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I need ideas for solve this improper integral, i know is hard and is a bonus for my analysis course, so i would really appreciate your help, thanks

$$\int_{1}^{\infty}\dfrac{x\sin(2x)}{x^2+3}dx$$

Hint: $$\begin{cases} \sin(\theta)\geq \dfrac{2\theta}{\pi},& 0\leq\theta\leq \dfrac{\pi}{2}\\\\\sin(\theta)\geq \dfrac{-2\theta}{\pi}+2,& \dfrac{\pi}{2}\leq\theta\leq{\pi}\end{cases}$$

In order to bound the integral

$$\int_{0}^{\pi} e^{-2R\sin(\theta)}d\theta$$

I don't know a nice and beauty approach in order to attack this properly to obtain a closed answer, so....

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1  
Your last integral is, with some minor changes, related to one of the the definitions of $I_0(\cdot)$ the modified Bessel function of the first kind and order $0$. See here for additional information. –  Dilip Sarwate Sep 14 '12 at 19:56
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Do you want a closed form or just to show convergence? –  Robert Israel Sep 14 '12 at 19:59
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From $0$ to $\infty$ it might have a nice solution. Are you sure you want to have the the bounds indicated in your question? –  Fabian Sep 14 '12 at 21:35
    
A closed form, is suitable for me –  Sebastian Griotberg Sep 14 '12 at 21:59
    
is there any alternative for this? –  Sebastian Griotberg Sep 15 '12 at 11:45

1 Answer 1

up vote 2 down vote accepted

Hint: Use $$ \int_1^\infty \frac{x \sin(2x)}{x^2+3} \mathrm{d} x = \Im \int_1^\infty \frac{x \mathrm{e}^{2 i x} }{x^2+3} \mathrm{d} x = \frac{1}{2} \Im \int_1^\infty \mathrm{e}^{2 i x} \left( \frac{1}{x - i \sqrt{3}} - \frac{1}{x + i \sqrt{3}} \right) \mathrm{d} x $$ Then check out the definition of the exponential integral.

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