Convergence of $\int_{0}^{+\infty}f(x)dx$ How can I check convergence of $\int_{0}^{+\infty}f(x)dx$ for the following $f$?


*

*$f(x)=\left(\frac{\sin x}{x}\right)^2$

*$f(x)=\left(\frac{\cos x}{x}\right)^2$

*$f(x)=\frac{1}{(1+x)^2}$

*$f(x)=\frac{1-e^{-x}}{x}$

*$f(x)=\frac{e^{x/2}}{x^{1/2}}$


What I've tried so far is write the improper integral as
$$
\int_{0}^1f+\int_{1}^{+\infty}f
$$
I think one of 1 and 2 can not be both convergent since $\int_0^{\infty}\frac{1}{x^2}$ is not convergent. One can calculate that 3 is convergent. How about 4 and 5? I think one may use the comparison test to conclude that 4 and 5 are not convergent for $\int_0^1f$ and $\int_1^{\infty}f$. But how about $\int_0^{\infty}f$? Am I on the right track?
 A: As you pointed out, given functions have singularities possible at $x = 0$ and $x = \infty$. Thus it suffices to investigate the behavior of each function near these singularity candidates.


*

*Since $\sin x = x + O(x^3)$ near $x = 0$, we have $\frac{\sin x}{x} = 1 + O(x^2)$. This proves that $x = 0$ is indeed a removable singularity of $f(x)$. It is also clear that $x = \infty$ is not a singularity since
$$ \int_{1}^{\infty} f(x) \, dx \leq \int_{1}^{\infty} \frac{dx}{x^2} = 1.$$
Therefore the improper integral of $f(x)$ over $(0, \infty)$ converges. Actually we can do more:
$$ \int_{0}^{\infty} \frac{\sin^2 x}{x^2} \, dx = \frac{\pi}{2}. $$

*Since $\cos x = 1 + O(x^2)$ near $x = 0$, we have $\frac{\cos x}{x} = \frac{1}{x} + O(x)$. This obviously shows that
$$ \int_{0}^{1} f(x) \, dx = \int_{0}^{1} \left( \frac{1}{x} + O(x) \right) \, dx = \infty$$
and the integral does not converge.

*I will skip this, as you already know the answer.

*Near $x = 0$, we have $e^{-x} = 1 - x + O(x^2)$, which implies that $f(x) = 1 + O(x)$ near $x = 0$. Thus it follows that $x = 0$ is a removable singularity. However, since $e^{-x} \leq e^{-1} < 1$ for $x \geq 1$, we can devise the following comparison:
$$ \int_{1}^{\infty} f(x) \, dx \geq \int_{1}^{\infty} \frac{1 - e^{-1}}{x} \,dx = \infty.$$
This proves that the integral diverges.

*$f(x) \geq \frac{1}{x^{1/2}} $. Need more explanation?
