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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.

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    $\begingroup$ Isn't the last integral the product of the sine Fourier transform of $f$ by the transform of $g$? $\endgroup$ – Henrique Augusto Souza Apr 28 '17 at 12:33
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    $\begingroup$ @HenriqueAugustoSouza, Not really. But it is, up to a constant. But this seems having nothing to do with my question. If $f$ is an even function on $\mathbb{R},$ then, in the derivation of Fourier cosine transform, I think the last integral is zero. $\endgroup$ – azc Apr 28 '17 at 12:37
  • $\begingroup$ I see.. so you`re interested in the conditions for the last integral to vanish? Those who would make the equality true? $\endgroup$ – Henrique Augusto Souza Apr 28 '17 at 12:48
  • $\begingroup$ That property of the transform (exchanging products with convolutions) is expected from the full Fourier transform, so I believe it isn't true in general when restricted to it's sine and cosine summands. Sure, if just so happens that one of the summands is zero (such as when $f$ and $g$ are even, then the sine transform is zero), we get the equality. But I`m not sure that it's a necessary condition, we may have a weaker hypothesis. $\endgroup$ – Henrique Augusto Souza Apr 28 '17 at 12:54
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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.

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