Hey thanks for stopping by , I would really appreciate your help in proving this:
If $f$ is continuous on $(-\infty,\infty)$ and if $\int_{-\infty}^\infty {f(x)dx}$ converges to A , prove that $$C.P.V. \int_{-\infty}^\infty {f(x)dx} = A$$ where C.P.V. stands for Cauchy Principal Value.
I thought of first taking the interval $(-s,s)$ which can later by varied to the original interval as $s$ approches $\infty$ and also thought of taking subdivisions in this interval so that i can discuss the integrability of $f$ in the component intervals of the subdivisions taken by using extreme value theoram in those component intervals , but all this is in my head and i don't think it is getting me anywhere . Please help me write the proof and also provide some pro-tip just to avoid rambling around in future. Thanks in Advance.
Suppose $\int_{-\infty}^{\infty} f(x) \, dx$ converges. That means that $\int_{-\infty}^{0} f(x) \, dx = B$ and $\int_{0}^{\infty} f(x) \, dx = C$ for some real numbers $B$ and $C$ such that $A=B+C$. This again means that for every $\epsilon >0$ there exit constants $N$ and $M$ such that
$$\bigg|\int_{s}^{0} f(x) \, dx - B \bigg| < \frac{\epsilon}{2} \quad \text{and} \quad \bigg|\int_{0}^{r} f(x) \, dx - C \bigg| < \frac{\epsilon}{2}$$
when $s \leq N$ and $r \geq M$. Now set $L = \max(|N|, |M|)$. For all $\rho \geq L$, we have
$$\bigg|\int_{-\rho}^{0} f(x) \, dx - B \bigg| < \frac{\epsilon}{2} \quad \text{and} \quad \bigg|\int_{0}^{\rho} f(x) \, dx - C \bigg| < \frac{\epsilon}{2}.$$
Furthermore, observere that
\begin{align*} \bigg| \int_{-\rho}^{\rho} f(x) \, dx -A \bigg| &= \bigg|\int_{-\rho}^{0} f(x) \, dx + \int_{0}^{\rho} f(x) \, dx - (B+C) \bigg| \\ &= \bigg|\int_{-\rho}^{0} f(x) \, dx - B + \int_{0}^{\rho} f(x) \, dx - C \bigg| \\ &\leq \bigg|\int_{-\rho}^{0} f(x) \, dx - B \bigg| + \bigg| \int_{0}^{\rho} f(x) \, dx - C \bigg| \\ &< \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon, \end{align*}
which shows that the Cauchy principal value converges to $A$.
However, as I pointed out in the comment section, this is easier to see by thinking about what convergence really means. The simple idea easily gets lost in this proof. The idea is that if the limit exits, you can take it in any way you want, i.e. you can approach $\infty$ and $-\infty$ at different rates. In particular, you can also approach them at the same rate, i.e. symmetrically, which is exactly what you do when you take the Cauchy principal value. Therefore the CVP has to converge to the integral.