Convergence of improper integral
$\int_{2}^{\infty} \frac{1}{log(t)}dt$
How do i start?
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1What have you done? – Christopher Carl Heckman Feb 10 '16 at 05:59
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@CarlHeckman give me a hint – Taylor Ted Feb 10 '16 at 05:59
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How are you supposed to handle integrals which have infinity as an endpoint? (Also, I'm not sure that that is the original problem; maybe the function was ${1\over t\ln t}$?) – Christopher Carl Heckman Feb 10 '16 at 06:01
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@CarlHeckman problem doesnot have t – Taylor Ted Feb 10 '16 at 06:02
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Do you know of some comparison theorem? What about the convergence of $\int_2^{+\infty}1/t,dt$? – mickep Feb 10 '16 at 06:06
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1There's no elementary antiderivative for that integral, so you need to either find a function $\displaystyle f(x)<{1\over \ln x}$ such that $\displaystyle \int_2^\infty f(x),dx$ diverges, or a function $\displaystyle f(x)>{1\over \ln x}$ such that $\displaystyle \int_2^\infty f(x),dx$ converges. – Christopher Carl Heckman Feb 10 '16 at 06:06
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1@mickep by comparison with 1\t it shows divergent. is this right – Taylor Ted Feb 10 '16 at 06:10
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I note that this is a very important integral in Number Theory, as the integral out to $n$ is asymptotic to the number of primes up to $n$. – Gerry Myerson Feb 10 '16 at 06:16