I want to evaluate the improper integral $$\displaystyle\lim_{\epsilon \rightarrow 0}\int_0^1{\left( \frac{\phi \left( x \right)}{x+i\epsilon}-\frac{\phi \left( x \right)}{x-i\epsilon} \right)}dx
$$ where $\phi(x) : [0, 1] \rightarrow \mathbb{R}$ is a continuous function.
What I’ve tried is just simplify integrand as $\displaystyle \frac{2i\epsilon \phi(x)}{x^2 + \epsilon^2}$.
I just guess it can be evaluated by some complex analytic method since the integrand involves $i$. But I can’t see how to use what I know(basic residue theorem, cauchy’s integral formula, etc.).
Could you give a direction to solve this? I appreciate your comment and answer.
Best Answer
Note that we can write for $\varepsilon>0$
$$\begin{align} \int_0^1 \phi(x)\frac{-2i\varepsilon}{x^2+\varepsilon^2}\,dx&=-2i\int_0^{1/\varepsilon} \phi(\varepsilon x)\frac{1}{x^2+1}\,dx\\\\ &=-2i\int_0^\infty \phi(\varepsilon x)\frac{1}{x^2+1}\,\xi_{[0,1/\varepsilon]}(x)\,dx\\\\ \end{align}$$
Since $\left|\frac{1}{x^2+1}\,\xi_{[0,1/\varepsilon]}(x)\right|\le g(x)= \max_{[0,1]}\{\phi\}\frac1{x^2+1}$ and $\int_0^\infty g(x)\,dx<\infty$, the Dominated Convergence Theorem guarantees that
$$\begin{align} \lim_{\varepsilon\to0^+}\int_0^1 \phi(x)\frac{-2i\varepsilon}{x^2+\varepsilon^2}\,dx&=-2i\int_0^\infty \lim_{\varepsilon\to0^+}\phi(\varepsilon x)\frac{1}{x^2+1}\,\xi_{[0,1/\varepsilon]}(x)\,dx\\\\&=-i\pi\phi(0) \end{align}$$