# Homework Help: Improper integral from -infinity to infinity

1. Mar 2, 2012

### RedReaper

What is the improper integral of sin[x]/(1+x^2) from -infinity to infinity equal to? It doesn't seem to be integrable in the real number system.

Is it 0, because sin[x] is an odd function and 1+x^2 is an even function, and an odd function divided by an even function is equal to an odd function which has a rule that f(x) + f(-x) = 0?

2. Mar 3, 2012

### jbunniii

Why do you say that? We have

$$\left|\frac{\sin(x)}{1 + x^2}\right| \leq \frac{1}{1 + x^2}$$

and the function on the right has a finite integral on the interval $(-\infty,\infty)$. (Indeed, the integral equals $\pi$.) Therefore

$$\left|\int_{-\infty}^{\infty}\frac{\sin(x)}{1 + x^2} dx \right| \leq \int_{-\infty}^{\infty} \left|\frac{\sin(x)}{1 + x^2}\right| dx \leq \pi$$

Not all odd functions have zero integral. E.g. $\int_{-\infty}^{\infty}\sin(x)dx$ doesn't converge. However, any odd function for which $\int_{0}^{\infty}f(x)dx$ is finite will satisfy

$$\int_{-\infty}^{\infty} f(x) dx = \int_{-\infty}^{0} f(x) dx + \int_{0}^{\infty}f(x)dx = \int_{0}^{\infty}f(x)dx - \int_{0}^{\infty}f(x)dx = 0$$

So see if you can show that $\int_{0}^{\infty}\sin(x)/(1 + x^2) dx$ is finite, using a similar argument to what I wrote above.