I am trying to solve/prove this Principal Value integral
$$ \mathscr{P}\int_{-\infty}^{+\infty}\dfrac{x^2\cdot \sin\left(x\right)}{x-a}{\rm{d}}x=\ -\ i\pi a^2\cdot\cos\left(a\right)\ ,$$
where $a$ takes a finite value.
The way I try to prove this is by noticing there pole is a first order pole at $x=a$, then by calculating the Residue at this value I find that
$$ \mathscr{P}\int_{-\infty}^{+\infty}\dfrac{x^2\cdot \sin\left(x\right)}{x-a}{\rm{d}}x =-i\pi \cdot Res\left(\dfrac{x^2\cdot \sin\left(x\right)}{x-a},a\right)= $$
$$ =-i\pi\cdot\lim_{x\rightarrow a}\left(x^2 \cdot \sin\left(x\right)\right) = -i\pi a^2 \cdot {\sin(a)}\ ,$$
which of course is not correct.
Hence, does anyone knows what I am doing wrong, and if yes could please explain it to me?
NOTE: Just to mention that i end up to this question because a made use of Green's function for quantum many-body system. So if anyone have a clue about this they might be able to help.
Thanks in advance.
******* EDIT:
I tried to keep it short, but the truth is that I need to prove this
$$-\dfrac{1}{a^3} \mathscr{P}\int_{-\infty}^{+\infty}\dfrac{x^3\cdot F(x)}{x-a}\dfrac{{\rm{d}}x}{2\pi} = G(a)\ ,$$
where
$$ F(x) = \frac{3}{2}\left[(1-y)\cdot \frac{\sin(x)}{x} + (1-3y)\cdot\left( \frac{cos(x)}{{x^2}} – \frac{sin(x)}{x^3}\right) \right]$$
and
$$ G(a) = \frac{3}{4}\left[-(1-y)\cdot \frac{\cos(a)}{a} + (1-3y)\cdot\left( \frac{sin(a)}{{a^2}} + \frac{cos(a)}{a^3}\right) \right],$$
where $y=const$.
Thanks again for the effort.
Best Answer
This integral is improper both at $a$ and at $\pm \infty$. We can attempt a principal value using $$ \lim_{\epsilon \to 0^+}\left(\int_{a+\epsilon}^{a+1} f(x)\;dx + \int_{a-1}^{a-\epsilon} f(x)\;dx\right) + \lim_{R \to +\infty}\left(\int_{a+1}^R f(x)\;dx + \int_{-R}^{a-1} f(x)\;dx\right) $$ the first limit exists. The second one diverges by oscillation. So I conclude this principal value does not exist.