# Thread: Checking whether integral exists

1. ## Checking whether integral exists

Okay, I have a fairly stupid question.

How do I check wheter integral
$\displaystyle \int_{-\infty}^{\infty} f(x) dx$ exists?
When I have
$\displaystyle \int_{a}^{b}f(x)dx$, I have to check that f is bounded and continuous almost everywhere. Do I have to do the same when I have $\displaystyle <-\infty, \infty>$? Is it enough to do the same?

Thank you.

2. If you are computing $\displaystyle \int_{- \infty}^{\infty} f(x)dx$, then the additional requirement of existence is that the following two limits exist and are finite: $\displaystyle \lim_{a \rightarrow -\infty} \int_{a}^{c} f(x)dx$ and $\displaystyle \lim_{b \rightarrow \infty} \int_{c}^{b} f(x)dx$, where c is some constant that is convenient for your function.

(Summed up from Wikipedia article on Improper Integrals)

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

If either integral on the right side diverges, then we say that

$\displaystyle \int_{-\infty}^{\infty}f(x)dx$ diverges.

Let f be continuous on some interval, say, [a,b]. With the exception of at

some point c satisfying a<c<b. f(x) becomes infinite as x approaches c from

the left or right. If the two improper integrals

$\displaystyle \int_{a}^{c}f(x)dx$ or $\displaystyle \int_{c}^{b}f(x)dx$

both converge, then we say that the improper integral

$\displaystyle \int_{a}^{b}f(x)dx$ converges. That way we define:

$\displaystyle \int_{a}^{b}f(x)dx=\int_{a}^{c}f(x)dx+\int_{c}^{b} f(x)dx$

See what I mean?. Is that what you were getting at?.

You can also say $\displaystyle \int_{a}^{b}f(x)dx=\lim_{L\to b^{-}}\int_{a}^{L}f(x)dx$

$\displaystyle \int_{a}^{+\infty}f(x)dx=\lim_{L\to {+\infty}}\int_{a}^{L}f(x)dx$

The thing is, check for convergence or divergence of the limit.

Take $\displaystyle \int_{-\infty}^{\infty}\frac{1}{1+x^{2}}dx$ for instance.

$\displaystyle \lim_{L\to {+\infty}}\int_{0}^{L}\frac{1}{1+x^{2}}dx$

$\displaystyle =\lim_{L\to {+\infty}}\left[tan^{-1}(x)\right]_{0}^{L}$

$\displaystyle =\lim_{l\to {+\infty}}tan^{-1}(L)=\frac{\pi}{2}$

The other side can be shown to be the same and we have $\displaystyle {\pi}$ as the solution.

But, we split the integral at x=0. We did not have to do that. We could have done it anywhere and not affected the convergence or divergence.

Does that help a wee bit?.

4. $\displaystyle \int_{-\infty}^{\infty} f(x) dx$

is equal with:

$\displaystyle \lim_{a\to\infty}\int_{-a}^c f(x) dx + \lim_{a\to\infty}\int_c^a f(x) dx$

5. Thank you all, you've been most helpful!
I think I got it. :-) The example galactus provided really sorted it out for me.