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show that the integral converges or that is diverges: $$\int_{1}^{\infty}{\frac{2 + \cos x}{x-1}dx} $$

i know that it does converge from the solution, but i have no idea how to arrive at the conclusion. we are probably supposed to use p-integrals, but not sure how.

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  • $\begingroup$ For any $L>1$, both the integrals $\int_1^L \frac{2+\cos(x)}{x-1}\,dx$ and $\int_L^\infty \frac{2+\cos(x)}{x-1}\,dx$ diverge. $\endgroup$ – Mark Viola Dec 6 '17 at 22:05
  • $\begingroup$ okei, but i am not sure how to show it. $\endgroup$ – Sander Dec 6 '17 at 22:07
  • $\begingroup$ I've posted a full solution. $\endgroup$ – Mark Viola Dec 6 '17 at 22:18
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Hint It diverges. How to see it?

$$\frac{2+\cos x} {x-1} \geq \frac{1}{x-1} $$

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  • $\begingroup$ hmm, i can not really see how that helps. what results/theorems should i use from here? $\endgroup$ – Sander Dec 6 '17 at 22:04
  • $\begingroup$ @JoakimHauger have you heard about comparison test? What are the methods to show convergence of integrals that you have studied? Comparison test must be one of them $\endgroup$ – Shashi Dec 6 '17 at 22:06
  • $\begingroup$ yes, but dosen't that require that i show that f(x) > (2+cosx) / (x-1 ). and then show that f(x) converges? $\endgroup$ – Sander Dec 6 '17 at 22:10
  • $\begingroup$ @JoakimHauger yes that is to show convergence. There is something similar for divergence of the integral. Do you remember? $\endgroup$ – Shashi Dec 6 '17 at 22:12
  • $\begingroup$ @JoakimHauger yes that is to show convergence. There is something similar for divergence of the integral. Do you remember? $\endgroup$ – Shashi Dec 6 '17 at 22:12
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For any $L>1$, both the integrals $\int_1^L \frac{2+\cos(x)}{x-1}\,dx$ and $\int_L^\infty \frac{2+\cos(x)}{x-1}\,dx$ diverge.

First, we write $\frac{2+\cos(x)}{x-1}=\frac{2-\cos(1)}{x-1}+\frac{\cos(x)-\cos(1)}{x-1}$. Therefore, we have for $\epsilon>0$

$$\begin{align} \int_{1+\epsilon}^L \frac{2+\cos(x)}{x-1}\,dx&=\int_{1+\epsilon}^L \frac{2-\cos(1)}{x-1}\,dx+\int_{1+\epsilon}^L \frac{\cos(x)-\cos(1)}{x-1}\,dx\tag1 \end{align}$$

The first integral on the right-hand side of $(1)$ is equal to $(2-\cos(1))\left(\log(L-1)-\log(\epsilon)\right)$, which diverges as $\epsilon\to 0$, while the second integral has a removable discontinuity at $x=1$ and is therefore integrable.


For the integral $\int_L^\infty \frac{2+\cos(x)}{x-1}\,dx$ we see that $\int_L^\infty \frac{2}{x-1}\,dx$ diverges logarithmically, while the integral $\int_L^\infty \frac{\cos(x)}{x-1}\,dx$ converges (integrate by parts with $u=\frac1{x-1}$ and $v=\sin(x)$). Hence, the sum of a convergent integral and a divergent integral is divergent.

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