Consider the Duhamel's convolution $f*g$ of $f$ and $g$ defined by \begin{gather*} f*g(x)=\int_{0}^{x}f(x-\tau)g(\tau) d \tau. \end{gather*} Let $\hat{f}_c$ be the Fourier cosine transform of $f,$ that is, \begin{gather*} \hat{f}_c(\lambda)=\sqrt{\frac{2}{\pi}}\int_{0}^{+\infty}f(x)\cos(\lambda x) d x,\qquad x>0. \end{gather*} The inverse Fourier cosine transform $\check{g}$ of $g(\lambda)$ is essentially the same, which is defined as \begin{gather*} \check{g}_c(x)=\sqrt{\frac{2}{\pi}}\int_{0}^{+\infty}g(\lambda)\cos(\lambda x) d\lambda. \end{gather*} My question is: Is it true that \begin{gather*} \sqrt{\frac{2}{\pi}}f*g(x)=\left(\hat{f}_c\cdot\hat{g}_c\right)^{\check{~}}_c \end{gather*} for every $f$ and $g$ who behavior well enough?
I thought it is true, so I tried to prove it as follows. \begin{align*} &\big(f*g(t)\big)^{\hat{ }}_c=\sqrt{\frac{2}{\pi}}\int_{0}^{+\infty}\int_{0}^{t}f(t-\eta)g(\eta)d \eta \cdot \cos(t\lambda)d t\\ =& \sqrt{\frac{2}{\pi}}\int_{0}^{+\infty}\int_{0}^{+\infty}f(s)\cos(\lambda\eta+\lambda s) d s\cdot g(\eta)d \eta \\ =&\sqrt{\frac{2}{\pi}}\int_{0}^{+\infty} f(s)\cos(\lambda s)d s\cdot \int_{0}^{+\infty}g(\eta)\cos(\lambda \eta)d\eta\\ &-\sqrt{\frac{2}{\pi}}\int_{0}^{+\infty}\int_{0}^{+\infty}f(s)\sin(\lambda s)g(\eta)\sin(\lambda \eta)d\eta d s. \end{align*} Thus, if $\sqrt{\frac{2}{\pi}}\int_{0}^{+\infty}\int_{0}^{+\infty}f(s)\sin(\lambda s)g(\eta)\sin(\lambda \eta)d\eta d s=0,$ then the desired result follows. But is it true that $\sqrt{\frac{2}{\pi}}\int_{0}^{+\infty}\int_{0}^{+\infty}f(s)\sin(\lambda s)g(\eta)\sin(\lambda \eta)d\eta d s=0?$
PS: This question comes from the possible alternative solution of the Cauchy problem \begin{gather*}\tag{ODE} \begin{cases} w''(t)+\mu w(t)=f(t), \quad t>0,\\ w(0)=0, w'(0)=b, \end{cases} \end{gather*} where $\mu>0.$ I have solved this problem by using Duhamel's principle, and the method of variation of parameters. Thus, I want to try to solve by using Fourier cosine transform. I've finished calculating \begin{gather*} w(t)=\frac{b\sin(\sqrt{\mu}t)}{\sqrt{\mu}}+\left(\frac{1}{\mu-\lambda^2}\hat{ f}_c(\lambda)\right)^{\check{~ }}_c, \end{gather*} and the inverse Fourier cosine transform of $\frac{1}{\mu-\lambda^2}$ is \begin{gather*} \sqrt{\frac{\pi}{2\mu}}\sin(\sqrt{\mu}t). \end{gather*} Hence, if the answer to my question is yes, then the desired solution follows.
Add: After some thinking, my previous guess is wrong! Indeed, to solve (ODE), it suffices to construct a particular solution of the corresponding inhomogeneous ODE $w''(x)+\mu w(x)=f(x),$ by applying Fourier cosine transform. Apparently, this method is cumbersome, comparing to, for example, Laplace transformation, or Duhamel's principle.