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I'm looking to determine whether the following function is unbounded or not: $$ F(x) = \int_1^x\left|\frac{\cos t}{t}\right|\text{d} t $$ I can't seem to do much with it because of the $|\cos(t)|$. I thought of using the fact that $\int |f| \ge |\int f|$, but the problem is that the integral of $\frac{\cos t}t$ (without the absolute values) is bounded, and so that doesn't prove that $F(x)$ is unbounded or bounded. I tried re-expressing this as a cosine integral (the function $\text{Ci}(x)$) but to no avail. I'm not sure where else to go with this; the main problem seems to be the fact that its very difficult to derive an inequality with the $|\cos(t)|$ without a $|\cos(t)|$ on the other side of the inequality (or at least some trig function).

Any help would be appreciated.

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    @Rustyn That doesn't prove anything. All it proves is that this function is less than an unbounded one, which doesn't say anything about whether or not $F(x)$ is bounded. – 2017-02-04
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    One of the methods is such: consider the graph of the $\cos t/t$ and "hunches" on which the $x$ axis subdivide this graph. The area of each next hunch is less then of the previous and the sign is the opposite. But the series with alternating signs and monotonically decreasing term converges. Of course, this need to be made rigorous, and it can be. – 2017-02-04
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    Sorry @user3002473, I didn't finish my comment. – 2017-02-04
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    @Rustyn but the point is to prove that it is bounded by a constant – 2017-02-04
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    @Wolfram could you make that rigorous in an answer please? – 2017-02-04
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    Maybe integration term by term the power expansion. – 2017-02-04
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    ok, I'll try that – 2017-02-04
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    @Masacroso The problem with that, which is another method I attempted, is the absolute value signs make it difficult to deal with, and a repeated application of the reverse triangle inequality yields too crude an inequality to be used to prove that it is unbounded. – 2017-02-04
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    Ah, stop, the module is inside integral... Then it is unbounded, sorry. I know how to prove that either, though. – 2017-02-04
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    @Wolfram, sorry, you do know how to prove it or no? – 2017-02-04
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    I'm typing an answer right now, yes. – 2017-02-04
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    Using Fourier series and integration by parts it is not difficult to show that $$ \int_{1}^{x}\left|\frac{\cos t}{t}\right|\,dt \sim \frac{2}{\pi}\,\log(x).$$ – 2017-02-04

1 Answers 1

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Hint:

Consider the harmonic series and

$$\int_{\pi/2 + k\pi}^{3\pi/2 + k\pi} \frac{| \cos t|}{t} \, dt \geqslant \frac{1}{3\pi/2 + k \pi}\int_{\pi/2 + k\pi}^{3\pi/2 + k\pi} |\cos t| \, dt = \frac{2}{3\pi/2 + k \pi}$$

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    Ohhhhhh clever. I completely forgot about that trick. Thanks for the answer! – 2017-02-04
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    @user3002473: You're welcome. – 2017-02-05