A problem when integrating $\int \frac{a}{x^2 + 1}\text{ } dx$

Question

I’m still not an expert in integration, and I’m still learning, so sorry if it’s a trivial question:

Here’s the general integral:

$$\int \frac{a}{x^2 + 1}\text{ } dx, a \in \mathbb R$$

I know that this antiderivative is $a \tan^{-1} (x)$.

Now, I want to compute

$$\int_{-\infty}^{\infty} \frac{a}{x^2 + 1}\text{ } dx = 2\int_{0}^{\infty} \frac{a}{x^2 + 1} \text{ } dx$$

This basically means

$$\left(a \lim\limits_{T\to \infty} \tan^{-1}(T)\right) - 0$$

I’ve heard a lot of people say it’s $\frac{a\pi}{2}$.

HOWEVER, the tangent also approaches infinity at any $x = \frac{a(2n+1)\pi}{2}$.

So really the entire area under that curve from negative to positive infinity, should be $a\pi$ but is equal to $a(2n+1)\pi = 2na\pi + a\pi$ by integrating algebraically.

What am I missing, and what should we do for other situations like this where the periodicity of the antiderivative can give multiple solutions?

Answer 1

$\tan^{-1}(x)$ is not just any $y$ with $\tan(y) = x$, it is specifically the one in the interval $(-\pi/2, \pi/2)$. So yes, $$ \int_0^\infty \frac{dx}{1+x^2} = \lim_{x \to +\infty} \tan^{-1}(x) = \frac{\pi}{2}$$

Answer 2

In ITF the range of $tan^{-1}x$ is widely considered as from $(\frac{-\pi}{2}, \frac{\pi}{2})$ even though like you said $tanx$ has multiple solutions for any $x = \alpha$. However for $tan^{-1}(\alpha)$ we only consider the solution that lies in the mentioned range. The reason is because when we choose to invert a function we make sure its bijective in the given domain. Since $tanx$ isnt bijective in its domain we just take a sub-set of its domain where its bijective and then find its inverse corresponding to that part only. The widely accepeted sub-set of the domain $tanx$ we used to find its inverse is $(\frac{-\pi}{2}, \frac{\pi}{2})$.

However for the issue your facing its not necessary to follow the strict range given above, but we must make sure its bijective. For example in an interval from $(\frac{(2n-1)\pi}{2},\frac{(2n+1)\pi}{2}$) $tanx$ is always bijective. So on integrating and applying the limits we get $2(\frac{(2n+1)\pi}{2} -n\pi)$ which is $\pi$.