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I should determine whether this is a convergent or divergent integral. I need to use the comparison test but I don't find any intgeral that helps me figure this. $$ \int_{0}^{\infty} \frac{\cos x}{x}dx $$

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  • $\begingroup$ i'm so sorry, but i made a mistake and the integral is from 0 to infinity, not 1 to infinity (like i wrote before). i fixed it. $\endgroup$ – user2637293 May 14 '14 at 9:04
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The proof is similar to the proof of conditional convergence of alternating series (monotonously convergent to 0 times alternating bounded = convergent). In fact, you can bound it by a discrete series:

$$a_n=\left|\int_{\pi (n-1/2)}^{\pi (n+1/2)}\frac{\cos x}{x}dx\right|\leq\frac{1}{\pi (n-1/2)}\left|\int_{\pi (n-1/2)}^{\pi (n+1/2)}\cos x\right|=\frac{2}{\pi(n-1/2)}$$

So the integral

$$\int_{\pi/2}^\infty\frac{\cos x}{x}dx=\sum_{n=1}^\infty (-1)^n a_n<\sum_{n=1}^\infty (-1)^n \frac{2}{\pi (n-1/2)}$$ converges.

There is a technicality of uniform convergence to consider (the question if the integral $\int^x$ for $x$ that is not a multiple of a period, doesn't wildly oscillate). However, I think that because the integrand is bounded from both sides by a function that monotonously converges to zero, it shouldn't be a problem.

The $\int_1^{\pi/2}$ part is finite and trivially convergent.

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The integral is conditionally convergent.

Hint: To show the convergence, use integration by parts and absolute convergence of $\sin t/t^2$ (which would imply its convergence)

To show the integral is not absolutely convergent, break the integral into many parts. $\int_1^{\pi}+\sum_{n=2}^{\infty} \int_{(n-1) \pi}^{n \pi}$ for n=2,3,4,..... Now, in the interval $\int_{(n-1) \pi}^{n \pi}$ $\lvert\frac{\cos y}{t}\rvert dt$ is greater than equal to $\int_{(n-1) \pi}^{n \pi}$ $\lvert\frac{\cos t}{k\pi}\rvert dt$ (the summation of which diverges)

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