Principal value integral of $\frac{f(x)}{x}$ if f is continuous in [-1,1]

Question

Suppose $f : [-1,1] \rightarrow \mathbb{R}$ is continuous, show p.v. $\int_{-1}^{1} \frac{f(x)}{x}dx$ exists.

I know that since $f(x)$ is continuous on the closed bounded interval $[1,1]$, f is then Riemann integrable on $[-1,1]$. Also since $\frac{1}{x}$ is continuous on $[-1,-\epsilon]\cup[\epsilon,1]$, so I'm guessing that $\frac{f(x)}{x}$ is continuous on the interval $[-1,1]$ except $0$. Given the definition of the principal value integral, I am not sure of how to use it and I am wondering whether I am heading in the right direction, or if I am, how I can phrase it rigourously.

Another thing I tried doing was integration by parts by the p.v. definition but it got me nowhere, please help I'm stuck.

Appreciate any kind of help, thank you.

Answer 1

HINT: You need to work with continuity of $f$ at $0$. Given $\eta>0$, there is $\epsilon_0>0$ so that for $|x|<\epsilon_0$ we have $|f(x)-f(0)|<\eta$. This means that $f(0)-\eta<f(x)<f(0)+\eta$ for $-\epsilon_0<x<\epsilon_0$. Now work with $\int_{-1}^{-\epsilon} + \int_\epsilon^1$ for $\epsilon<\epsilon_0$.

Answer 2

It is kinda weird but, if you look at 1/X on [-b,-a] u [a,b] (having symetrical intervals is important there), it does have an integral equal to 0 (primitives are easy there) and the limit when a->0 can then give a sense to your integral i believe

Answer 3

Most theorems concerning Riemann integrals have equivalent theorems concerning the corresponding improper Riemann integrals. See pp. 266-267 of "Mathematical Analysis" (2nd ed.) by T. Apostol, for example.