# Improper integral

1. Mar 15, 2017

### Mr Davis 97

1. The problem statement, all variables and given/known data
Evaluate $\displaystyle \int_{1}^{\infty} \frac{dx}{(x+a)\sqrt{x-1}}$

2. Relevant equations

3. The attempt at a solution
First I make the substitution $u = \sqrt{x-1}$, which ends up giving me $\displaystyle \int_{0}^{\infty} \frac{2u}{u(u^2 + 1 + a)}du$. Here is where I am stuck. I am not sure if I am able to divide the u's since 0 is within the interval of the integral.

Last edited: Mar 15, 2017
2. Mar 15, 2017

### Ray Vickson

By definition:
$$\int_0^{\infty} \frac{2u}{u(u^2+a+1)} \, du = \lim_{r \to 0+, s \to \infty} \int_r^s \frac{2u}{u(u^2+a+1)} \, du,$$
and $u$ cancels between the numerator and denominator in the integral $\int_r^s.$

3. Mar 15, 2017

### Mr Davis 97

So since it's a limit we are allowed to cancel the u's, since at no point they actually equal 0?

Also, how long should I go in my computation until I explicitly write write the limits? It seems that if I carried is through the entire computation it would be very unwieldy.

4. Mar 15, 2017

### JoePhysics

The limit is in r and s, not the variable of integration. You can cancel out the u's just like you can cancel out the x in x/x^2 = 1/x.

5. Mar 15, 2017

### Staff: Mentor

The factors of u in the numerator and denominator represent a removable discontinuity at u = 0 -- a single point on the graph of the integrand at which the integrand is undefined. A "hole" in the graph, if you like. Integrating across the hole makes no difference in the result, unlike in the integral in another post of yours -- $\int_0^8 \frac{dx}{x - 2}$, where there is a nonremovable discontinuity at x = 2.

6. Mar 15, 2017

### Mr Davis 97

So you're saying I can cancel them because the area under the curve of the function with the removable discontinuity will be the same as the area of the function with this discontinuity removed? Is there a formal statement of this fact?

7. Mar 15, 2017

### Staff: Mentor

Yes. You can convince yourself by evaluating $\int_0^4 \frac{x - 2}{x^2 - 4}dx$. There is a removable discontinuity at x = 2.
Break up the integral as $\lim_{a \to 2^-}\int_0^a \frac{x - 2}{x^2 - 4}dx + \lim_{b \to 2^+}\int_b^4 \frac{x - 2}{x^2 - 4}dx$
I'm sure there is a formal statement about it somewhere, but the gist is that a function is integrable on a closed interval [a, b] if it is continuous on (a, b) or if there are only a finite number of jump discontinuities or removable discontinuities.