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I'm going to show this:

$$\lim_{x\rightarrow ∞}\,\,\,\int_0^x \! \frac{\arctan{t}}{t} \, \mathrm{d} t = ∞$$

We're working on L.Hopital and integrals... I guess it's very easy. But I struggle... Don't need to solve it for me if you give just give me guidelines. Thank you so much!

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    $$\arctan\,u=\frac{\pi}{2}-\arctan\frac1{u}$$ might be useful...2011-10-27

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Hint: $\arctan(t)\ge\frac{\pi}{4}$ when $t\ge1$

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Here's one possible hint: $\arctan t \to \pi/2$, so $\arctan t \ge 1$ from some point on (say for $t\ge a$). What can you say about $\lim_{x\to\infty} \int_a^x \frac{dt}{t}$?

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    Thank you! I still very unsure. arctan(t) goes to pi/2 when x->infinity. 1/t goes to 0 when x->infinity. But the expression will never actually become 0. So the area under the graph will -> infinity. But I have no idea how to prove it. I guess it's not enought just writing that...?2011-10-27
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    @IQlessThan70 That the function never becomes 0 is not enough to show that the integral diverges. But you can easily find $\int_0^x \mathrm{constant}/t\;dt$.2011-10-27
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    Seriously? Can I just set that arctan t to a constant?! ln to doesn't work. Can I change the integral to: $$\lim_{x\rightarrow ∞}\,\,\,\int_1^x \! \frac{c}{t} \, \mathrm{d} t = ∞$$ ? Thank you so much!2011-10-27
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    @IQlessThan70 Just to be clear, you're not really setting $\arctan(t)$ to a constant, you're using the fact that eventually, for x large enough (say $x>b$), $\arctan(t)>1$. Now you have that $arctan(t) / t > 1/t$, and you can integrate both sides from $b$ to $x$. At this point, you're using the comparison theorem to show that since $\int_1^{\infty} \frac {dt} t$ diverges, the integral you're interested in also diverges. Geometrically this is because the area under $\arctan(t)/t$ has larger area than the area under $1/t$ between $b$ and any $x>b$.2011-10-27
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    I'm pretty sure we haven't learned anything about comparison theorem... I'm afraid if they see me use something like that they may feel like I'm cheating...? Or what do you think? But what you wrote was very evident! Thank you :)2011-10-28
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    @IQlessThan70 You can also prove it rigorously, "from the ground up". Taking off from what I said in my earlier comment, you integrate the inequality from $b$ to $x$: $$\int_b^x \frac {dt} t < \int_b^x \frac {\arctan(t)} t dt$$ $$\implies \log (x) - \log(b) < \int_b^x \frac {\arctan(t)} t dt$$ Since the left side is unbounded above as $x\to\infty$, you can use the definition of limit and prove the result by contradiction.2011-10-28