Exact expression of the resolvent kernel of $\frac{\partial^2}{\partial x^2}$? Let $\frac{\partial^2}{\partial x^2}$ be the Laplacian operator on $\mathbb R$ and denoted by $R(\lambda,t)$ its resolvent kernel. I would like to know the exact expression of $R(\lambda,t)$. 
Is $R(\lambda,t)= -\frac{1}{2i\lambda} e^{i t \lambda}$ or $R(\lambda,t)= \frac{1}{2i\lambda} e^{i t \lambda}$  or $R(\lambda,t) = \frac{1}{2\lambda} e^{-t \lambda}$  ?
Thank you in advance
 A: Assuming the underlying space in $L^2(\mathbb{R})$, the resolvent is $R(\lambda)f=(\frac{d^2}{dx^2}-\lambda I)^{-1}$, which means that, given $f\in L^2$, the function $g=R(\lambda)f$ is the unique solution of
$$
                  g''-\lambda g = f,\;\;\; g,g'\in L^2.
$$
You can construct the resolvent by solving for $h\in L^2$ such that
$$
                        h''(t)-\lambda h(t) = \delta_{x}(t).
$$
That means $h''-\lambda h=0$ for $t < x$ and for $t > x$. The function $h$ must be continuous at $t=x$, but a jump discontinuity in the derivative of $h$ of $+1$ at $t=x$ is required for the $\delta_{x}$ behavior. And $h$ must be chosen to decay as $t\rightarrow\pm\infty$. Using a branch cut for $\sqrt{\lambda}$ on the negative real axis, gives $\Re\lambda > 0$ for $\lambda\notin(-\infty,0]$. Then the square integrability condition gives
$$
             h_x(t) =\left\{\begin{array}{ll}
                           Ce^{\sqrt{\lambda}(t-x)}, & t < x \\
                           Ce^{-\sqrt{\lambda}(t-x)}, & t > x.
                          \end{array}\right.
$$
The constant $C$ must be chosen so that
$$
            1=h_{x}'(x+0)-h_{x}'(x-0)=-C\sqrt{\lambda}-C\sqrt{\lambda} \\
                  C = -2/\sqrt{\lambda}
$$
The solution $g=R(\lambda)f$ is
$$
          R(\lambda)f = \int_{-\infty}^{\infty}h_{x}(t)f(t)dt \\
   (R(\lambda)f)(x) = -\frac{2e^{-\sqrt{\lambda}x}}{\sqrt{\lambda}}\int_{-\infty}^{x}e^{\sqrt{\lambda}t}f(t)dt-\frac{2e^{\sqrt{\lambda}x}}{\sqrt{\lambda}}\int_{x}^{\infty}e^{\sqrt{-\lambda}t}f(t)dt.
$$
Unless I messed up a sign, you can directly verify that
$$
            (R(\lambda)f)''-\lambda R(\lambda)f = f.
$$
The singularities of the resolvent lie on the negative real axis, where the branch cut of $\sqrt{\lambda}$ is found. This is expected because $\frac{d^2}{dx^2}$ is a negative operator, while $-\frac{d^2}{dx^2}$ is positive.
