$$\int_{0}^{\infty}\frac{\sin(x)}{x(x+1)} \mathrm{dx}$$

In a homework for a standard Calculus I class, we are given this improper integral.

The problem is to prove that it exists. However I am clueless as how to do it. Any hints are appreciated.

My only idea is to apply general convergence criteria for functions or sequences directly.

But this seems hopeless because I can't find a way to calculate the anti-derivative.

Integration by parts only makes this integral even (seemingly) harder and there appears to be no obvious substitution to make.

  • 1
    $\begingroup$ The integrand is bounded as $x\to0$ and is $O(1/x^2)$ as $x\to\infty$. $\endgroup$ May 25, 2019 at 20:13
  • 11
    $\begingroup$ $|\int_0^\infty \frac{\sin(x)}{x(x+1)} dx| \le \int_0^1 \frac{1}{x+1} dx + \int_1^\infty \frac{1}{x(x+1)} dx$ $\endgroup$
    – Tom Chen
    May 25, 2019 at 20:15
  • $\begingroup$ To distill a general hint out of the existing comments: using the Comparison Test is another way to prove convergence of improper integrals. Often you'll be able to choose an integral to compare to where the methods you describe trying work far better. $\endgroup$ May 25, 2019 at 20:21

1 Answer 1


$$ -1/p\leq-\ln(1+p^{-1})\leq\int_p^\infty\frac{\sin x}{x(x+1)}dx\leq\ln(1+p^{-1})\leq 1/p,\\ 0\leq\int_0^s\frac{\sin x}{x(x+1)}dx\leq\ln(1+s)\leq s, $$ so $$ \int_p^\infty\frac{\sin x}{x(x+1)}dx\to 0, \int_0^s\frac{\sin x}{x(x+1)}dx\to 0 $$ as $p\to \infty,s\to 0$.

It should be clear now that the integral converges.

We alos have $$ \int_0^\infty\frac{\sin x}{x(x+1)}dx=\int_0^1\frac{\sin x}{x(x+1)}dx+\int_1^\infty\frac{\sin x}{x(x+1)}dx\\ \leq\int_0^1\frac{x}{x(x+1)}dx+\int_1^\infty\frac{1}{x(x+1)}dx\\ =[\ln(x+1)]^1_0+\left[\ln\frac{x}{x+1}\right]^\infty_1 =\ln2+\ln 2=\ln 4. $$ Similarly, $$ \int_0^\infty\frac{\sin x}{x(x+1)}dx=\int_0^1\frac{\sin x}{x(x+1)}dx+\int_1^\infty\frac{\sin x}{x(x+1)}dx\\ \geq\int_0^1\frac{0}{x(x+1)}dx+\int_1^\infty\frac{-1}{x(x+1)}dx\\ =-\ln 2. $$

  • 1
    $\begingroup$ I only wanted hints, but since Tom Chen's comment already helped me solve the problem, I will accept your answer to close this question. $\endgroup$
    – lol
    May 26, 2019 at 11:14

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