Is this relation true about Fourier cosine transform and Duhamel's convolution 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.
 A: The reason convolution works is that is a way of collecting objects of like kind. For example,
$$
              \sum_{n=0}^{\infty}a_n x^n\sum_{n=0}^{\infty}b_n x^n = \sum_{n=0}^{\infty}\left(\sum_{j+k=n}a_j b_k\right)x^n \\
     \sum_{j+k=n}a_j b_k = \sum_{l=0}^{n}a_{l}b_{n-l}
$$
Similarly,
$$
     \int_{0}^{\infty}e^{st}f(t)dt\int_{0}^{\infty}e^{st}g(t)dt= \int_{0}^{\infty}\left(\int_{0}^{t}f(u)g(t-u)du\right)e^{st}dt
$$
Cosines are a little trickier. Let $\hat{f}$ denote the Fourier cosine transform. The question is:
$$
     \int_{0}^{\infty}\hat{f}(s)\cos(st)ds\int_{0}^{\infty}\hat{g}(s')\cos(s't)ds' = ?
$$
Gathering like terms requires use of the identity
$$
    \cos(st)\cos(s't)=\frac{1}{2}\{\cos((s+s')t)+\cos((s-s')t)\}.
$$
Collecting $s+s'=constant$ gives a convolution. The second terms require you to collect $|s-s'|=constant$, the absolute value coming from the fact that $cos(-a)=cos(a)$. So
$$
    ? = \frac{1}{2}\int_{0}^{\infty}\left(\int_{s'+s''=s}\hat{f}(s')\hat{g}(s'')+\int_{|s'-s''|=s}\hat{f}(s')\hat{g}(s'')\right)\cos(st)ds.
$$
As before,
$$
       \int_{s'+s''=s}\hat{f}(s')\hat{g}(s'') = \int_{0}^{s}\hat{f}(s')\hat{g}(s-s')ds'.
$$
And it would appear to me that
$$
    \int_{|s'-s''|=s}\hat{f}(s')\hat{g}(s'') = \int_{s'-s''=s}\hat{f}(s')\hat{g}(s'')+\int_{s''-s'=s}\hat{f}(s')\hat{g}(s'') \\
    = \int_{0}^{\infty}\hat{f}(s''+s)\hat{g}(s'')ds''+\int_{0}^{\infty}\hat{f}(s')\hat{g}(s'+s)ds'
$$
So my guess is that convolution is replaced by
$$
    \hat{f}\star_{c}\hat{g} = \frac{1}{2}\left[\int_{0}^{s}\hat{f}(s')\hat{g}(s-s')ds'+\int_{0}^{\infty}\hat{f}(s''+s)\hat{g}(s'')ds''+\int_{0}^{\infty}\hat{f}(s')\hat{g}(s'+s)ds'\right]
$$
Then there is reason to suspect that
$$
      f(t)g(t) = \int_{0}^{\infty}(\hat{f}\star_{c}\hat{g})(s)\cos(st)ds.
$$
But that's just my guess.
