# Evaluating integral from -∞ to ∞

1. Mar 17, 2013

### wahaj

When evaluating an integral $$\int^∞_{-∞}$$ do I have to split the integral up like $$\int^∞_0 + \int^0_{-∞}$$ and evaluate it this way or can I simply keep the integral in it's original form and plug in infinity and negative infinity into the expression and evaluate it?

2. Mar 17, 2013

### mathman

It depends on the expression you get for the indefinite integral. In practice most of the integrals of this nature require more than getting an antiderivative.

3. Mar 17, 2013

### wahaj

Well right now I am just learning how to solve improper integrals. I was working on $$\int^∞_{-∞} \frac{1}{2}tan^{-1} x^2$$. I don't know if I am doing this right but splitting up and keeping the integral together gets me two different answers. Splitting it up gets me the right answer

4. Mar 17, 2013

### Prove It

Since it's an even function, the property $$\displaystyle \int_{-a}^a{ f(x)\,dx } = 2\int_0^a{f(x)\,dx}$$ holds.

So that means you can evaluate it as $$\displaystyle 2\int_0^{\infty}{\frac{1}{2}\arctan{\left( x^2 \right)} \,dx}$$.

5. Mar 18, 2013

### HallsofIvy

You should know that $\int_0^\infty f(x)dx$ is defined as $\lim_{a\to\infty}\int_0^a f(x) dx$ while $\int_{-\infty}^\infty f(x)dx$ is defined as $\lim_{a\to -\infty}\lim_{b\to\infty} \int_a^b f(x)dx$.

In particular, you cannot use the same variable, and the same limits for the two limits of integration. That is, the "Cauchy Principle Value",
$$\lim_{a\to\infty} \int_{-a}^a f(x)dx$$
may exist when the integral itself does not.

6. Mar 19, 2013

### wahaj

OK that's why I have been getting the wrong answer. I was using one variable as both limits. Thanks for the answer.