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$$\int \limits_{-\infty}^{\infty}\frac{e^{-|x|}}{|1-\sin x|^{\frac{1}{4}}} \,dx$$

Any advice and comments will be appreciated

  • 0
    It might be worth noting that the range for $\sin x$ is $-1\leq \sin x \leq 1$. Therefore $|1−\sin x|$ is always positive and so you can remove the absolute value in the denominator.2012-02-16
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    This is the integral on $(0,2\pi)$ of the same function without the absolute values, times $2/(1-\mathrm e^{-2\pi})$.2012-02-16
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    To see what Didier Piau means you might divide the interval into intervals of length $2\pi$ and use periodicity of the denominator together with the fact that $\exp(-x-2n\pi)= \exp(-2n\pi)\exp(-x)$ (giving a geometrical series).2012-02-16
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    @DidierPiau I see what you mean, but not if it helps..2012-02-16
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    Is this homework? For what class? Are you supposed to know complex analysis?2012-02-16

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