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could anyone tell me how to solve this problem? I am bit bad in integration, in the middle of some problem I come here and stuck

$$\int_{R=0}^{\infty}\int_{\theta=0}^{2\pi}R\sin\theta \cos(gR\sin\theta)dRd\theta$$

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The integral is not absolutely convergent (i.e., does not exist in the Lebesgue sense). On the other hand, for any finite disk $R\leq R_0$ you get $0$, because the integrand is an odd $2\pi$-periodic function of $\theta$. –  Christian Blatter Sep 22 '12 at 13:05
    
how to show that it is not absolutely convergent? –  miosaki Sep 22 '12 at 13:41

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The integral is not absolutely convergent: For ${\pi\over6}\leq \theta\leq{\pi\over3}$ one has $\sin\theta\geq{1\over2}\geq {1\over2}\cos\theta$. Therefore $$\eqalign{\int_0^{2\pi}|\sin\theta||\cos(\lambda\sin\theta)|\ d\theta& \geq \int_{\pi/6}^{\pi/3}\sin\theta\ |\cos(\lambda\sin\theta)|\ d\theta\cr &\geq {1\over2} \int_{\pi/6}^{\pi/3}\cos\theta\ |\cos(\lambda\sin\theta)|\ d\theta \cr &={1\over2} \int_{1/2}^{\sqrt{3}/2}|\cos(\lambda u)|\ du\ .\cr}$$ For $\lambda\gg1$ the right side has order of magnitude $${1\over2}\Bigl({\sqrt{3}\over2}-{1\over2}\Bigr){2\over\pi}=:c\doteq0.1165\ ,$$ independently of $\lambda$. So in the end we are faced with the integral $\int_0^\infty c\, R\ dR$, which is certainly divergent.

On the other hand, for any finite disk $R≤R_0$ the original integral has value $0$, because the integrand is an odd $2\pi$-periodic function of $\theta$.

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