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Prove the following equation:

\begin{equation} \frac{2}{\pi} \int_0^\infty \frac{\cos kr - ak \sin kr}{k^2a^2 +1} \left (\int_0^\infty \cos kr' \left [u(r')-au'(r') \right] dr' \right ) dk =u(r) . \end{equation} where $u,u' \in L^1(\mathbb{R}^+)$ and $a,r >0$.

I have found the following not rigorous proof. By reversing the order of integration (this is the not rigorous part) one obtains: $$ \ldots = \frac{2}{\pi} \int_0^\infty \left ( \int_0^\infty \frac{\cos kr\cos kr' - ak \sin kr \cos kr'}{k^2a^2 +1} dk \right ) [u(r')-au'(r') ] dr'; $$ By using the Werner formulas, the integral in $dk$ becomes: \begin{align*} & \int_0^\infty \frac{\cos k(r + r') + \cos k(r - r') - ak \sin k(r+r') - ak \sin k(r-r')}{2(k^2a^2 +1)} dk. \end{align*} Given the following integrals: $$ \int_0^\infty \frac{\cos kx}{a^2 k^2 +1} dk = \frac{\pi}{2} \frac{e^{-|x/a|}}{a}; \; \; \int_0^\infty \frac{ak \sin kx}{a^2 k^2 +1} dk = \frac{\pi}{2} \frac{\text{sgn}(x)e^{-|x/a|}}{a}; $$ the complete formula becomes: \begin{align*} &\ldots = \int_0^\infty \Theta(r'-r) \frac{e^{-(r'-r)/a}}{a}[u(r')-au'(r') ] dr' = \\ & - \int_r^\infty \partial_{r'}[ e^{-(r'-r)/a}u(r') ]dr'= - e^{-(r'-r)/a}u(r') \Big |_{r'=r}^\infty = u(r). \end{align*}

I think that reversing the order of integration is not allowed, because the conditions for Fubini are not satisfied. In fact, probably:

\begin{equation} \int_0^\infty \left |\frac{\cos kr\cos kr' - ak \sin kr \cos kr'}{k^2a^2 +1} \right | dk = \infty. \end{equation}

Can somebody give a rigorous proof of the equation?

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  • $\begingroup$ If I could prove it with the additional assumption that $r\mapsto ru(r)$ and $r\mapsto ru'(r)\in L^1(\mathbb R^+)$ would it be interesting for you? $\endgroup$ – Sebastien B Jul 10 '13 at 15:58
  • $\begingroup$ Yes, it would be. $\endgroup$ – BGA Jul 11 '13 at 22:10

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