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I want to show that $\int_{0}^{\infty}{\dfrac{1}{t}}dt$ is divergent. Now we have a problem of boundedness of $f(t)=\dfrac{1}{t}$ on $t=0$ and we have a problem of boundedness of the domain $(0,\infty)$ So we treat each problem apart by writing $$\int_{0}^{\infty}{\dfrac{1}{t}}dt=\int_{0}^{1}{\dfrac{1}{t}}dt+\int_{1}^{\infty}{\dfrac{1}{t}}dt$$ Now the two parts are Riemann integrals such that $\int_{0}^{1}{\dfrac{1}{t}}dt$ is divergent and $\int_{1}^{\infty}{\dfrac{1}{t}}dt$ is also divergent. The sum of two divergent integrals may or may not converge, so how do we treat this problem. Thank you for your help!!

palio
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  • If one of the limits diverges, the integral diverges. No matter what about the sum. – LeeNeverGup Dec 01 '13 at 13:09
  • You are integrating non-negative functions then the integral converges iff both partial integrals converge. You might want to check this from your definition. – Listing Dec 01 '13 at 13:10
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    Just use the fact that $$\int_0^{\infty} \frac{1}{t}dt \geq \int_1^{\infty} \frac{dt}{t} $$ – Prahlad Vaidyanathan Dec 01 '13 at 13:10
  • @Listing I thought it was similar to infinite series where the sum of two divergent series can converge. But here you are saying that in a sum of two integrals, if one diverges then the sum diverges no matter of the nature of the second integral.. is this only for sum of integrals of two positive functions? what about the general case? – palio Dec 01 '13 at 13:17
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    @palio for infinite series if both series are non-negative and diverge their sum also diverges. For non-positive functions it is generally not true, look at functions who add up to zero.. – Listing Dec 01 '13 at 13:20

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