I am working on fundamental solutions and I need to solve the distributional/weak derivative of the following which I have written in the formula for weak derivative,
$\int_{-\infty}^{\infty}\frac{1}{k}(sin(k(x-\xi))H(x-\xi)\varphi''(x)\,dx$ $\,$-$\,$ $\int_{-\infty}^{\infty}{k}(sin(k(x-\xi))H(x-\xi)\varphi(x)\,dx$
The above integral should equal the test function $\varphi(\xi)=\int_{\mathbb{R}}\delta(x-\xi)\varphi(x)\,dx$. Ive made it so far as to split the first term into two parts,
$\int_{-\infty}^{\xi}\frac{1}{k}(sin(k(x-\xi))H(x-\xi)\varphi''(x)\,dx$ and $\int_{\xi}^{\infty}\frac{1}{k}(sin(k(x-\xi))H(x-\xi)\varphi''(x)\,dx$
and for each term I'm trying integration by parts where
$u=sin(k(x-\xi))H(x-\xi),\,du=sin(k(x-\xi))\delta(x-\xi)\,+\,2H(x-\xi)cos(k(x-\xi))\,dx$
$v=\varphi'(x)\,,\,dv=\varphi''(x)\,dx$
From here, I'm only making the integration by parts worse for the other terms, can anyone help me in helping solve this integration... thanks in advance!
Here's part of the calculations. You can probably complete it yourself: $$ \int_{-\infty}^{\infty}\frac{1}{k}(\sin(k(x-\xi))H(x-\xi)\varphi''(x)\,dx =\int_{\xi}^{\infty}\frac{1}{k}(\sin(k(x-\xi))\varphi''(x)\,dx \\ = \left[\frac{1}{k}(\sin(k(x-\xi))\varphi'(x)\right]_{\xi}^{\infty} - \int_{\xi}^{\infty}(\cos(k(x-\xi))\varphi'(x)\,dx \\ = -\frac{1}{k}(\sin(k(\xi-\xi))\varphi'(\xi) - \int_{\xi}^{\infty}(\cos(k(x-\xi))\varphi'(x)\,dx \\ = - \int_{\xi}^{\infty}(\cos(k(x-\xi))\varphi'(x)\,dx \\ = -\left[(\cos(k(x-\xi))\varphi(x)\right]_{\xi}^{\infty} + \int_{\xi}^{\infty}(-k\sin(k(x-\xi))\varphi(x)\,dx \\ = (\cos(k(\xi-\xi))\varphi(\xi) - k \int_{\xi}^{\infty}(\sin(k(x-\xi))\varphi(x)\,dx \\ = \varphi(\xi) - k \int_{\xi}^{\infty}(\sin(k(x-\xi))\varphi(x)\,dx \\ $$