# Improper integral

## Homework Statement

Evaluate ##\displaystyle \int_{1}^{\infty} \frac{dx}{(x+a)\sqrt{x-1}}##

## 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:

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

Evaluate ##\displaystyle \int_{1}^{\infty} \frac{dx}{(x+a)\sqrt{x-1}}##

## 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)}dx##. 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.

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. ##

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. ##
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.

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.
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.

Mark44
Mentor
So since it's a limit we are allowed to cancel the u's, since at no point they actually equal 0?
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.
Mr Davis 97 said:
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.

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.
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?

Mark44
Mentor
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?
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##
Mr Davis 97 said:
Is there a formal statement of this fact?
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.