I've searched and found almost nothing about how to integrate the integral containing $\delta^{(n)}$. I wanted to calculate $$\delta^{(n)}*\mathcal{F}_x(J_\alpha(x))$$ where $J_\alpha$ is the Bessell function type 2. I have calculated the Fourier transform of the $J_\alpha$ function and I wanted to calculate the convolution of one of its parts which is $$\delta^{(n)}*\lgroup\int\limits_0^\pi cos((n+1/2)\tau)(\delta(\omega-sin\tau))d\tau\rgroup$$ I searched in https://en.wikipedia.org/wiki/Dirac_delta_function#Distributional_derivatives and I thought that if we use the point mentioned we obtain again $$\int\limits_0^\pi cos((n+1/2)\tau)(\delta^{(n)}(\omega-sin\tau))d\tau$$ which contains $\delta^{(n)}$. I can't move forward and stuck here. Can anyone help me?
How to calculate an integral containing $\delta^{(n)}$?
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integration
distribution-theory
dirac-delta
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1The only way forward seems to be to integrate by parts $n$-times, to 'transfer' the $n$-th derivative from the delta function to the remaining chunk of the integrand... – 2017-02-11
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1@PierpaoloVivo I tried to integrate by parts as you said. $$\delta^{(n-1)}cos(...)|_0^\pi+\delta^{(n-2)}sin(...)|_0^\pi-\delta^{(n-3)}cos(...)|_0^\pi+...$$ is it right? (I mean for the Dirac delta function can we integrate by parts like this)? And also for the first term for example how can we calculate $\delta^{(n-1)}cos(...)|_0^\pi$? Can we say the $\delta^{(n)}$ is equal to $\delta$ at 0 and use $\delta$ instead of $\delta^{(n)}$? – 2017-02-11