds\right|\leq\sqrt{\frac{1}{2}\int_{0}^{+\infty}\frac{\left(\sin(1)-\cos(1)s\right)^2}{(s^2+1)^2}\,ds}=\sqrt{\frac{\pi-2\sin 2}{8}}. $$

Jack D'Aurizio
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