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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... –  J. M. Oct 27 '11 at 15:05

2 Answers 2

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...? –  IQlessThan70 Oct 27 '11 at 16:08
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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 $\int_0^x \mathrm{constant}/t\;dt$. –  Michael Hardy Oct 27 '11 at 17:09
    
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! –  IQlessThan70 Oct 27 '11 at 18:57
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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 $x>b$. –  process91 Oct 27 '11 at 23:17
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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. –  process91 Oct 28 '11 at 2:36

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