# Degree choice in improper integrals resulting in trigonometric functions

people. I have a question regarding the following improper integral, and others like it: $$\int_{-\infty}^\infty \frac{dx}{1+x^2}$$ The end result of that are the two limits: $$\lim_{a\to -\infty} \big(\arctan(0)-\arctan(a)\big)$$ and $$\lim_{b\to \infty} \big(\arctan(b)-\arctan(0)\big)$$ Now, $$\arctan(-\infty)$$ is $$\frac{3\pi}{2}$$ and $$\arctan(\infty)$$ is $$\frac{\pi}{2}$$.

Applying those values to the formula would give a result of $$-\pi$$, but that makes no sense. In this particular case, the issue is resolved by rewriting $$\frac{3\pi}{2}$$ as $$-\frac{\pi}{2}$$.

My question is, is there a rule of thumb to picking radian values when dealing with integrals like this one or do simply pick whatever fits the specific problem to avoid ending up with a negative end value?

• In general the value of $\arctan(-\infty)$ is taken as $-\frac{\pi}{2}$ but it is an interesting question. Jan 21, 2019 at 0:07
• It is okay as soon as you choose $\arctan$ to be continuous on $\mathbb{R}$. Jan 21, 2019 at 0:13
• Arctan function is defined as the inverse of $\tan:\left]-\frac{\pi}{2},\frac{\pi}{2}\right[ \rightarrow \mathbb{R}$ (a monotonic continuous function). In particular, $\lim_{x\rightarrow \infty}\arctan x=\frac{\pi}{2}$ and $\lim_{x\rightarrow -\infty}\arctan x=-\frac{\pi}{2}$.
– FDP
Jan 21, 2019 at 13:45

As a rule of thumb, do not treat the numbers as if they are on the unit circle. In other words, instead of using $$\text{mod}(-\pi/2,2\pi)=3\pi/2$$, just use $$-\pi/2=-\pi/2$$. Or better yet note that $$\frac1{t^2+1}$$ is symmetric about $$t=0$$, so $$\int_{-\infty}^{\infty}\frac{\mathrm dt}{t^2+1}=2\int_0^\infty\frac{\mathrm dt}{t^2+1}=2\cdot\frac\pi2=\pi$$ Which gives the answer without any confusion over radians