Proving divergence of $\int_1^{\infty}e^{-\arctan(x)}\ \mathrm{d}x$

Question

I wish to show that the integral: $$\int_1^{\infty}e^{-\arctan(x)}\ \mathrm{d}x$$

diverges.

I tried to use the different comparison tests but couldn't get to anything.

I would like to get a hint, and hopefully continue from there on my own.

Answer 1

Note that\begin{align}\lim_{x\to+\infty}e^{-\arctan x}&=e^{-\lim_{x\to+\infty}\arctan x}\\&=e^{-\frac\pi2}.\end{align}So, take $M>0$ such that $M<e^{-\frac\pi2}$. Then, compare your integral with $\int_1^{+\infty}M\,\mathrm dx$.

Answer 2

The substitution $x=\tan t$ rewrites the integral as $$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}e^{-t}\sec^2(t)\ \mathrm{d}t\ge e^{-\pi/2}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\sec^2(t)\ \mathrm{d}t=\infty.$$