Evaluate $\int_{-\infty}^{\infty}\frac{dx}{1+x^2}$

Question

$$\int_{-\infty}^{\infty}\frac{dx}{1+x^2}=\lim_{t\to \infty}\int_{-t}^{t}\frac{dx}{1+x^2}=\lim_{t\to \infty}\tan^{-1}(t)-\tan^{-1}(-t)=\pi$$

I can not do the following $\int_{-\infty}^{\infty}\frac{dx}{1+x^2}=\lim_{t\to \infty}\int_{-t}^{0}\frac{dx}{1+x^2}+\int_{0}^{t}\frac{dx}{1+x^2} $ as I do not know if the integral converges?

Is there a way to evaluate $\lim_{t\to \infty}\tan^{-1}(t)$? (I just know the graph)

Answer 1

$$I=\displaystyle\int \limits^{\infty }_{-\infty }\frac{1}{1+x^{2}} dx=2\int \limits^{\infty }_{0}\frac{1}{1+x^{2}} dx=2\times \arctan \left( x\right) ^{\infty }_{0}=2\times \frac{\pi }{2} =\pi $$

Answer 2

$\displaystyle\lim_{x\to(\pi/2)^-}\tan(x)=\infty$, so $\displaystyle\lim_{x\to\infty}\tan^{-1}(x)=\frac\pi2$.

Answer 3

First of all be careful. It is not necessary true that:

$$\int_{-\infty}^{\infty} f(x)dx=\lim_{t \to \infty} \int_{-t}^{t} f(x) dx$$

We need to integrate from left to right:

$$\int_{-\infty}^{\infty} f(x) dx=\lim_{s \to -\infty}\int_{s}^{a} f(x)+\lim_{t \to \infty} \int_{a}^{t} f(x) dx$$

To be convinced that your approach is incorrect as it gives an incorrect answer to some integral so, consider $\int_{-\infty}^{\infty} x^3 dx$.

Try filling that are by taking a interval $x \in [-t,t]$ for your integral bounds and letting $t \to \infty$. Now try taking $x \in [-t,2t]$.

Next to address your question:

$$\lim_{x \to \infty} \arctan(x)$$

The way I view $\arctan(x)$ is a function that given some slope of a ray that points in a direction of the first quadrant and that crosses the origin by, it will return the angle created by the positive $x$ axis and the line. To see this just draw a line, make a right angle, and do some trig. You just just view it as a function that given some slope of a line returns the angle with the positive $x$ axis and the line, but a ray whose endpoint and is at the origin will simplify things. Anyways as $x \to \infty$ that means the slope is approaching $\infty$, i.e. It's line is becoming vertical (in the positive quadrant) and thus $\arctan(x)$ approaching $\frac{\pi}{2}$.

Ignore anything in this picture not in the first quadrant. Picture from geoan.com: http://www.geoan.com/recta/pendiente.html.