Solve $\displaystyle \int_{-\infty}^{\infty} dx \ \delta'' (x) \ \text{exp} \left( \dfrac{2}{1-x} \right)$.
Hi. I'm asked to solve the previous integral. I've tried to integrate by parts twice:
$\displaystyle \left. \int_{-\infty}^{\infty} dx \ \delta'' (x) \ \text{exp} \left( \dfrac{2}{1-x} \right) = \delta' (x) \ \text{exp} \left( \dfrac{2}{1-x} \right) \right|_{-\infty}^{\infty} - \int_{-\infty}^{\infty} \delta' (x) \dfrac{d}{dx} \text{exp} \left( \dfrac{2}{1-x} \right)$.
If the first integral goes to 0, then the result is
$\dfrac{d^2}{dx^2} \text{exp} \left( \dfrac{2}{1-x} \right) (0)$,
which is what I want to obtain. Therefore, I'm not sure if the first term of the integration by parts is, in fact, 0.