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I have a Convolution integral $$ \int_{t_0}^{t} \int_{t_0}^{\tau} C(t-t')C(\tau -t'') \delta(t''-t') dt'' dt'=\int_{t_0}^{t} C(t-t')C(\tau -t') dt'= ? $$

I do not know how to proceed any further, $\int_{t_0}^{t} C(t-t')C(\tau -t') dt'= ?$ Anyone who can guide me about it? Thanks in advance.

$$C(t)=\int_{-\infty}^{\infty} d\omega e^{i\omega t} f(\omega) J(\omega)$$

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  • $\begingroup$ If $t=0,$ then obviously your integral is just a convolution. What do you know about $C?$. $\endgroup$ – Matematleta Dec 11 '18 at 16:04
  • $\begingroup$ C is given by $C(t)=\int_{-\infty}^{\infty} d\omega e^{i\omega t} f(\omega) J(\omega)$ $\endgroup$ – Skeptical Khan Dec 11 '18 at 17:12
  • $\begingroup$ The problem is that your integral is a function of both $t$ and $\tau.$ You have $C(t)=\mathcal F^{-1}(f\cdot J)(t)$ and so you want to calculate $\int_{t_0}^{t}\mathcal F^{-1}(f\cdot J)(t-t')\mathcal F^{-1}(f\cdot J)(\tau-t') dt'=\int_{0}^{t-t_0}\mathcal F^{-1}(f\cdot J)(z)\mathcal F^{-1}(f\cdot J)((\tau-t)-z)) dz.$ If $t$ and $\tau$ are related by $\tau=2t-t_0$, then you have a convolution of inverse Fourier transforms, and you can use standard formulas to simplify. Otherwise, I do not see how this expression can be simplified. $\endgroup$ – Matematleta Dec 12 '18 at 16:51
  • $\begingroup$ let me see, Thanks for your response. $\endgroup$ – Skeptical Khan Dec 13 '18 at 6:19
  • $\begingroup$ It should not be $$(-) \int_{t-t_0}^{0}\mathcal F^{-1}(f\cdot J)(z)\mathcal F^{-1}(f\cdot J)((\tau-t)+z)) dz.$$ because if $z=t-t'$ then $dz=-dt'$, for $t' \rightarrow 0$ $ z=t-t_0$ and if $ t' \rightarrow t $ then $ z=0$ $\endgroup$ – Skeptical Khan Dec 13 '18 at 7:13
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\begin{align*} \int_{t_0}^{t} \int_{t_0}^{\tau} C^\sigma(t-t') C^\sigma(\tau -t'') \langle \nu_{it'}\nu_{jt''} \rangle dt'' dt' \\& = \int_{t_0}^{t} \int_{t_0}^{\tau} C^\sigma(t-t') C^\sigma(\tau -t'') \delta_{ij}\delta(t'-t'') dt'' dt' \end{align*} For $i=j, \delta_{ij}=1 $ \begin{align*} = \int_{t_0}^{t} C^\sigma(t-t') C^\sigma (\tau -t') dt' \qquad\qquad \end{align*} Here, $C^\sigma(t) = \int_{-\infty}^{\infty} d\omega f^\sigma(\omega) J(\omega) e^{i\sigma\omega t}$. Solving the integral for $C^\sigma$ \begin{align*} \int_{t_0}^{t} C^{\sigma}(t-t')C^{\sigma}(\tau -t')dt' &= \int_{t_0}^{t} (\int_{-\infty}^{\infty} d\omega f^\sigma(\omega) J(\omega) e^{i\sigma\omega (t-t')})(\int_{-\infty}^{\infty} d\omega' f^\sigma(\omega') J(\omega') e^{i\sigma\omega' (\tau -t')}) dt' \\& = C^\sigma(t)C^{\sigma'}(\tau) \int_{t_0}^{t} e^{-i\sigma(\omega+\omega') t'}dt' \\& = C^\sigma(t)C^{\sigma'}(\tau) \frac{1}{-i\sigma(\omega+\omega')}(e^{-i\sigma(\omega+\omega') t} -e^{-i\sigma(\omega+\omega') t_0} ) \end{align*}

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  • $\begingroup$ The answer cannot depend on the bound variables $\omega$ and $\omega'$. $\endgroup$ – Maxim Dec 24 '18 at 14:59

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