## peggiepenguin Group Title does the integral of 1/x+ srqtx from 1 to infinity converge? one year ago one year ago

1. satellite73 Group Title

i doubt it

2. satellite73 Group Title

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

3. electrokid Group Title

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

4. peggiepenguin Group Title

the second one

5. electrokid Group Title

@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 Group Title

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

7. satellite73 Group Title

oh no!!

8. electrokid Group Title

oh yes! :D

9. satellite73 Group Title

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 Group Title

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

11. satellite73 Group Title

neither does the one above

12. electrokid Group Title

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