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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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    @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