Find domain of function whose independent variable is the upper limit of an integral? I'm trying to answer the following question:
Find the domain of $F(x)$ where $$F(x)=\int_{5}^{x}\frac{1}{1-t^2}dt$$
Are we supposed to use some type of convergence test? I've heard of that for series, but not for integrals. When I search the Net, I only get results for improper integrals. However, this is not an improper integral, right?
Also, if I just evaluate the integral using the formula:
$$\int\frac{1}{1-t^2}dt=\frac{1}{2}\ln\bigg|\frac{t+1}{t-1}\bigg|+C$$
then I get the domain as $x\neq\pm1$, which is wrong. How can you explain that? Thanks!
 A: I figured it out. Let me answer my second question first. FTC2 only applies if the integrand is a continuous function on $[a,b]$. But since we have vertical asymptotes at $x=\pm1$, we cannot use $F(b)-F(a)$.
This is, in fact an improper integral. Integrals with an infinite upper or lower limit only make up one type of improper integral. But any integral with an unbounded region is improper. The graph of the integrand looks like this:

So we have three unbounded regions, and we need to test whether they converge or diverge. Let's start by testing the region on the right, by setting $x$ to $1$.
Since vertical asymptotes usually occur in rational functions, as in this example, we can sometimes test for convergence using a combination of a p-series test and comparison test. Let's try it. We need to factor out a negative, giving us: $$F(1)=-\int_{5}^{1}\frac{1}{t^2-1}dt$$
The idea here is to compare with the p-series $\frac{1}{t^2}$. However, since $\frac{1}{t^2-1}$ is larger than $\frac{1}{t^2}$, the comparison test doesn't apply and we have to try something else.
Let's try to evaluate the integral using partial fraction decomposition. After applying PFD, we get: $$F(1)=\frac{1}{2}\int_{5}^{1}\frac{1}{1-t}+\frac{1}{1+t}dt$$
Evaluating the integral gives us: $$\frac{1}{2}[\ln(1+t)-\ln(1-t)]_5^1$$
When we substitute the $1$, we end up with a $\ln(0)$ term, which is undefined. Therefore, the region on the right diverges.
Since the original function is defined with the lower limit of integration at $t=5$, which is to the right of the unbounded regions, and we have a divergence at $t=1$, then the domain is $\large \textbf{x>1}$.
