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?

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    $\begingroup$ In general the value of $\arctan(-\infty)$ is taken as $-\frac{\pi}{2}$ but it is an interesting question. $\endgroup$
    – Henry Lee
    Commented Jan 21, 2019 at 0:07
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    $\begingroup$ It is okay as soon as you choose $\arctan$ to be continuous on $\mathbb{R}$. $\endgroup$ Commented Jan 21, 2019 at 0:13
  • $\begingroup$ 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 $\lim_{x\rightarrow -\infty}\arctan x=-\frac{\pi}{2}$. $\endgroup$
    – FDP
    Commented Jan 21, 2019 at 13:45

1 Answer 1


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


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