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How can I evaluate this integral (without complex analysis)?

$$\int_{-\infty}^\infty\sinh [x(1-b)] \exp(iax) dx\qquad a, b\in \mathbb R$$

Thanks.

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    The way you have written this, it looks like $\sinh[a(1-b)]$ is just a constant which you can take outside of the integral. In this case the integral $\int_{-\infty}^{\infty} \exp(iax) dx$ does not converge. Maybe you meant to write something else.2011-11-17
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    Did you intend to have an $x$ somewhere in the $\sinh$? Yeah, what Aleks said ;-)2011-11-17
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    @AleksVlasev: Thanks, very well spotted! It's late and I'm losing it, anyway, I have edited it now.2011-11-17
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    @robjohn: You are very right! :-)2011-11-17
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    $\sinh(x)\to\pm\infty$ as $x\to\pm\infty$ so the integral doesn't converge, no complex analysis needed :-).2011-11-17
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    @robjohn: Ah, thanks!2011-11-17
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    @AleksVlasev $\large\int_{-\infty}^{\infty}{\rm e}^{{\rm i}ax}\,{\rm d}x = 2\pi\,\delta\left(a\right)$ ---> [See this link](http://en.wikipedia.org/wiki/Dirac_delta_function)2014-01-28

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This is the Fourier transform of $\sinh$, which can be broken down as a sum of exponentials. Because the Fourier transform is a linear operator, your integral is a sum of Fourier transforms of exponentials, that is a sum of Lorentzian functions (as it's shown for instance here).