## peggiepenguin 2 years ago does the integral of 1/x+ srqtx from 1 to infinity converge?

1. satellite73

i doubt it

2. satellite73

the degree of the denominator has to be larger than the degree of the numerator by more than 1

3. electrokid

is it $\int_1^\infty\left({1\over x}+\sqrt{x}\right)dx$ OR $\int_1^\infty{dx\over x+\sqrt{x}}$

4. peggiepenguin

the second one

5. electrokid

@satellite73 aha... @peggiepenguin next time please use the "equation editor on this website" or "TEX" to enter your equations.. so we know what you are talking about!!

6. electrokid

so, it DOES converge coz, the denominator increases continuously and hence, the function decreases continuously

7. satellite73

oh no!!

8. electrokid

oh yes! :D

9. satellite73

just because the function decreases does not mean the integral converges that is a necessary but not a sufficient condition it must decrease fast enough

10. satellite73

simple example $\int_1^{\infty}\frac{dx}{x}$ does not converge

11. satellite73

neither does the one above

12. electrokid

aah.. jumping to conclusions too soon. $u=\sqrt{x}\implies u^2=x\implies 2udu=dx\\ \int_1^\infty\frac{2u}{u^2+u}du\\ =2\int_1^\infty {du\over u+1}=2[\ln (u+1)]_1^\infty=\boxed{\infty}$