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?