anonymous
  • anonymous
does the integral of 1/x+ srqtx from 1 to infinity converge?
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
i doubt it
anonymous
  • anonymous
the degree of the denominator has to be larger than the degree of the numerator by more than 1
anonymous
  • anonymous
is it \[\int_1^\infty\left({1\over x}+\sqrt{x}\right)dx\] OR \[\int_1^\infty{dx\over x+\sqrt{x}} \]

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anonymous
  • anonymous
the second one
anonymous
  • anonymous
@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!!
anonymous
  • anonymous
so, it DOES converge coz, the denominator increases continuously and hence, the function decreases continuously
anonymous
  • anonymous
oh no!!
anonymous
  • anonymous
oh yes! :D
anonymous
  • anonymous
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
anonymous
  • anonymous
simple example \[\int_1^{\infty}\frac{dx}{x}\] does not converge
anonymous
  • anonymous
neither does the one above
anonymous
  • anonymous
aah.. jumping to conclusions too soon. \[ u=\sqrt{x}\implies u^2=x\implies 2udu=dx\\ \int_1^\infty\frac{2u}{u^2+u}du\\ =2\int_1^\infty {du\over u+1}=2[\ln (u+1)]_1^\infty=\boxed{\infty} \]

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