For what values of K, is the integral improper?

Question

For what values of $K$ ($K > 0$), is the following integral improper?

$$\int_{0}^{K}\frac{x}{x^2-2}$$

Now, I know that the function is undefined at $x=\sqrt{2}$. I also figured out that the integral is $\frac{\ln(x^2-2)}{2}$ from $0$ to $K$. The answer is that it is improper in $[\sqrt{2},\infty]$. Can someone help me understand why this is so?

Answer 1

You normally can't integrate over a function that is not defined at a certain point in the area you are integrating over. Since $f(x)=\frac{x}{x^2-2}$ is not defined at $x=\sqrt{2}$, because we then divide by zero, you can't integrate over any interval that contains $x=\sqrt{2}$.

That means that if $K\geq \sqrt{2}$, the integral is improper.

If $K=2$, then we still have to pass $x=\sqrt{2}$, which makes the integral improper.

Answer 2

Within the scopes of Riemann integration theory, you are allowed to integrate a function $f \colon [a,b] \to \mathbb{R}$ that is bounded on the interval $[a,b]$. Otherwise, $\int_a^b f(x) \, dx$ is called "improper".

In your case, whenever $K >\sqrt{2}$ the function $f(x) = x/(x^2-2)$ is unbounded on $[0,K]$, and hence the integral becomes improper.

Answer 3

Also if $K$ is infinite the integral is said to be improper. As for your doubt of $K=2$, according to Riemann integral the range of integration must remain finite. When you have $K=2$ you pass $K= \sqrt 2$, and for that point the function which you are integrating becomes infinite, so the integral is improper for $K\geq \sqrt 2$.