$$\int_{0}^{\infty}\frac{\sin(x)}{x(x+1)} \mathrm{dx}$$
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.
Best Answer
$$ -1/p\leq-\ln(1+p^{-1})\leq\int_p^\infty\frac{\sin x}{x(x+1)}dx\leq\ln(1+p^{-1})\leq 1/p,\\ 0\leq\int_0^s\frac{\sin x}{x(x+1)}dx\leq\ln(1+s)\leq s, $$ 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, $$ \int_0^\infty\frac{\sin x}{x(x+1)}dx=\int_0^1\frac{\sin x}{x(x+1)}dx+\int_1^\infty\frac{\sin x}{x(x+1)}dx\\ \geq\int_0^1\frac{0}{x(x+1)}dx+\int_1^\infty\frac{-1}{x(x+1)}dx\\ =-\ln 2. $$