# Improper integral from -infinity to infinity

• RedReaper
In summary, the improper integral of sin[x]/(1+x^2) from -infinity to infinity is equal to 0 because the function on the right has a finite integral on the interval (-\infty,\infty) and the integral of an odd function over a symmetric interval is equal to 0.
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?

RedReaper said:
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.

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

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

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?

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.

## What is an improper integral from -infinity to infinity?

An improper integral from -infinity to infinity is an integral where one or both of the limits of integration are infinite. This means that the function being integrated does not have a finite area under the curve, and the integral must be evaluated using a different method.

## How do you evaluate an improper integral from -infinity to infinity?

To evaluate an improper integral from -infinity to infinity, you need to use a specific technique called "limit of integration". This involves taking the limit as the upper and/or lower bound of the integral approaches infinity or negative infinity. If the limit exists, then that will be the value of the improper integral.

## Can an improper integral from -infinity to infinity be convergent?

Yes, an improper integral from -infinity to infinity can be convergent, meaning that it has a finite value. This can happen if the function being integrated approaches zero fast enough as x approaches infinity, or if it has oscillations that cancel out over the infinite interval.

## How do you determine if an improper integral from -infinity to infinity is convergent or divergent?

To determine if an improper integral from -infinity to infinity is convergent or divergent, you need to evaluate the limit of integration. If the limit exists and is a finite number, then the integral is convergent. If the limit does not exist or is infinite, then the integral is divergent.

## What are some real-world applications of improper integrals from -infinity to infinity?

Improper integrals from -infinity to infinity have various applications in physics, engineering, and statistics. For example, they can be used to calculate the total energy of a system, the center of mass of an object, or the probability of an event occurring within a given range. They are also commonly used in the study of infinite series and the behavior of functions at infinity.

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