Checking the convergence of the improper integral

Question

So I have the following function to which i should check the convergence: $$\int_{1}^{+\infty} \frac{\arctan(-e^{x})}{\sin(\frac{1}x{})}$$

I thought that one way to do it was with a comparison test. I thought that the values of the function $\arctan(-e^x)$ lie between $\frac{\pi}{2}$ and $0$. Therefore:

$$ \int_{1}^{+\infty} \frac{\arctan(-e^{x})}{\sin(\frac{1}x{})} \leq \int_{1}^{+\infty} \frac{0}{\sin(\frac{1}{x})} \leq 0 $$

And since the integral of $0$ converges than by comparison, our integral, which is smaller, must also converge. Would this logic make any sense?

Thank you in advance for your help,

Annalisa

Answer 1

The comparison test would say if $0 \le a(x) \le b(x)$ and $\int_1^\infty b(x)\; dx$ converges then so does $\int_1^\infty a(x)\; dx$. But in this case $a(x) < 0$.

Answer 2

If you take into account @Robert Israel's comment $$I_p=\int_{1}^{p} \frac{\arctan(-e^{x})}{\sin \left(\frac{1}{x}\right)}\,dx\sim -\frac \pi 2 \int_{1}^{p} x \,dx=-\frac \pi 4 (p^2-1)\tag 1$$ Pushing the expansion of $$\sin \left(\frac{1}{x}\right)=x+\frac{1}{6 x}+\frac{7}{360 x^3}+O\left(\frac{1}{x^5}\right)$$ you would get $$I_p\sim -\frac{\pi \left(360 p^4-353 p^2+120 p^2 \log (p)-7\right)}{1440 p^2} \tag 2$$