# Convergence of infinite series of sinusoid with product of two improper integrals in its argument

I'm dealing with an Analysis I test, and online I found some of the tests our teacher handed out in the past. I tried this exercise, but I desisted immediately:

$$\sum^{\infty}_{n=3} \sin\left[ 2\pi n^2 + \left( \int^{∞}_{\ln^a n} \arctan\left(t\right)\sin^3\left(\frac{1}{t}\right) \, dt \right) \left( \int^{∞}_{\ln n} \frac{\arctan\left(t^2\right)}{e^t+1}\,dt \right) \right]$$

I'm dealing with Analysis I test and I found online some of the old tests our teacher handed out in the past, and I'm basically stone dead for how impossible they look like. Then I tried to have a try on this exercise, but I desisted immediately:

And I'm required to find the values of $a$ for which that monster converges.

I'm pretty sure the integrals are not expressible in terms of known functions, so I can't even get it to a "traditional" thing.

The first thing I thought is that at least the $\left( \int^{∞}_{\ln^a n} \arctan\left(t\right)\sin^3\left(\frac{1}{t}\right) \, dt \right) \left( \int^{∞}_{\ln n} \frac{\arctan\left(t^2\right)}{e^t+1}\,dt \right)$ part might give the monster a chance to converge, must be and integer value, so that $\lim_{n\to\infty} \sin\left( 2πn^2 + \text{integer}\right) = 0$ and the general term is infinitesimal.

Any suggestion on how to go on?

• $\sin x$ can be obtained by typing "\sin x" between dollar signs. Same for $\ln x, \arctan x$
– zhw.
Sep 7, 2017 at 16:19
• You need to asymptotically estimate the integrals. For instance the first with a lower bound of b for large b is on the order of $1/b^2$: why is that?
– Ian
Sep 7, 2017 at 16:34
• @Ian: For the $\sin^3(\frac{1}{t})$ I feel ok, cause when $x->∞, \frac{1}{x} -> 0$, and so $\sin^3(\frac{1}{t}) = (\sin(\frac{1}{t})) ^3 ∼ (\frac{1}{x})^3$ On the other hand I'm not sure that $arctg(x) ∼ x$ when $x->∞$, as it should be $x->0$ Sep 7, 2017 at 16:40
• By the way, here's a MathJax tutorial Sep 7, 2017 at 19:24
• @ChaseRyanTaylor: Out of topic comment... Sep 7, 2017 at 21:10

Hint: The problem is designed to terrify you at first, but it's not nearly as hard as it looks.

First, you can ignore the $2\pi n^2$ by periocity. Second, consider that first integral. Since $\arctan t \to \pi/2$ as $t\to \infty,$ and $\sin^3(1/t)\sim 1/t^3$ for large $t,$ this integral looks very much like

$$(\pi/2)\int_{\ln^a n}^\infty \frac{1}{t^3}\, dt$$

for large $n.$ That integral can be done easilty. The second integral will look like

$$(\pi/2)\int_{\ln n}^\infty e^{-t}\, dt$$

for large $n.$

Thinking along these lines should help clear up this problem for you.

• How come can I safely take out from the integral $\frac{\pi}{2}$ ignoring $\ln^a(n)$? Sep 7, 2017 at 16:44
• All you need is that there exist $c,C>0$ such that the first integral lies between $$c\int_{\ln^a n}^\infty \frac{1}{t^3}\, dt,\,\,\,\, C\int_{\ln^a n}^\infty \frac{1}{t^3}\, dt$$
– zhw.
Sep 7, 2017 at 17:26